Chapter 3 Trigonometric Functions

Given

The angle is $40^\circ 20'$.

To Find

Convert the given angle into radian measure.

Solution

We are given the angle in degrees and minutes. First, we convert the minutes part into degrees.

We know that $1^\circ = 60$ minutes ($60'$).

Therefore, $1' = \left(\frac{1}{60}\right)^\circ$.

So, $20' = \left(20 \times \frac{1}{60}\right)^\circ = \left(\frac{20}{60}\right)^\circ = \left(\frac{1}{3}\right)^\circ$.

Now, we add this to the degree part:

$40^\circ 20' = 40^\circ + \left(\frac{1}{3}\right)^\circ = \left(40 + \frac{1}{3}\right)^\circ$

Combining the terms:

$\left(\frac{40 \times 3 + 1}{3}\right)^\circ = \left(\frac{120 + 1}{3}\right)^\circ = \left(\frac{121}{3}\right)^\circ$.

Next, we convert degrees to radians. The conversion formula is: $1^\circ = \frac{\pi}{180}$ radians.

To convert $\left(\frac{121}{3}\right)^\circ$ to radians, we multiply by $\frac{\pi}{180}$:

$\left(\frac{121}{3}\right)^\circ = \frac{121}{3} \times \frac{\pi}{180}$ radians.

Multiplying the fractions:

$\frac{121 \times \pi}{3 \times 180}$ radians

$\frac{121\pi}{540}$ radians.

Thus, $40^\circ 20' = \frac{121\pi}{540}$ radians.

The final answer is:

$40^\circ 20' = \frac{121\pi}{540}$ radians.

Given

The angle is 6 radians.

To Find

Convert 6 radians into degree measure.

Solution

We know the relationship between radian and degree measure:

$\pi \text{ radians} = 180^\circ$

From this, we can find the degree measure of 1 radian:

$1 \text{ radian} = \left(\frac{180}{\pi}\right)^\circ$

To convert 6 radians to degrees, we multiply 6 by the conversion factor $\left(\frac{180}{\pi}\right)^\circ$:

$6 \text{ radians} = 6 \times \left(\frac{180}{\pi}\right)^\circ$

Multiplying the numbers:

$6 \times 180 = 1080$

So, 6 radians is equal to:

$\left(\frac{1080}{\pi}\right)^\circ$

This is the exact value in degrees. If an approximate value is required, we can use an approximate value for $\pi$, such as $\frac{22}{7}$ or $3.14159$. Using $\pi \approx \frac{22}{7}$:

$6 \text{ radians} \approx \left(\frac{1080}{22/7}\right)^\circ = \left(1080 \times \frac{7}{22}\right)^\circ$

$\left(\frac{1080 \times 7}{22}\right)^\circ = \left(\frac{\cancel{1080}^{540} \times 7}{\cancel{22}_{11}}\right)^\circ = \left(\frac{540 \times 7}{11}\right)^\circ$

$\left(\frac{3780}{11}\right)^\circ$

So, $\frac{3780}{11} = 343$ with a remainder of 7.

Thus, $\left(\frac{3780}{11}\right)^\circ = 343^\circ + \left(\frac{7}{11}\right)^\circ$.

To convert the fractional part of a degree into minutes, we multiply by 60:

$\left(\frac{7}{11}\right)^\circ = \frac{7}{11} \times 60'$

$\frac{420}{11}'$

So, $\frac{420}{11}' = 38'$ with a remainder of 2.

Thus, $\frac{420}{11}' = 38' + \left(\frac{2}{11}\right)'$.

To convert the fractional part of a minute into seconds, we multiply by 60:

$\left(\frac{2}{11}\right)' = \frac{2}{11} \times 60'' = \frac{120}{11}''$

$\frac{120}{11}'' \approx 10.9''$

So, 6 radians is approximately $343^\circ 38' 11''$ (rounding to the nearest second).

However, the question asks for the conversion, and $\left(\frac{1080}{\pi}\right)^\circ$ is the exact conversion.

The final answer (exact value) is:

$6 \text{ radians} = \left(\frac{1080}{\pi}\right)^\circ$.

The final answer (approximate value using $\pi \approx \frac{22}{7}$) is:

$6 \text{ radians} \approx 343^\circ 38' 11''$.

Given

Central angle, $\theta = 60^\circ$.

Arc length, $l = 37.4 \text{ cm}$.

Use $\pi = \frac{22}{7}$.

To Find

The radius of the circle, $r$.

Solution

First, we need to convert the central angle from degrees to radians, as the formula relating arc length, radius, and central angle requires the angle to be in radians.

The conversion formula from degrees to radians is to multiply by $\frac{\pi}{180}$.

So, $\theta = 60^\circ = 60 \times \frac{\pi}{180}$ radians.

Simplifying the fraction $\frac{60}{180}$:

$60 \times \frac{\pi}{180} = \frac{\cancel{60}^{1} \times \pi}{\cancel{180}_{3}} = \frac{\pi}{3}$ radians.

So, the central angle in radians is $\theta = \frac{\pi}{3}$.

The formula for the arc length ($l$) intercepted by a central angle ($\theta$ in radians) in a circle with radius ($r$) is:

We want to find the radius $r$, so we rearrange the formula (i) to solve for $r$:

Now, we substitute the given values: $l = 37.4$ cm, $\theta = \frac{\pi}{3}$, and $\pi = \frac{22}{7}$.

Inverting the denominator fraction and multiplying:

Substitute the value of $\pi = \frac{22}{7}$:

Again, invert the denominator fraction and multiply:

Calculate $37.4 \times 3 = 112.2$.

Now, we can simplify $\frac{112.2}{22}$.

$\frac{112.2}{22} = \frac{1122 \div 10}{22} = \frac{1122}{22 \times 10}$.

We know that $1122 = 22 \times 51$.

So, $\frac{1122}{22 \times 10} = \frac{\cancel{22} \times 51}{\cancel{22} \times 10} = \frac{51}{10} = 5.1$.

Thus, $\frac{112.2}{22} = 5.1$.

The calculation becomes:

Finally, multiply $5.1$ by $7$:

$r = 5.1 \times 7 \text{ cm} = 35.7 \text{ cm}$.

The radius of the circle is $35.7$ cm.

The final answer is:

The radius of the circle is $35.7$ cm.

Given

Length of the minute hand (radius of the circle traced by the tip), $r = 1.5 \text{ cm}$.

Time duration, $t = 40 \text{ minutes}$.

Use $\pi = 3.14$.

To Find

The distance the tip of the minute hand moves in 40 minutes (arc length).

Solution

The tip of the minute hand moves along the circumference of a circle with radius equal to the length of the minute hand, which is $r = 1.5$ cm.

In 60 minutes, the minute hand completes one full revolution, which corresponds to an angle of $360^\circ$ or $2\pi$ radians.

The angle moved by the minute hand in 1 minute is:

$\frac{360^\circ}{60 \text{ minutes}} = 6^\circ \text{ per minute}$.

The angle moved by the minute hand in 40 minutes is:

Angle in degrees $= 40 \text{ minutes} \times 6^\circ/\text{minute} = 240^\circ$.

Now, we need to convert this angle from degrees to radians, because the formula relating arc length, radius, and central angle requires the angle in radians.

The conversion factor from degrees to radians is $\frac{\pi}{180}$.

Angle in radians, $\theta = 240^\circ \times \frac{\pi}{180} \text{ radians}$.

$\theta = \frac{240}{180}\pi \text{ radians} = \frac{24}{18}\pi \text{ radians} = \frac{\cancel{6} \times 4}{\cancel{6} \times 3}\pi \text{ radians} = \frac{4\pi}{3} \text{ radians}$.

The arc length ($l$) is given by the formula:

Substitute the values of $r = 1.5$ cm and $\theta = \frac{4\pi}{3}$ radians into formula (i):

Perform the multiplication:

$l = \frac{1.5 \times 4}{3}\pi \text{ cm} = \frac{6.0}{3}\pi \text{ cm} = 2\pi \text{ cm}$.

Now, substitute the given value of $\pi = 3.14$:

$l = 6.28 \text{ cm}$.

The distance the tip of the minute hand moves in 40 minutes is 6.28 cm.

The final answer is:

The tip of the minute hand moves $6.28$ cm in 40 minutes.

Given

Let the radius of the first circle be $r_1$ and the central angle subtended by the arc be $\theta_1 = 65^\circ$.

Let the radius of the second circle be $r_2$ and the central angle subtended by the arc be $\theta_2 = 110^\circ$.

The arcs in both circles have the same length. Let this length be $l$. So, $l_1 = l_2 = l$.

To Find

The ratio of the radii of the two circles, i.e., $\frac{r_1}{r_2}$.

Solution

We know that the arc length ($l$) of a sector of a circle with radius ($r$) and central angle ($\theta$ in radians) is given by the formula $l = r\theta$.

First, we must convert the given angles from degrees to radians.

We use the conversion factor: $1^\circ = \frac{\pi}{180}$ radians.

For the first circle, the angle in radians is:

$\theta_1 = 65^\circ = 65 \times \frac{\pi}{180} \text{ radians}$

$\theta_1 = \frac{65\pi}{180} = \frac{\cancel{65}^{13}\pi}{\cancel{180}_{36}} = \frac{13\pi}{36} \text{ radians}$.

For the second circle, the angle in radians is:

$\theta_2 = 110^\circ = 110 \times \frac{\pi}{180} \text{ radians}$

$\theta_2 = \frac{110\pi}{180} = \frac{\cancel{110}^{11}\pi}{\cancel{180}_{18}} = \frac{11\pi}{18} \text{ radians}$.

Since the arc lengths are equal, we have $l_1 = l_2$.

Using the formula $l = r\theta$ for both circles:

Substitute the values of $\theta_1$ and $\theta_2$ (in radians) into equation (i):

We want to find the ratio $\frac{r_1}{r_2}$. Divide both sides by $r_2$ and by $\frac{13\pi}{36}$:

Simplify the complex fraction by multiplying the numerator by the reciprocal of the denominator:

Cancel out $\pi$ and simplify the numerical fraction:

$\frac{r_1}{r_2} = \frac{11 \times 2}{1 \times 13} = \frac{22}{13}$.

The ratio of the radii is $22:13$.

The final answer is:

The ratio of the radii of the two circles is $22:13$.

To convert degree measure to radian measure, we use the formula: Radians = Degrees $\times \frac{\pi}{180}$.

(i) 25°

We need to convert $25^\circ$ to radians.

Radian measure = $25 \times \frac{\pi}{180}$

We can simplify the fraction $\frac{25}{180}$ by dividing both numerator and denominator by their greatest common divisor, which is 5.

So, Radian measure = $\frac{5\pi}{36}$.

Thus, $25^\circ = \frac{5\pi}{36}$ radians.

(ii) – 47°30′

First, convert the angle completely into degrees. We know that $1^\circ = 60'$. So, $30' = \left(\frac{30}{60}\right)^\circ = \left(\frac{1}{2}\right)^\circ = 0.5^\circ$.

The angle is $-47^\circ 30' = -(47^\circ + 30') = -\left(47 + \frac{1}{2}\right)^\circ = -\left(\frac{94+1}{2}\right)^\circ = -\left(\frac{95}{2}\right)^\circ = -47.5^\circ$.

Now, convert $-47.5^\circ$ to radians.

Radian measure = $-47.5 \times \frac{\pi}{180}$

Radian measure = $-\frac{95}{2} \times \frac{\pi}{180} = -\frac{95\pi}{2 \times 180} = -\frac{95\pi}{360}$.

We can simplify the fraction $\frac{95}{360}$ by dividing both numerator and denominator by 5.

So, Radian measure = $-\frac{19\pi}{72}$.

Thus, $-47^\circ 30' = -\frac{19\pi}{72}$ radians.

(iii) 240°

We need to convert $240^\circ$ to radians.

Radian measure = $240 \times \frac{\pi}{180}$

Simplify the fraction $\frac{240}{180}$ by dividing both numerator and denominator by their greatest common divisor, which is 60.

So, Radian measure = $\frac{4\pi}{3}$.

Thus, $240^\circ = \frac{4\pi}{3}$ radians.

(iv) 520°

We need to convert $520^\circ$ to radians.

Radian measure = $520 \times \frac{\pi}{180}$

Simplify the fraction $\frac{520}{180}$ by dividing both numerator and denominator by their greatest common divisor, which is 20.

So, Radian measure = $\frac{26\pi}{9}$.

Thus, $520^\circ = \frac{26\pi}{9}$ radians.

To convert radian measure to degree measure, we use the formula: Degrees = Radians $\times \frac{180}{\pi}$.

We are given to use $\pi = \frac{22}{7}$.

(i) $\frac{11}{16}$

We need to convert $\frac{11}{16}$ radians to degrees.

Degree measure = $\frac{11}{16} \times \frac{180}{\pi}$

Substitute $\pi = \frac{22}{7}$:

Degree measure = $\frac{11}{16} \times \frac{180}{\frac{22}{7}} = \frac{11}{16} \times 180 \times \frac{7}{22}$

Simplify the expression:

$\frac{\cancel{11}^{1}}{16} \times \frac{180 \times 7}{\cancel{22}_{2}} = \frac{1}{16} \times \frac{1260}{2} = \frac{1}{16} \times 630$

Degree measure = $\frac{315}{8} \times \frac{1}{1} = \frac{315}{8}$ degrees.

We can write this as a mixed number: $\frac{315}{8} = 39 \frac{3}{8}$ degrees.

Convert the fractional part of the degree to minutes ($1^\circ = 60'$):

$\left(\frac{3}{8}\right)^\circ = \frac{3}{8} \times 60' = \frac{180}{8}' = \frac{90}{4}' = \frac{45}{2}'$

$\frac{45}{2}' = 22 \frac{1}{2}' = 22' + \frac{1}{2}'$.

Convert the fractional part of the minute to seconds ($1' = 60''$):

$\left(\frac{1}{2}\right)' = \frac{1}{2} \times 60'' = 30''$.

So, $39 \frac{3}{8}^\circ = 39^\circ 22' 30''$.

Thus, $\frac{11}{16}$ radians = $39^\circ 22' 30''$.

(ii) – 4

We need to convert $-4$ radians to degrees.

Degree measure = $-4 \times \frac{180}{\pi}$

Substitute $\pi = \frac{22}{7}$:

Degree measure = $-4 \times \frac{180}{\frac{22}{7}} = -4 \times 180 \times \frac{7}{22}$

Simplify the expression:

Degree measure = $-2 \times 180 \times \frac{7}{11} = -\frac{360 \times 7}{11} = -\frac{2520}{11}$ degrees.

We can write this as a mixed number: $-\frac{2520}{11} = -229 \frac{1}{11}$ degrees.

Convert the fractional part of the degree to minutes:

$\left(\frac{1}{11}\right)^\circ = \frac{1}{11} \times 60' = \frac{60}{11}'$

$\frac{60}{11}' = 5 \frac{5}{11}' = 5' + \frac{5}{11}'$.

Convert the fractional part of the minute to seconds:

$\left(\frac{5}{11}\right)' = \frac{5}{11} \times 60'' = \frac{300}{11}''$

$\frac{300}{11}'' \approx 27.27''$. Rounding to the nearest second, we get $27''$.

So, $-229 \frac{1}{11}^\circ \approx -229^\circ 5' 27''$.

Thus, $-4$ radians $\approx -229^\circ 5' 27''$.

(iii) $\frac{5\pi}{3}$

We need to convert $\frac{5\pi}{3}$ radians to degrees.

Degree measure = $\frac{5\pi}{3} \times \frac{180}{\pi}$

Simplify the expression:

Degree measure = $5 \times 60 = 300$ degrees.

Thus, $\frac{5\pi}{3}$ radians = $300^\circ$.

(iv) $\frac{7\pi}{6}$

We need to convert $\frac{7\pi}{6}$ radians to degrees.

Degree measure = $\frac{7\pi}{6} \times \frac{180}{\pi}$

Simplify the expression:

Degree measure = $7 \times 30 = 210$ degrees.

Thus, $\frac{7\pi}{6}$ radians = $210^\circ$.

Given

Number of revolutions made by the wheel in one minute = 360.

To Find

The angle in radians through which the wheel turns in one second.

Solution

The wheel makes 360 revolutions in one minute.

We need to find the angle turned in one second. First, convert minutes to seconds.

1 minute = 60 seconds.

Number of revolutions in 60 seconds = 360.

Number of revolutions in 1 second = $\frac{\text{Total revolutions}}{\text{Total time in seconds}} = \frac{360}{60} = 6$ revolutions per second.

One complete revolution corresponds to an angle of $360^\circ$ or $2\pi$ radians.

The angle turned in one revolution = $2\pi$ radians.

Since the wheel makes 6 revolutions in one second, the total angle turned in one second is:

Angle in one second = (Number of revolutions in one second) $\times$ (Angle per revolution)

Calculate the value:

Angle in one second = $12\pi$ radians.

The wheel turns through $12\pi$ radians in one second.

The final answer is:

The wheel turns through $12\pi$ radians in one second.

Given

Radius of the circle, $r = 100 \text{ cm}$.

Length of the arc, $l = 22 \text{ cm}$.

Use $\pi = \frac{22}{7}$.

To Find

The degree measure of the angle subtended at the centre by the arc.

Solution

The relationship between arc length ($l$), radius ($r$), and the central angle ($\theta$) in radians is given by the formula:

We are given $l = 22$ cm and $r = 100$ cm. We can use formula (i) to find the angle $\theta$ in radians.

Rearranging the formula for $\theta$:

Substitute the given values:

$\theta = \frac{22}{100} = \frac{11}{50}$ radians.

Now, we need to convert this radian measure to degree measure. The conversion formula is: Degrees = Radians $\times \frac{180}{\pi}$.

Degree measure = $\frac{11}{50} \times \frac{180}{\pi}$.

We are given to use $\pi = \frac{22}{7}$. Substitute this value:

Degree measure = $\frac{11}{50} \times \frac{180}{\frac{22}{7}} = \frac{11}{50} \times 180 \times \frac{7}{22}$.

Simplify the expression:

Degree measure = $\frac{1}{5} \times 9 \times 7 = \frac{63}{5}$ degrees.

We can express this in degrees, minutes, and seconds. $\frac{63}{5}$ degrees = $12 \frac{3}{5}$ degrees.

Convert the fractional part of the degree to minutes ($1^\circ = 60'$):

$\left(\frac{3}{5}\right)^\circ = \frac{3}{5} \times 60' = \frac{180}{5}' = 36'$.

So, $\frac{63}{5}$ degrees = $12^\circ 36'$.

The angle subtended at the centre is $12^\circ 36'$.

The final answer is:

The degree measure of the angle is $12^\circ 36'$.

Given

Diameter of the circle = 40 cm.

Length of the chord = 20 cm.

To Find

The length of the minor arc of the chord.

Solution

The radius of the circle is half of the diameter.

Radius, $r = \frac{\text{Diameter}}{2} = \frac{40 \text{ cm}}{2} = 20 \text{ cm}$.

Let the centre of the circle be O. Let the chord be AB. The length of the chord AB is 20 cm.

The lines connecting the endpoints of the chord to the centre are radii. So, OA and OB are radii of the circle.

Consider the triangle $\triangle OAB$. The lengths of its sides are OA = 20 cm, OB = 20 cm, and AB = 20 cm.

Since all three sides of $\triangle OAB$ are equal, it is an equilateral triangle.

In an equilateral triangle, each angle is $60^\circ$. The angle subtended by the chord AB at the centre O is $\angle AOB$.

So, the central angle $\theta = \angle AOB = 60^\circ$.

To find the arc length, we need to convert the central angle from degrees to radians.

The conversion formula is: Radians = Degrees $\times \frac{\pi}{180}$.

$\theta = 60^\circ = 60 \times \frac{\pi}{180}$ radians.

Simplify the fraction:

$\theta = \frac{\pi}{3}$ radians.

The length of the minor arc ($l$) is related to the radius ($r$) and the central angle ($\theta$ in radians) by the formula:

Substitute the values $r = 20$ cm and $\theta = \frac{\pi}{3}$ radians into formula (i):

The length of the minor arc is $\frac{20\pi}{3}$ cm.

The final answer is:

The length of the minor arc is $\frac{20\pi}{3}$ cm.

Given

Let the two circles have radii $r_1$ and $r_2$.

The angles subtended at the centre are $\theta_1 = 60^\circ$ and $\theta_2 = 75^\circ$.

The arcs have the same length. Let this length be $l$. So, $l_1 = l_2 = l$.

To Find

The ratio of the radii, $\frac{r_1}{r_2}$.

Solution

The formula relating arc length ($l$), radius ($r$), and central angle ($\theta$) is $l = r\theta$, where $\theta$ is in radians.

First, convert the given angles from degrees to radians.

The conversion factor is $\frac{\pi}{180}$ radians per degree.

For the first circle:

$\theta_1 = 60^\circ = 60 \times \frac{\pi}{180}$ radians.

$\theta_1 = \frac{60}{180}\pi = \frac{1}{3}\pi = \frac{\pi}{3}$ radians.

For the second circle:

$\theta_2 = 75^\circ = 75 \times \frac{\pi}{180}$ radians.

Simplify the fraction $\frac{75}{180}$:

$\frac{\cancel{75}^{5}}{\cancel{180}_{12}}\pi = \frac{5\pi}{12}$ radians.

$\theta_2 = \frac{5\pi}{12}$ radians.

Now, use the arc length formula $l = r\theta$ for each circle.

For the first circle: $l_1 = r_1 \theta_1 = r_1 \left(\frac{\pi}{3}\right)$.

For the second circle: $l_2 = r_2 \theta_2 = r_2 \left(\frac{5\pi}{12}\right)$.

Since the arc lengths are equal, $l_1 = l_2$:

To find the ratio $\frac{r_1}{r_2}$, rearrange equation (i):

Simplify the complex fraction by multiplying the numerator by the reciprocal of the denominator:

Cancel common terms:

$\frac{r_1}{r_2} = \frac{5 \times 1}{4 \times 1} = \frac{5}{4}$.

The ratio of their radii is $5:4$.

The final answer is:

The ratio of the radii is $5:4$.

Given

Length of the pendulum = 75 cm. The pendulum swings, so its length is the radius of the circle traced by the tip. So, the radius $r = 75$ cm.

To Find

The angle (in radians) through which the pendulum swings for the given arc lengths.

Solution

The tip of the pendulum describes an arc of a circle with radius equal to the length of the pendulum. The angle through which it swings is the central angle subtended by this arc.

The relationship between arc length ($l$), radius ($r$), and the central angle ($\theta$ in radians) is given by the formula:

We need to find the angle $\theta$, so we rearrange the formula (i):

Here, $r = 75$ cm. We will use the given arc lengths ($l$) for each part.

(i) Arc length $l = 10$ cm

Using the formula $\theta = \frac{l}{r}$:

$\theta = \frac{10 \text{ cm}}{75 \text{ cm}} = \frac{10}{75}$ radians.

Simplify the fraction $\frac{10}{75}$ by dividing both numerator and denominator by 5.

$\frac{\cancel{10}^{2}}{\cancel{75}_{15}} = \frac{2}{15}$.

So, $\theta = \frac{2}{15}$ radians.

The angle is $\frac{2}{15}$ radians.

(ii) Arc length $l = 15$ cm

Using the formula $\theta = \frac{l}{r}$:

$\theta = \frac{15 \text{ cm}}{75 \text{ cm}} = \frac{15}{75}$ radians.

Simplify the fraction $\frac{15}{75}$ by dividing both numerator and denominator by 15.

$\frac{\cancel{15}^{1}}{\cancel{75}_{5}} = \frac{1}{5}$.

So, $\theta = \frac{1}{5}$ radians.

The angle is $\frac{1}{5}$ radians.

(iii) Arc length $l = 21$ cm

Using the formula $\theta = \frac{l}{r}$:

$\theta = \frac{21 \text{ cm}}{75 \text{ cm}} = \frac{21}{75}$ radians.

Simplify the fraction $\frac{21}{75}$ by dividing both numerator and denominator by 3.

$\frac{\cancel{21}^{7}}{\cancel{75}_{25}} = \frac{7}{25}$.

So, $\theta = \frac{7}{25}$ radians.

The angle is $\frac{7}{25}$ radians.

The final answers are:

(i) The angle is $\frac{2}{15}$ radians.

(ii) The angle is $\frac{1}{5}$ radians.

(iii) The angle is $\frac{7}{25}$ radians.

Given

$\cos x = -\frac{3}{5}$.

$x$ lies in the third quadrant.

To Find

The values of $\sin x$, $\tan x$, $\text{cosec } x$, $\sec x$, and $\cot x$.

Solution

We are given $\cos x = -\frac{3}{5}$ and that $x$ is in the third quadrant.

In the third quadrant, sine and cosine values are negative, while tangent and cotangent values are positive. Secant is the reciprocal of cosine (negative), and cosecant is the reciprocal of sine (negative).

First, let's find $\sin x$ using the fundamental trigonometric identity:

Substitute the given value of $\cos x$:

$\sin^2 x + \frac{9}{25} = 1$

$\sin^2 x = 1 - \frac{9}{25} = \frac{25 - 9}{25} = \frac{16}{25}$.

Taking the square root of both sides:

$\sin x = \pm\sqrt{\frac{16}{25}} = \pm\frac{4}{5}$.

Since $x$ lies in the third quadrant, $\sin x$ is negative.

Now, we can find the other trigonometric functions:

$\tan x = \frac{\sin x}{\cos x}$

$\tan x = \frac{-\frac{4}{5}}{-\frac{3}{5}} = \frac{-\frac{4}{5}}{\text{-}\frac{3}{5}} = \frac{4}{5} \times \frac{5}{3} = \frac{4}{\cancel{5}} \times \frac{\cancel{5}}{3} = \frac{4}{3}$.

$\sec x = \frac{1}{\cos x}$

$\sec x = \frac{1}{-\frac{3}{5}} = -\frac{5}{3}$.

$\text{cosec } x = \frac{1}{\sin x}$

$\text{cosec } x = \frac{1}{-\frac{4}{5}} = -\frac{5}{4}$.

$\cot x = \frac{1}{\tan x}$

$\cot x = \frac{1}{\frac{4}{3}} = \frac{3}{4}$.

The values of the other five trigonometric functions are:

$\sin x = -\frac{4}{5}$

$\tan x = \frac{4}{3}$

$\text{cosec } x = -\frac{5}{4}$

$\sec x = -\frac{5}{3}$

$\cot x = \frac{3}{4}$

Given

$\cot x = -\frac{5}{12}$.

$x$ lies in the second quadrant.

To Find

The values of $\sin x$, $\cos x$, $\tan x$, $\sec x$, and $\text{cosec } x$.

Solution

We are given $\cot x = -\frac{5}{12}$ and that $x$ is in the second quadrant.

In the second quadrant, $\sin x$ and $\text{cosec } x$ are positive, while $\cos x$, $\sec x$, $\tan x$, and $\cot x$ are negative.

First, let's find $\tan x$ using the reciprocal identity:

$\tan x = \frac{1}{-\frac{5}{12}} = -\frac{12}{5}$.

Next, we find $\text{cosec } x$ using the identity $\text{cosec}^2 x = 1 + \cot^2 x$:

$\text{cosec}^2 x = 1 + \frac{25}{144} = \frac{144 + 25}{144} = \frac{169}{144}$.

Taking the square root of both sides:

$\text{cosec } x = \pm\sqrt{\frac{169}{144}} = \pm\frac{13}{12}$.

Since $x$ lies in the second quadrant, $\text{cosec } x$ is positive.

Now, find $\sin x$ using the reciprocal identity:

$\sin x = \frac{1}{\frac{13}{12}} = \frac{12}{13}$.

Next, find $\cos x$. We can use the identity $\tan x = \frac{\sin x}{\cos x}$, which means $\cos x = \frac{\sin x}{\tan x}$.

$\cos x = \frac{\frac{12}{13}}{-\frac{12}{5}} = \frac{12}{13} \times \left(-\frac{5}{12}\right) = -\frac{\cancel{12}^{1}}{13} \times \frac{5}{\cancel{12}^{1}} = -\frac{5}{13}$.

Finally, find $\sec x$ using the reciprocal identity:

$\sec x = \frac{1}{-\frac{5}{13}} = -\frac{13}{5}$.

The values of the other five trigonometric functions are:

$\sin x = \frac{12}{13}$

$\cos x = -\frac{5}{13}$

$\tan x = -\frac{12}{5}$

$\text{cosec } x = \frac{13}{12}$

$\sec x = -\frac{13}{5}$

Given

The expression is $\sin \frac{31\pi}{3}$.

To Find

The value of $\sin \frac{31\pi}{3}$.

Solution

We need to find the value of $\sin \frac{31\pi}{3}$. The angle $\frac{31\pi}{3}$ is greater than $2\pi$. We can find a coterminal angle by subtracting multiples of $2\pi$ from $\frac{31\pi}{3}$.

Divide 31 by 3 to find how many multiples of $2\pi$ are present. $\frac{31}{3} = 10$ with a remainder of 1. So, $\frac{31}{3} = 10 + \frac{1}{3}$.

Thus, $\frac{31\pi}{3} = \left(10 + \frac{1}{3}\right)\pi = 10\pi + \frac{\pi}{3}$.

The angle $10\pi$ is an even multiple of $\pi$ ($10\pi = 5 \times 2\pi$). Adding an even multiple of $\pi$ to an angle results in a coterminal angle with the same trigonometric values.

We know that $\sin(2n\pi + \theta) = \sin \theta$ for any integer $n$.

In this case, $2n\pi = 10\pi$ (so $n=5$) and $\theta = \frac{\pi}{3}$.

Therefore, $\sin \left(\frac{31\pi}{3}\right) = \sin \left(10\pi + \frac{\pi}{3}\right)$.

We know the exact value of $\sin \left(\frac{\pi}{3}\right)$. The angle $\frac{\pi}{3}$ radians is equivalent to $60^\circ$.

$\sin \left(\frac{\pi}{3}\right) = \sin(60^\circ) = \frac{\sqrt{3}}{2}$.

So, $\sin \left(\frac{31\pi}{3}\right) = \frac{\sqrt{3}}{2}$.

The final answer is:

$\sin \frac{31\pi}{3} = \frac{\sqrt{3}}{2}$.

Given

The expression is $\cos(-1710^\circ)$.

To Find

The value of $\cos(-1710^\circ)$.

Solution

We need to find the value of $\cos(-1710^\circ)$.

First, we use the property that $\cos(-\theta) = \cos \theta$.

Now, we need to find the value of $\cos(1710^\circ)$. The angle $1710^\circ$ is greater than $360^\circ$. We can find a coterminal angle by subtracting multiples of $360^\circ$ from $1710^\circ$.

Divide 1710 by 360 to see how many full rotations are involved.

$\frac{1710}{360} = \frac{171}{36}$. Divide both by 9: $\frac{19}{4} = 4$ with a remainder of 3. So, $1710 = 4 \times 360 + 270$.

Alternatively, $1710^\circ = 4 \times 360^\circ + 270^\circ$.

We know that $\cos(360^\circ n + \theta) = \cos \theta$ for any integer $n$.

In this case, $360^\circ n = 4 \times 360^\circ$ (so $n=4$) and $\theta = 270^\circ$.

Therefore, $\cos(1710^\circ) = \cos(4 \times 360^\circ + 270^\circ)$.

We know the exact value of $\cos(270^\circ)$. The angle $270^\circ$ is a quadrantal angle. On the unit circle, the point corresponding to $270^\circ$ is $(0, -1)$. The cosine value is the x-coordinate of this point.

$\cos(270^\circ) = 0$.

So, $\cos(-1710^\circ) = \cos(1710^\circ) = \cos(270^\circ) = 0$.

The final answer is:

$\cos (-1710^\circ) = 0$.

Given

$\cos x = -\frac{1}{2}$.

$x$ lies in the third quadrant.

To Find

The values of $\sin x$, $\tan x$, $\text{cosec } x$, $\sec x$, and $\cot x$.

Solution

We are given that $\cos x = -\frac{1}{2}$ and the angle $x$ lies in the third quadrant.

In the third quadrant, sine is negative, cosine is negative, and tangent and cotangent are positive.

We use the fundamental trigonometric identity $\sin^2 x + \cos^2 x = 1$ to find $\sin x$:

Square the value of $\cos x$:

$\sin^2 x + \frac{1}{4} = 1$

Subtract $\frac{1}{4}$ from both sides:

Take the square root of both sides to find $\sin x$:

$\sin x = \pm\sqrt{\frac{3}{4}} = \pm\frac{\sqrt{3}}{2}$.

Since $x$ lies in the third quadrant, the value of $\sin x$ is negative.

Now, we find $\tan x$ using the definition $\tan x = \frac{\sin x}{\cos x}$:

When dividing by a fraction, we multiply by its reciprocal:

$\tan x = -\frac{\sqrt{3}}{2} \times \left(-\frac{2}{1}\right) = \frac{\sqrt{3}}{\cancel{2}} \times \frac{\cancel{2}}{1} = \sqrt{3}$.

Next, we find the reciprocal trigonometric functions using the values of $\sin x$, $\cos x$, and $\tan x$:

$\sec x = \frac{1}{\cos x}$

$\text{cosec } x = \frac{1}{\sin x}$

$\cot x = \frac{1}{\tan x}$

The values of the other five trigonometric functions are:

$\sin x = -\frac{\sqrt{3}}{2}$

$\tan x = \sqrt{3}$

$\text{cosec } x = -\frac{2}{\sqrt{3}}$

$\sec x = -2$

$\cot x = \frac{1}{\sqrt{3}}$

Given

$\sin x = \frac{3}{5}$.

$x$ lies in the second quadrant.

To Find

The values of $\cos x$, $\tan x$, $\text{cosec } x$, $\sec x$, and $\cot x$.

Solution

We are given $\sin x = \frac{3}{5}$ and that the angle $x$ lies in the second quadrant.

In the second quadrant, sine is positive, cosine is negative, and tangent and cotangent are negative.

We use the fundamental trigonometric identity $\sin^2 x + \cos^2 x = 1$ to find $\cos x$:

Square the value of $\sin x$:

$\frac{9}{25} + \cos^2 x = 1$

Subtract $\frac{9}{25}$ from both sides:

Take the square root of both sides to find $\cos x$:

$\cos x = \pm\sqrt{\frac{16}{25}} = \pm\frac{4}{5}$.

Since $x$ lies in the second quadrant, the value of $\cos x$ is negative.

Now, we find $\tan x$ using the definition $\tan x = \frac{\sin x}{\cos x}$:

$\tan x = \frac{3}{5} \times \left(-\frac{5}{4}\right) = -\frac{3}{\cancel{5}} \times \frac{\cancel{5}}{4} = -\frac{3}{4}$.

Next, we find the reciprocal trigonometric functions using the values of $\sin x$, $\cos x$, and $\tan x$:

$\text{cosec } x = \frac{1}{\sin x}$

$\sec x = \frac{1}{\cos x}$

$\cot x = \frac{1}{\tan x}$

The values of the other five trigonometric functions are:

$\cos x = -\frac{4}{5}$

$\tan x = -\frac{3}{4}$

$\text{cosec } x = \frac{5}{3}$

$\sec x = -\frac{5}{4}$

$\cot x = -\frac{4}{3}$

Given

$\cot x = \frac{3}{4}$.

$x$ lies in the third quadrant.

To Find

The values of $\sin x$, $\cos x$, $\tan x$, $\sec x$, and $\text{cosec } x$.

Solution

We are given that $\cot x = \frac{3}{4}$ and the angle $x$ lies in the third quadrant.

In the third quadrant, $\sin x$ and $\cos x$ are negative, while $\tan x$ and $\cot x$ are positive.

First, we find $\tan x$ using the reciprocal identity:

Substitute the given value of $\cot x$:

$\tan x = \frac{1}{\frac{3}{4}} = \frac{4}{3}$.

Next, we find $\text{cosec } x$ using the identity $\text{cosec}^2 x = 1 + \cot^2 x$:

Square the value of $\cot x$:

$\text{cosec}^2 x = 1 + \frac{9}{16} = \frac{16 + 9}{16} = \frac{25}{16}$.

Take the square root of both sides:

$\text{cosec } x = \pm\sqrt{\frac{25}{16}} = \pm\frac{5}{4}$.

Since $x$ lies in the third quadrant, the value of $\text{cosec } x$ is negative.

Now, find $\sin x$ using the reciprocal identity:

$\sin x = \frac{1}{-\frac{5}{4}} = -\frac{4}{5}$.

Next, find $\cos x$. We can use the definition $\cot x = \frac{\cos x}{\sin x}$, which implies $\cos x = \cot x \times \sin x$.

Multiply the values:

$\cos x = -\frac{3 \times \cancel{4}^{1}}{\cancel{4}^{1} \times 5} = -\frac{3}{5}$.

Finally, find $\sec x$ using the reciprocal identity:

$\sec x = \frac{1}{-\frac{3}{5}} = -\frac{5}{3}$.

The values of the other five trigonometric functions are:

$\sin x = -\frac{4}{5}$

$\cos x = -\frac{3}{5}$

$\tan x = \frac{4}{3}$

$\text{cosec } x = -\frac{5}{4}$

$\sec x = -\frac{5}{3}$

Given

$\sec x = \frac{13}{5}$.

$x$ lies in the fourth quadrant.

To Find

The values of $\sin x$, $\cos x$, $\tan x$, $\text{cosec } x$, and $\cot x$.

Solution

We are given that $\sec x = \frac{13}{5}$ and the angle $x$ lies in the fourth quadrant.

In the fourth quadrant, cosine and secant are positive, while sine, cosecant, tangent, and cotangent are negative.

First, we find $\cos x$ using the reciprocal identity:

Substitute the given value of $\sec x$:

$\cos x = \frac{1}{\frac{13}{5}} = \frac{5}{13}$.

Next, we find $\sin x$ using the identity $\sin^2 x + \cos^2 x = 1$:

Square the value of $\cos x$:

$\sin^2 x + \frac{25}{169} = 1$

Subtract $\frac{25}{169}$ from both sides:

Take the square root of both sides to find $\sin x$:

$\sin x = \pm\sqrt{\frac{144}{169}} = \pm\frac{12}{13}$.

Since $x$ lies in the fourth quadrant, the value of $\sin x$ is negative.

Now, we find $\tan x$ using the definition $\tan x = \frac{\sin x}{\cos x}$:

$\tan x = -\frac{12}{13} \times \frac{13}{5} = -\frac{12}{\cancel{13}} \times \frac{\cancel{13}}{5} = -\frac{12}{5}$.

Next, we find the remaining reciprocal trigonometric functions:

$\text{cosec } x = \frac{1}{\sin x}$

$\cot x = \frac{1}{\tan x}$

The values of the other five trigonometric functions are:

$\sin x = -\frac{12}{13}$

$\cos x = \frac{5}{13}$

$\tan x = -\frac{12}{5}$

$\text{cosec } x = -\frac{13}{12}$

$\cot x = -\frac{5}{12}$

Given

$\tan x = -\frac{5}{12}$.

$x$ lies in the second quadrant.

To Find

The values of $\sin x$, $\cos x$, $\text{cosec } x$, $\sec x$, and $\cot x$.

Solution

We are given that $\tan x = -\frac{5}{12}$ and the angle $x$ lies in the second quadrant.

In the second quadrant, $\sin x$ and $\text{cosec } x$ are positive, while $\cos x$, $\sec x$, $\tan x$, and $\cot x$ are negative.

First, we find $\cot x$ using the reciprocal identity:

Substitute the given value of $\tan x$:

$\cot x = \frac{1}{-\frac{5}{12}} = -\frac{12}{5}$.

Next, we find $\sec x$ using the identity $\sec^2 x = 1 + \tan^2 x$:

Square the value of $\tan x$:

$\sec^2 x = 1 + \frac{25}{144} = \frac{144 + 25}{144} = \frac{169}{144}$.

Take the square root of both sides:

$\sec x = \pm\sqrt{\frac{169}{144}} = \pm\frac{13}{12}$.

Since $x$ lies in the second quadrant, the value of $\sec x$ is negative.

Now, find $\cos x$ using the reciprocal identity:

$\cos x = \frac{1}{-\frac{13}{12}} = -\frac{12}{13}$.

Next, find $\sin x$. We can use the definition $\tan x = \frac{\sin x}{\cos x}$, which implies $\sin x = \tan x \times \cos x$.

Multiply the values:

$\sin x = \frac{5}{\cancel{12}^{1}} \times \frac{\cancel{12}^{1}}{13} = \frac{5}{13}$.

Finally, find $\text{cosec } x$ using the reciprocal identity:

$\text{cosec } x = \frac{1}{\frac{5}{13}} = \frac{13}{5}$.

The values of the other five trigonometric functions are:

$\sin x = \frac{5}{13}$

$\cos x = -\frac{12}{13}$

$\text{cosec } x = \frac{13}{5}$

$\sec x = -\frac{13}{12}$

$\cot x = -\frac{12}{5}$

Given

The expression is $\sin 765^\circ$.

To Find

The value of $\sin 765^\circ$.

Solution

We need to find the value of $\sin 765^\circ$. The angle $765^\circ$ is greater than $360^\circ$. We can find a coterminal angle by subtracting multiples of $360^\circ$ from $765^\circ$.

Divide 765 by 360 to see how many full rotations are involved.

$\frac{765}{360} = \frac{153}{72}$. Divide both by 9: $\frac{17}{8} = 2$ with a remainder of 1. So, $765 = 2 \times 360 + 45$.

Alternatively, $765^\circ = 2 \times 360^\circ + 45^\circ$.

We know that $\sin(360^\circ n + \theta) = \sin \theta$ for any integer $n$.

In this case, $360^\circ n = 2 \times 360^\circ$ (so $n=2$) and $\theta = 45^\circ$.

Therefore, $\sin(765^\circ) = \sin(2 \times 360^\circ + 45^\circ)$.

We know the exact value of $\sin(45^\circ)$.

$\sin(45^\circ) = \frac{1}{\sqrt{2}}$.

So, $\sin(765^\circ) = \frac{1}{\sqrt{2}}$.

The final answer is:

$\sin 765^\circ = \frac{1}{\sqrt{2}}$.

Given

The expression is $\text{cosec } (-1410^\circ)$.

To Find

The value of $\text{cosec } (-1410^\circ)$.

Solution

We need to find the value of $\text{cosec } (-1410^\circ)$.

First, we use the property that $\text{cosec } (-\theta) = -\text{cosec } \theta$.

Now, we need to find the value of $\text{cosec }(1410^\circ)$. The angle $1410^\circ$ is greater than $360^\circ$. We can find a coterminal angle by subtracting multiples of $360^\circ$ from $1410^\circ$.

Divide 1410 by 360 to see how many full rotations are involved.

$\frac{1410}{360} = \frac{141}{36}$. Divide both by 3: $\frac{47}{12} = 3$ with a remainder of 11. So, $1410 = 3 \times 360 + 330$.

Alternatively, $1410^\circ = 3 \times 360^\circ + 330^\circ$.

We know that $\text{cosec }(360^\circ n + \theta) = \text{cosec } \theta$ for any integer $n$.

In this case, $360^\circ n = 3 \times 360^\circ$ (so $n=3$) and $\theta = 330^\circ$.

Therefore, $\text{cosec }(1410^\circ) = \text{cosec }(3 \times 360^\circ + 330^\circ)$.

The angle $330^\circ$ is in the fourth quadrant. We can write $330^\circ$ as $360^\circ - 30^\circ$.

We know that $\text{cosec }(360^\circ - \theta) = -\text{cosec } \theta$.

We know the exact value of $\text{cosec }(30^\circ)$.

$\text{cosec }(30^\circ) = 2$.

So, $\text{cosec }(330^\circ) = -2$.

Now, substitute this back into the original expression:

$\text{cosec } (-1410^\circ) = 2$.

The final answer is:

$\text{cosec } (-1410^\circ) = 2$.

Given

The expression is $\tan \frac{19\pi}{3}$.

To Find

The value of $\tan \frac{19\pi}{3}$.

Solution

We need to find the value of $\tan \frac{19\pi}{3}$. The angle $\frac{19\pi}{3}$ is greater than $2\pi$. We can find a coterminal angle by subtracting multiples of $2\pi$ from $\frac{19\pi}{3}$.

Divide 19 by 3 to find how many multiples of $2\pi$ are present. $\frac{19}{3} = 6$ with a remainder of 1. So, $\frac{19}{3} = 6 + \frac{1}{3}$.

Thus, $\frac{19\pi}{3} = \left(6 + \frac{1}{3}\right)\pi = 6\pi + \frac{\pi}{3}$.

The angle $6\pi$ is an even multiple of $\pi$ ($6\pi = 3 \times 2\pi$). Adding an even multiple of $\pi$ to an angle results in a coterminal angle with the same trigonometric values.

We know that $\tan(n\pi + \theta) = \tan \theta$ for any integer $n$. (Also $\tan(2n\pi + \theta) = \tan \theta$ as $2n\pi$ is a multiple of $\pi$).

In this case, $n\pi = 6\pi$ (so $n=6$) and $\theta = \frac{\pi}{3}$.

Therefore, $\tan \left(\frac{19\pi}{3}\right) = \tan \left(6\pi + \frac{\pi}{3}\right)$.

We know the exact value of $\tan \left(\frac{\pi}{3}\right)$. The angle $\frac{\pi}{3}$ radians is equivalent to $60^\circ$.

$\tan \left(\frac{\pi}{3}\right) = \tan(60^\circ) = \sqrt{3}$.

So, $\tan \left(\frac{19\pi}{3}\right) = \sqrt{3}$.

The final answer is:

$\tan \frac{19\pi}{3} = \sqrt{3}$.

Given

The expression is $\sin \left(-\frac{11\pi}{3}\right)$.

To Find

The value of $\sin \left(-\frac{11\pi}{3}\right)$.

Solution

We need to find the value of $\sin \left(-\frac{11\pi}{3}\right)$.

First, we use the property that $\sin(-\theta) = -\sin \theta$.

Now, we need to find the value of $\sin \left(\frac{11\pi}{3}\right)$. The angle $\frac{11\pi}{3}$ is greater than $2\pi$. We can find a coterminal angle by subtracting multiples of $2\pi$ from $\frac{11\pi}{3}$.

Divide 11 by 3 to find how many multiples of $2\pi$ are present. $\frac{11}{3} = 3$ with a remainder of 2. So, $\frac{11}{3} = 3 + \frac{2}{3}$.

Thus, $\frac{11\pi}{3} = \left(3 + \frac{2}{3}\right)\pi = 3\pi + \frac{2\pi}{3}$.

This form $3\pi + \frac{2\pi}{3}$ is not directly in the form $2n\pi + \theta$. We can write $3\pi$ as $2\pi + \pi$.

$\frac{11\pi}{3} = 2\pi + \pi + \frac{2\pi}{3} = 2\pi + \left(\pi + \frac{2\pi}{3}\right) = 2\pi + \left(\frac{3\pi + 2\pi}{3}\right) = 2\pi + \frac{5\pi}{3}$.

So, the coterminal angle is $\frac{5\pi}{3}$.

Using the property $\sin(2n\pi + \theta) = \sin \theta$, where $2n\pi = 2\pi$ and $\theta = \frac{5\pi}{3}$:

The angle $\frac{5\pi}{3}$ is in the fourth quadrant ($\frac{3\pi}{2} < \frac{5\pi}{3} < 2\pi$). We can write $\frac{5\pi}{3}$ as $2\pi - \frac{\pi}{3}$.

We know that $\sin(2\pi - \theta) = -\sin \theta$.

We know the exact value of $\sin \left(\frac{\pi}{3}\right)$. The angle $\frac{\pi}{3}$ radians is equivalent to $60^\circ$.

$\sin \left(\frac{\pi}{3}\right) = \sin(60^\circ) = \frac{\sqrt{3}}{2}$.

So, $\sin \left(\frac{5\pi}{3}\right) = -\frac{\sqrt{3}}{2}$.

Now, substitute this back into the original expression:

$\sin \left(-\frac{11\pi}{3}\right) = \frac{\sqrt{3}}{2}$.

Alternatively, we can add multiples of $2\pi$ to the negative angle to get a coterminal angle between 0 and $2\pi$.

$-\frac{11\pi}{3}$. We need to add $2n\pi$ such that $-\frac{11\pi}{3} + 2n\pi$ is in $[0, 2\pi)$.

$-\frac{11}{3} = -3.66...$. We need to add a multiple of 2 that is greater than 3.66... The smallest such multiple is 4 (since $2 \times 2 = 4$).

Add $4\pi = \frac{12\pi}{3}$ to $-\frac{11\pi}{3}$:

$-\frac{11\pi}{3} + 4\pi = -\frac{11\pi}{3} + \frac{12\pi}{3} = \frac{-\pi + 12\pi}{3} = \frac{\pi}{3}$.

So, $-\frac{11\pi}{3}$ is coterminal with $\frac{\pi}{3}$.

Thus, $\sin \left(-\frac{11\pi}{3}\right) = \sin \left(\frac{\pi}{3}\right)$.

$\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$.

The final answer is:

$\sin \left(-\frac{11\pi}{3}\right) = \frac{\sqrt{3}}{2}$.

Given

The expression is $\cot \left(-\frac{15\pi}{4}\right)$.

To Find

The value of $\cot \left(-\frac{15\pi}{4}\right)$.

Solution

We need to find the value of $\cot \left(-\frac{15\pi}{4}\right)$.

First, we use the property that $\cot(-\theta) = -\cot \theta$.

Now, we need to find the value of $\cot \left(\frac{15\pi}{4}\right)$. The angle $\frac{15\pi}{4}$ is greater than $2\pi$. We can find a coterminal angle by subtracting multiples of $2\pi$ from $\frac{15\pi}{4}$.

Divide 15 by 4 to find how many multiples of $2\pi$ are present. $\frac{15}{4} = 3$ with a remainder of 3. So, $\frac{15}{4} = 3 + \frac{3}{4}$.

Thus, $\frac{15\pi}{4} = \left(3 + \frac{3}{4}\right)\pi = 3\pi + \frac{3\pi}{4}$.

This form $3\pi + \frac{3\pi}{4}$ is not directly in the form $2n\pi + \theta$. We can write $3\pi$ as $2\pi + \pi$.

$\frac{15\pi}{4} = 2\pi + \pi + \frac{3\pi}{4} = 2\pi + \left(\pi + \frac{3\pi}{4}\right) = 2\pi + \left(\frac{4\pi + 3\pi}{4}\right) = 2\pi + \frac{7\pi}{4}$.

So, the coterminal angle is $\frac{7\pi}{4}$.

Using the property $\cot(2n\pi + \theta) = \cot \theta$, where $2n\pi = 2\pi$ and $\theta = \frac{7\pi}{4}$:

The angle $\frac{7\pi}{4}$ is in the fourth quadrant ($\frac{3\pi}{2} < \frac{7\pi}{4} < 2\pi$). We can write $\frac{7\pi}{4}$ as $2\pi - \frac{\pi}{4}$.

We know that $\cot(2\pi - \theta) = -\cot \theta$.

We know the exact value of $\cot \left(\frac{\pi}{4}\right)$. The angle $\frac{\pi}{4}$ radians is equivalent to $45^\circ$.

$\cot \left(\frac{\pi}{4}\right) = \cot(45^\circ) = 1$.

So, $\cot \left(\frac{7\pi}{4}\right) = -1$.

Now, substitute this back into the original expression:

$\cot \left(-\frac{15\pi}{4}\right) = 1$.

Alternatively, we can add multiples of $2\pi$ to the negative angle to get a coterminal angle between 0 and $2\pi$.

$-\frac{15\pi}{4}$. We need to add $2n\pi$ such that $-\frac{15\pi}{4} + 2n\pi$ is in $[0, 2\pi)$.

$-\frac{15}{4} = -3.75$. We need to add a multiple of 2 that is greater than 3.75. The smallest such multiple is 4 (since $2 \times 2 = 4$).

Add $4\pi = \frac{16\pi}{4}$ to $-\frac{15\pi}{4}$:

$-\frac{15\pi}{4} + 4\pi = -\frac{15\pi}{4} + \frac{16\pi}{4} = \frac{-15\pi + 16\pi}{4} = \frac{\pi}{4}$.

So, $-\frac{15\pi}{4}$ is coterminal with $\frac{\pi}{4}$.

Thus, $\cot \left(-\frac{15\pi}{4}\right) = \cot \left(\frac{\pi}{4}\right)$.

$\cot \left(\frac{\pi}{4}\right) = 1$.

The final answer is:

$\cot \left(-\frac{15\pi}{4}\right) = 1$.

To Prove

$3\sin\frac{\pi}{6}\sec\frac{\pi}{3} - 4\sin\frac{5\pi}{6}\cot\frac{\pi}{4}=1$

Proof

We need to evaluate the Left Hand Side (LHS) of the equation and show that it is equal to 1.

LHS $= 3\sin\frac{\pi}{6}\sec\frac{\pi}{3} - 4\sin\frac{5\pi}{6}\cot\frac{\pi}{4}$

Let's find the values of each trigonometric term:

$\frac{\pi}{6}$ radians = $30^\circ$. $\sin \frac{\pi}{6} = \sin 30^\circ = \frac{1}{2}$.

$\frac{\pi}{3}$ radians = $60^\circ$. $\sec \frac{\pi}{3} = \sec 60^\circ = \frac{1}{\cos 60^\circ} = \frac{1}{1/2} = 2$.

$\frac{\pi}{4}$ radians = $45^\circ$. $\cot \frac{\pi}{4} = \cot 45^\circ = 1$.

For $\sin \frac{5\pi}{6}$, we notice that $\frac{5\pi}{6}$ is in the second quadrant ($\frac{\pi}{2} < \frac{5\pi}{6} < \pi$). We can write $\frac{5\pi}{6}$ as $\pi - \frac{\pi}{6}$.

Using the reduction formula $\sin(\pi - \theta) = \sin \theta$:

$\sin \frac{5\pi}{6} = \sin 30^\circ = \frac{1}{2}$.

Now substitute these values back into the LHS expression:

LHS $= 3 \times \left(\frac{1}{2}\right) \times (2) - 4 \times \left(\frac{1}{2}\right) \times (1)$

Perform the multiplications:

LHS $= 3 \times \frac{1}{\cancel{2}} \times \cancel{2} - \cancel{4}^2 \times \frac{1}{\cancel{2}_{1}} \times 1$

LHS $= 3 \times 1 \times 1 - 2 \times 1 \times 1$

LHS $= 3 - 2$

LHS $= 1$.

The Right Hand Side (RHS) of the equation is 1.

Since LHS = RHS (1 = 1), the equation is proven.

Hence Proved.

Given

The expression is $\sin 15^\circ$.

To Find

The value of $\sin 15^\circ$.

Solution

We need to find the value of $\sin 15^\circ$. We can express $15^\circ$ as the difference of two standard angles whose trigonometric values are known. For example, $15^\circ = 45^\circ - 30^\circ$ or $15^\circ = 60^\circ - 45^\circ$.

Let's use the difference of angles formula for sine:

Let $A = 45^\circ$ and $B = 30^\circ$. Then $A - B = 45^\circ - 30^\circ = 15^\circ$.

Substitute these values into the formula:

$\sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ$

Substitute the known standard values:

$\sin 45^\circ = \frac{1}{\sqrt{2}}$

$\cos 30^\circ = \frac{\sqrt{3}}{2}$

$\cos 45^\circ = \frac{1}{\sqrt{2}}$

$\sin 30^\circ = \frac{1}{2}$

So, $\sin 15^\circ = \left(\frac{1}{\sqrt{2}}\right) \left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{2}\right)$

$\sin 15^\circ = \frac{1 \times \sqrt{3}}{\sqrt{2} \times 2} - \frac{1 \times 1}{\sqrt{2} \times 2}$

$\sin 15^\circ = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}}$

Combine the terms with a common denominator:

We can rationalize the denominator by multiplying the numerator and the denominator by $\sqrt{2}$:

$\sin 15^\circ = \frac{\sqrt{3} \times \sqrt{2} - 1 \times \sqrt{2}}{2 \times (\sqrt{2})^2} = \frac{\sqrt{6} - \sqrt{2}}{2 \times 2} = \frac{\sqrt{6} - \sqrt{2}}{4}$.

Alternatively, using $15^\circ = 60^\circ - 45^\circ$

$\sin 15^\circ = \sin(60^\circ - 45^\circ) = \sin 60^\circ \cos 45^\circ - \cos 60^\circ \sin 45^\circ$

Substitute the known standard values:

$\sin 60^\circ = \frac{\sqrt{3}}{2}$

$\cos 45^\circ = \frac{1}{\sqrt{2}}$

$\cos 60^\circ = \frac{1}{2}$

$\sin 45^\circ = \frac{1}{\sqrt{2}}$

So, $\sin 15^\circ = \left(\frac{\sqrt{3}}{2}\right) \left(\frac{1}{\sqrt{2}}\right) - \left(\frac{1}{2}\right) \left(\frac{1}{\sqrt{2}}\right)$

$\sin 15^\circ = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}}$

Combine the terms:

$\sin 15^\circ = \frac{\sqrt{3} - 1}{2\sqrt{2}} = \frac{\sqrt{6} - \sqrt{2}}{4}$.

The final answer is:

$\sin 15^\circ = \frac{\sqrt{6} - \sqrt{2}}{4}$.

Given

The expression is $\tan \frac{13\pi}{12}$.

To Find

The value of $\tan \frac{13\pi}{12}$.

Solution

We need to find the value of $\tan \frac{13\pi}{12}$. The angle $\frac{13\pi}{12}$ is slightly greater than $\pi$. We can write it as $\pi + \frac{\pi}{12}$.

We know the reduction formula $\tan(\pi + \theta) = \tan \theta$.

Now, we need to find the value of $\tan \frac{\pi}{12}$. The angle $\frac{\pi}{12}$ radians is equivalent to $\frac{180^\circ}{12} = 15^\circ$. So, we need to find $\tan 15^\circ$.

We can express $15^\circ$ as the difference of two standard angles, e.g., $45^\circ - 30^\circ$.

Use the difference of angles formula for tangent:

Let $A = 45^\circ$ and $B = 30^\circ$. Then $A - B = 45^\circ - 30^\circ = 15^\circ$.

Substitute these values into the formula:

$\tan 15^\circ = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ}$

Substitute the known standard values:

$\tan 45^\circ = 1$

$\tan 30^\circ = \frac{1}{\sqrt{3}}$

So, $\tan 15^\circ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + (1) \left(\frac{1}{\sqrt{3}}\right)} = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}}$.

Simplify the complex fraction:

$\tan 15^\circ = \frac{\sqrt{3} - 1}{\cancel{\sqrt{3}}} \times \frac{\cancel{\sqrt{3}}}{\sqrt{3} + 1} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$.

Rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is $\sqrt{3} - 1$:

$\tan 15^\circ = \frac{(\sqrt{3} - 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - 1^2}$

Expand the numerator and simplify the denominator:

$\tan 15^\circ = \frac{(\sqrt{3})^2 - 2(\sqrt{3})(1) + 1^2}{3 - 1} = \frac{3 - 2\sqrt{3} + 1}{2} = \frac{4 - 2\sqrt{3}}{2}$.

Factor out 2 from the numerator:

So, $\tan \frac{\pi}{12} = 2 - \sqrt{3}$.

Therefore, $\tan \frac{13\pi}{12} = \tan \frac{\pi}{12} = 2 - \sqrt{3}$.

The final answer is:

$\tan \frac{13\pi}{12} = 2 - \sqrt{3}$.

To Prove

$\frac{\sin\;(x \;+\; y)}{\sin\;(x \;-\; y)} = \frac{\tan\;x \;+\; \tan\;y}{\tan\;x \;-\; \tan\;y}$

Proof

We will start with the Left Hand Side (LHS) of the equation and transform it into the Right Hand Side (RHS).

LHS $= \frac{\sin\;(x \;+\; y)}{\sin\;(x \;-\; y)}$

Use the sum and difference formulas for sine:

$\sin(A + B) = \sin A \cos B + \cos A \sin B$

$\sin(A - B) = \sin A \cos B - \cos A \sin B$

Apply these formulas to the LHS with $A=x$ and $B=y$:

We want the RHS in terms of tangent. Recall that $\tan \theta = \frac{\sin \theta}{\cos \theta}$. To introduce $\tan x$ and $\tan y$, we can divide the numerator and the denominator of the LHS by $\cos x \cos y$.

Divide each term in the numerator and denominator by $\cos x \cos y$:

Simplify each term by cancelling common factors:

$\frac{\sin x \cos y}{\cos x \cos y} = \frac{\sin x}{\cos x} \times \frac{\cos y}{\cos y} = \tan x \times 1 = \tan x$

$\frac{\cos x \sin y}{\cos x \cos y} = \frac{\cos x}{\cos x} \times \frac{\sin y}{\cos y} = 1 \times \tan y = \tan y$

Substitute these simplified terms back into the expression for LHS:

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven.

Hence Proved.

To Show

$\tan 3x \tan 2x \tan x = \tan 3x – \tan 2x – \tan x$

Proof

We can write $3x$ as the sum of $2x$ and $x$.

Take the tangent of both sides of this equation:

Use the sum of angles formula for tangent on the right side: $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.

Let $A = 2x$ and $B = x$.

Multiply both sides of the equation by $(1 - \tan 2x \tan x)$:

Distribute $\tan 3x$ on the left side:

Rearrange the terms to match the desired equation. Move the term $\tan 3x \tan 2x \tan x$ to the right side and move $\tan 2x + \tan x$ to the left side.

This is the same as the equation we were asked to show, just with the sides swapped.

Thus, we have shown that $\tan 3x \tan 2x \tan x = \tan 3x – \tan 2x – \tan x$.

Hence Proved.

To Prove

$\cos\left( \frac{\pi}{4}+x \right) + \cos\left( \frac{\pi}{4}-x \right)= \sqrt{2} \;\cos \;x$

Proof

We will start with the Left Hand Side (LHS) of the equation and simplify it.

LHS $= \cos\left( \frac{\pi}{4}+x \right) + \cos\left( \frac{\pi}{4}-x \right)$

We can use the sum and difference formulas for cosine:

$\cos(A + B) = \cos A \cos B - \sin A \sin B$

$\cos(A - B) = \cos A \cos B + \sin A \sin B$

Apply these formulas to the terms in the LHS with $A = \frac{\pi}{4}$ and $B = x$:

$\cos\left( \frac{\pi}{4}+x \right) = \cos \frac{\pi}{4} \cos x - \sin \frac{\pi}{4} \sin x$

$\cos\left( \frac{\pi}{4}-x \right) = \cos \frac{\pi}{4} \cos x + \sin \frac{\pi}{4} \sin x$

Substitute these expansions back into the LHS:

LHS $= (\cos \frac{\pi}{4} \cos x - \sin \frac{\pi}{4} \sin x) + (\cos \frac{\pi}{4} \cos x + \sin \frac{\pi}{4} \sin x)$

Combine like terms. Notice that the terms $-\sin \frac{\pi}{4} \sin x$ and $+\sin \frac{\pi}{4} \sin x$ cancel out:

LHS $= \cos \frac{\pi}{4} \cos x + \cos \frac{\pi}{4} \cos x$

LHS $= 2 \cos \frac{\pi}{4} \cos x$.

We know the value of $\cos \frac{\pi}{4}$. The angle $\frac{\pi}{4}$ radians is equivalent to $45^\circ$.

$\cos \frac{\pi}{4} = \cos 45^\circ = \frac{1}{\sqrt{2}}$.

Substitute this value into the expression for LHS:

Simplify the expression. We know that $2 = \sqrt{2} \times \sqrt{2}$.

LHS $= \sqrt{2} \cos x$.

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven.

Alternatively, using the sum-to-product formula:

The sum-to-product formula for $\cos A + \cos B$ is:

Let $A = \frac{\pi}{4} + x$ and $B = \frac{\pi}{4} - x$.

Then $A + B = \left(\frac{\pi}{4} + x\right) + \left(\frac{\pi}{4} - x\right) = \frac{\pi}{4} + \frac{\pi}{4} + x - x = \frac{2\pi}{4} = \frac{\pi}{2}$.

And $\frac{A + B}{2} = \frac{\pi/2}{2} = \frac{\pi}{4}$.

Also, $A - B = \left(\frac{\pi}{4} + x\right) - \left(\frac{\pi}{4} - x\right) = \frac{\pi}{4} + x - \frac{\pi}{4} + x = x + x = 2x$.

And $\frac{A - B}{2} = \frac{2x}{2} = x$.

Substitute these into the sum-to-product formula:

Substitute the value of $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$:

Simplify $2 \times \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

This is equal to the RHS.

Since LHS = RHS, the identity is proven.

Hence Proved.

To Prove

$\frac{\cos\;7x \;+\; \cos \;5x}{\sin \;7x \;-\; \sin \;5x} = \cot x$

Proof

We will start with the Left Hand Side (LHS) of the equation and simplify it using the sum and difference formulas for cosine and sine in product form.

Use the sum-to-product formula for the numerator ($\cos A + \cos B$):

Let $A = 7x$ and $B = 5x$.

Numerator: $\cos 7x + \cos 5x = 2 \cos \left(\frac{7x + 5x}{2}\right) \cos \left(\frac{7x - 5x}{2}\right)$

$= 2 \cos \left(\frac{12x}{2}\right) \cos \left(\frac{2x}{2}\right) = 2 \cos (6x) \cos (x)$.

Use the difference-to-product formula for the denominator ($\sin A - \sin B$):

Let $A = 7x$ and $B = 5x$.

Denominator: $\sin 7x - \sin 5x = 2 \cos \left(\frac{7x + 5x}{2}\right) \sin \left(\frac{7x - 5x}{2}\right)$

$= 2 \cos \left(\frac{12x}{2}\right) \sin \left(\frac{2x}{2}\right) = 2 \cos (6x) \sin (x)$.

Now substitute these expressions back into the LHS:

Cancel the common terms ($2$ and $\cos (6x)$):

Recall the definition of $\cot x = \frac{\cos x}{\sin x}$.

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven (under the assumption that $\cos(6x) \neq 0$ and $\sin(x) \neq 0$).

Hence Proved.

To Prove

$\frac{\sin\;5x \;-\; 2\sin \;3x \;+\; \sin \;x}{\cos \;5x \;-\; \cos \;x} = \tan x$

Proof

We will start with the Left Hand Side (LHS) of the equation and simplify it.

LHS $= \frac{\sin\;5x \;-\; 2\sin \;3x \;+\; \sin \;x}{\cos \;5x \;-\; \cos \;x}$

Rearrange the terms in the numerator:

Numerator $= (\sin 5x + \sin x) - 2\sin 3x$

Apply the sum-to-product formula for $\sin A + \sin B = 2 \sin \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right)$ to the term $(\sin 5x + \sin x)$:

Let $A = 5x$ and $B = x$.

$\frac{A + B}{2} = \frac{5x + x}{2} = 3x$

$\frac{A - B}{2} = \frac{5x - x}{2} = 2x$

So, $\sin 5x + \sin x = 2 \sin (3x) \cos (2x)$.

Substitute this back into the numerator expression:

Numerator $= 2 \sin (3x) \cos (2x) - 2 \sin 3x$

Factor out the common term $2 \sin 3x$ from the numerator:

Now, apply the difference-to-product formula for the denominator $\cos A - \cos B = -2 \sin \left(\frac{A + B}{2}\right) \sin \left(\frac{A - B}{2}\right)$ to the term $(\cos 5x - \cos x)$:

Let $A = 5x$ and $B = x$.

$\frac{A + B}{2} = \frac{5x + x}{2} = 3x$

$\frac{A - B}{2} = \frac{5x - x}{2} = 2x$

So, $\cos 5x - \cos x = -2 \sin (3x) \sin (2x)$.

Substitute the simplified expressions for the numerator and denominator back into the LHS:

Cancel the common term $2 \sin 3x$ from the numerator and the denominator (assuming $\sin 3x \neq 0$):

Use the double angle identities: $\cos 2x = 1 - 2\sin^2 x$ and $\sin 2x = 2 \sin x \cos x$.

From $\cos 2x = 1 - 2\sin^2 x$, we have $\cos 2x - 1 = -2\sin^2 x$.

Substitute these identities into the LHS expression:

Cancel the negative signs and the common term $2 \sin x$ from the numerator and the denominator (assuming $\sin x \neq 0$ and $\cos x \neq 0$):

Recall the definition of $\tan x = \frac{\sin x}{\cos x}$.

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven (provided that the denominators are not zero, i.e., $\sin 3x \neq 0$ and $\cos x \neq 0$).

Hence Proved.

To Prove

$\sin^2\frac{\pi}{6}+\cos^2\frac{\pi}{3}-\tan^2\frac{\pi}{4} = -\frac{1}{2}$

Proof

We will evaluate the Left Hand Side (LHS) of the equation and show that it is equal to $-\frac{1}{2}$.

LHS $= \sin^2\frac{\pi}{6}+\cos^2\frac{\pi}{3}-\tan^2\frac{\pi}{4}$

First, find the values of the trigonometric functions for the given angles:

$\frac{\pi}{6}$ radians = $30^\circ$. $\sin \frac{\pi}{6} = \sin 30^\circ = \frac{1}{2}$.

$\frac{\pi}{3}$ radians = $60^\circ$. $\cos \frac{\pi}{3} = \cos 60^\circ = \frac{1}{2}$.

$\frac{\pi}{4}$ radians = $45^\circ$. $\tan \frac{\pi}{4} = \tan 45^\circ = 1$.

Now, substitute these values into the LHS expression and square them:

Calculate the squares:

LHS $= \frac{1^2}{2^2} + \frac{1^2}{2^2} - 1^2$

LHS $= \frac{1}{4} + \frac{1}{4} - 1$

Add the fractions:

LHS $= \frac{1 + 1}{4} - 1 = \frac{2}{4} - 1 = \frac{1}{2} - 1$.

Perform the subtraction:

The Right Hand Side (RHS) of the given equation is $-\frac{1}{2}$.

Since LHS = RHS ($-\frac{1}{2} = -\frac{1}{2}$), the equation is proven.

Hence Proved.

To Prove

$2\sin^2\frac{\pi}{6}+\text{cosec}^2 \;\frac{7\pi}{6}\;\cos^2\frac{\pi}{3} = \frac{3}{2}$

Proof

We will evaluate the Left Hand Side (LHS) of the equation and show that it is equal to $\frac{3}{2}$.

LHS $= 2\sin^2\frac{\pi}{6}+\text{cosec}^2 \;\frac{7\pi}{6}\;\cos^2\frac{\pi}{3}$

First, find the values of the trigonometric functions for the given angles:

$\frac{\pi}{6}$ radians = $30^\circ$. $\sin \frac{\pi}{6} = \sin 30^\circ = \frac{1}{2}$.

$\frac{\pi}{3}$ radians = $60^\circ$. $\cos \frac{\pi}{3} = \cos 60^\circ = \frac{1}{2}$.

For $\text{cosec } \frac{7\pi}{6}$, we first evaluate the angle. $\frac{7\pi}{6}$ is in the third quadrant ($\pi < \frac{7\pi}{6} < \frac{3\pi}{2}$). We can write $\frac{7\pi}{6}$ as $\pi + \frac{\pi}{6}$.

Using the reduction formula $\text{cosec }(\pi + \theta) = -\text{cosec } \theta$:

$\text{cosec } \frac{\pi}{6} = \text{cosec } 30^\circ = \frac{1}{\sin 30^\circ} = \frac{1}{1/2} = 2$.

So, $\text{cosec } \frac{7\pi}{6} = -2$.

Now, substitute these values back into the LHS expression and square them:

Calculate the squares:

LHS $= 2 \times \frac{1}{4} + 4 \times \frac{1}{4}$

Perform the multiplications:

LHS $= \frac{2}{4} + \frac{4}{4} = \frac{1}{2} + 1$.

Perform the addition:

The Right Hand Side (RHS) of the given equation is $\frac{3}{2}$.

Since LHS = RHS ($\frac{3}{2} = \frac{3}{2}$), the equation is proven.

Hence Proved.

To Prove

$\cot^2\frac{\pi}{6}+\text{cosec } \;\frac{5\pi}{6}+3\tan^2\frac{\pi}{6} = 6$

Proof

We will evaluate the Left Hand Side (LHS) of the equation and show that it is equal to 6.

LHS $= \cot^2\frac{\pi}{6}+\text{cosec } \;\frac{5\pi}{6}+3\tan^2\frac{\pi}{6}$

First, find the values of the trigonometric functions for the given angles:

$\frac{\pi}{6}$ radians = $30^\circ$. $\cot \frac{\pi}{6} = \cot 30^\circ = \sqrt{3}$.

$\frac{\pi}{6}$ radians = $30^\circ$. $\tan \frac{\pi}{6} = \tan 30^\circ = \frac{1}{\sqrt{3}}$.

For $\text{cosec } \frac{5\pi}{6}$, we first evaluate the angle. $\frac{5\pi}{6}$ is in the second quadrant ($\frac{\pi}{2} < \frac{5\pi}{6} < \pi$). We can write $\frac{5\pi}{6}$ as $\pi - \frac{\pi}{6}$.

Using the reduction formula $\text{cosec }(\pi - \theta) = \text{cosec } \theta$:

$\text{cosec } \frac{\pi}{6} = \text{cosec } 30^\circ = \frac{1}{\sin 30^\circ} = \frac{1}{1/2} = 2$.

So, $\text{cosec } \frac{5\pi}{6} = 2$.

Now, substitute these values back into the LHS expression and square the squared terms:

Calculate the squares:

LHS $= 3 + 2 + 3 \times \frac{1}{3}$

Perform the multiplication:

LHS $= 3 + 2 + \frac{\cancel{3}^{1}}{\cancel{3}_{1}} = 3 + 2 + 1$.

Perform the addition:

The Right Hand Side (RHS) of the given equation is 6.

Since LHS = RHS (6 = 6), the equation is proven.

Hence Proved.

To Prove

$2\sin^2\frac{3\pi}{4}+2\cos^2\frac{\pi}{4}+2\sec^2\frac{\pi}{3} = 10$

Proof

We will evaluate the Left Hand Side (LHS) of the equation and show that it is equal to 10.

LHS $= 2\sin^2\frac{3\pi}{4}+2\cos^2\frac{\pi}{4}+2\sec^2\frac{\pi}{3}$

First, find the values of the trigonometric functions for the given angles:

$\frac{\pi}{4}$ radians = $45^\circ$. $\cos \frac{\pi}{4} = \cos 45^\circ = \frac{1}{\sqrt{2}}$.

$\frac{\pi}{3}$ radians = $60^\circ$. $\sec \frac{\pi}{3} = \sec 60^\circ = \frac{1}{\cos 60^\circ} = \frac{1}{1/2} = 2$.

For $\sin \frac{3\pi}{4}$, we first evaluate the angle. $\frac{3\pi}{4}$ is in the second quadrant ($\frac{\pi}{2} < \frac{3\pi}{4} < \pi$). We can write $\frac{3\pi}{4}$ as $\pi - \frac{\pi}{4}$.

Using the reduction formula $\sin(\pi - \theta) = \sin \theta$:

$\sin \frac{\pi}{4} = \sin 45^\circ = \frac{1}{\sqrt{2}}$.

So, $\sin \frac{3\pi}{4} = \frac{1}{\sqrt{2}}$.

Now, substitute these values back into the LHS expression and square the squared terms:

Calculate the squares:

LHS $= 2 \times \frac{1}{2} + 2 \times \frac{1}{2} + 2 \times 4$

Perform the multiplications:

LHS $= \cancel{2} \times \frac{1}{\cancel{2}} + \cancel{2} \times \frac{1}{\cancel{2}} + 8$

LHS $= 1 + 1 + 8$.

Perform the addition:

The Right Hand Side (RHS) of the given equation is 10.

Since LHS = RHS (10 = 10), the equation is proven.

Hence Proved.

(i) Find the value of sin 75°

We can express $75^\circ$ as the sum of two standard angles whose trigonometric values are known. For example, $75^\circ = 45^\circ + 30^\circ$.

Use the sum of angles formula for sine:

Let $A = 45^\circ$ and $B = 30^\circ$. Then $A + B = 45^\circ + 30^\circ = 75^\circ$.

Substitute these values into the formula:

$\sin 75^\circ = \sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ$

Substitute the known standard values:

$\sin 45^\circ = \frac{1}{\sqrt{2}}$

$\cos 30^\circ = \frac{\sqrt{3}}{2}$

$\cos 45^\circ = \frac{1}{\sqrt{2}}$

$\sin 30^\circ = \frac{1}{2}$

So, $\sin 75^\circ = \left(\frac{1}{\sqrt{2}}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{2}\right)$

$\sin 75^\circ = \frac{1 \times \sqrt{3}}{\sqrt{2} \times 2} + \frac{1 \times 1}{\sqrt{2} \times 2}$

$\sin 75^\circ = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}}$

Combine the terms with a common denominator:

We can rationalize the denominator by multiplying the numerator and the denominator by $\sqrt{2}$:

$\sin 75^\circ = \frac{\sqrt{3} \times \sqrt{2} + 1 \times \sqrt{2}}{2 \times (\sqrt{2})^2} = \frac{\sqrt{6} + \sqrt{2}}{2 \times 2} = \frac{\sqrt{6} + \sqrt{2}}{4}$.

(ii) Find the value of tan 15°

We can express $15^\circ$ as the difference of two standard angles whose trigonometric values are known. For example, $15^\circ = 45^\circ - 30^\circ$.

Use the difference of angles formula for tangent:

Let $A = 45^\circ$ and $B = 30^\circ$. Then $A - B = 45^\circ - 30^\circ = 15^\circ$.

Substitute these values into the formula:

$\tan 15^\circ = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ}$

Substitute the known standard values:

$\tan 45^\circ = 1$

$\tan 30^\circ = \frac{1}{\sqrt{3}}$

So, $\tan 15^\circ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + (1) \left(\frac{1}{\sqrt{3}}\right)} = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}}$.

Simplify the complex fraction:

$\tan 15^\circ = \frac{\sqrt{3} - 1}{\cancel{\sqrt{3}}} \times \frac{\cancel{\sqrt{3}}}{\sqrt{3} + 1} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$.

Rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is $\sqrt{3} - 1$:

$\tan 15^\circ = \frac{(\sqrt{3} - 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - 1^2}$

Expand the numerator and simplify the denominator:

$\tan 15^\circ = \frac{(\sqrt{3})^2 - 2(\sqrt{3})(1) + 1^2}{3 - 1} = \frac{3 - 2\sqrt{3} + 1}{2} = \frac{4 - 2\sqrt{3}}{2}$.

Factor out 2 from the numerator:

The final answers are:

(i) $\sin 75^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}$.

(ii) $\tan 15^\circ = 2 - \sqrt{3}$.

To Prove

$\cos\left( \frac{\pi}{4}-x \right) \cos\left( \frac{\pi}{4}-y \right) - \sin\left( \frac{\pi}{4}-x \right) \sin\left( \frac{\pi}{4}-y \right) = \sin \;(x + y)$

Proof

We will start with the Left Hand Side (LHS) of the equation and simplify it.

LHS $= \cos\left( \frac{\pi}{4}-x \right) \cos\left( \frac{\pi}{4}-y \right) - \sin\left( \frac{\pi}{4}-x \right) \sin\left( \frac{\pi}{4}-y \right)$

This expression has the form $\cos A \cos B - \sin A \sin B$, which is the expansion of $\cos(A + B)$.

Let $A = \frac{\pi}{4}-x$ and $B = \frac{\pi}{4}-y$.

Then the LHS is equal to $\cos(A + B)$.

Simplify the angle inside the cosine function:

LHS $= \cos\left( \frac{\pi}{4} - x + \frac{\pi}{4} - y \right)$

LHS $= \cos\left( \left(\frac{\pi}{4} + \frac{\pi}{4}\right) - (x + y) \right)$

LHS $= \cos\left( \frac{2\pi}{4} - (x + y) \right)$

Now, use the complementary angle identity $\cos\left(\frac{\pi}{2} - \theta\right) = \sin \theta$.

Let $\theta = x + y$.

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven.

Hence Proved.

To Prove

$\frac{\tan \;\left( \frac{\pi}{4} \;+\; x \right)}{\tan \;\left( \frac{\pi}{4} \;-\; x \right)}=\left( \frac{1\;+\;\tan \;x}{1\;-\;\tan \;x} \right)^{2}$

Proof

We will start with the Left Hand Side (LHS) of the equation and simplify it.

LHS $= \frac{\tan \;\left( \frac{\pi}{4} \;+\; x \right)}{\tan \;\left( \frac{\pi}{4} \;-\; x \right)}$

Use the sum and difference of angles formulas for tangent:

$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$

Apply these formulas to the numerator and the denominator of the LHS. Here, $A = \frac{\pi}{4}$ and $B = x$.

We know that $\tan \frac{\pi}{4} = \tan 45^\circ = 1$.

Numerator: $\tan\left( \frac{\pi}{4} + x \right) = \frac{\tan \frac{\pi}{4} + \tan x}{1 - \tan \frac{\pi}{4} \tan x} = \frac{1 + \tan x}{1 - (1)\tan x} = \frac{1 + \tan x}{1 - \tan x}$.

Denominator: $\tan\left( \frac{\pi}{4} - x \right) = \frac{\tan \frac{\pi}{4} - \tan x}{1 + \tan \frac{\pi}{4} \tan x} = \frac{1 - \tan x}{1 + (1)\tan x} = \frac{1 - \tan x}{1 + \tan x}$.

Substitute these back into the LHS expression:

To simplify the complex fraction, multiply the numerator by the reciprocal of the denominator:

Multiply the two fractions:

We can write this as a single square:

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven (provided that $\tan(\frac{\pi}{4} - x) \neq 0$).

Hence Proved.

To Prove

$\frac{\cos\;(\pi\;+\;x) \; \cos\;(-\;x)}{\sin\;(\pi\;-\;x) \; \cos\;\left( \frac{\pi}{2}\;+\;x \right)}=\cot^2x$

Proof

We will start with the Left Hand Side (LHS) of the equation and simplify it using reduction formulas.

LHS $= \frac{\cos\;(\pi\;+\;x) \; \cos\;(-\;x)}{\sin\;(\pi\;-\;x) \; \cos\;\left( \frac{\pi}{2}\;+\;x \right)}$

Let's evaluate each trigonometric term in the expression:

$\cos(\pi + x)$: The angle $(\pi + x)$ is in the third quadrant. In the third quadrant, cosine is negative. The reference angle is $x$. So, $\cos(\pi + x) = -\cos x$.

$\cos(-x)$: Cosine is an even function, so $\cos(-x) = \cos x$.

$\sin(\pi - x)$: The angle $(\pi - x)$ is in the second quadrant. In the second quadrant, sine is positive. The reference angle is $x$. So, $\sin(\pi - x) = \sin x$.

$\cos\left(\frac{\pi}{2} + x\right)$: The angle $\left(\frac{\pi}{2} + x\right)$ is in the second quadrant. In the second quadrant, cosine is negative. When the angle is $\frac{\pi}{2} + \theta$, cosine changes to sine. So, $\cos\left(\frac{\pi}{2} + x\right) = -\sin x$.

Now substitute these simplified terms back into the LHS expression:

Multiply the terms in the numerator and the denominator:

Cancel the negative signs:

Recall the definition of $\cot x = \frac{\cos x}{\sin x}$. Therefore, $\cot^2 x = \left(\frac{\cos x}{\sin x}\right)^2 = \frac{\cos^2 x}{\sin^2 x}$.

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven (provided that $\sin(\pi - x) \neq 0$ and $\cos(\frac{\pi}{2} + x) \neq 0$).

Hence Proved.

To Prove

$\cos \left( \frac{3\pi}{2}+x \right) \cos (2\pi+x)\left[ \cot\left( \frac{3\pi}{2}-x \right)+ \cot (2\pi+x) \right]=1$

Proof

We will start with the Left Hand Side (LHS) of the equation and simplify it using reduction formulas.

LHS $= \cos \left( \frac{3\pi}{2}+x \right) \cos (2\pi+x)\left[ \cot\left( \frac{3\pi}{2}-x \right)+ \cot (2\pi+x) \right]$

Let's evaluate each trigonometric term using reduction formulas:

$\cos\left(\frac{3\pi}{2} + x\right)$: The angle $\left(\frac{3\pi}{2} + x\right)$ is in the fourth quadrant. In the fourth quadrant, cosine is positive. When the angle is $\frac{3\pi}{2} + \theta$, cosine changes to sine. So, $\cos\left(\frac{3\pi}{2} + x\right) = \sin x$.

$\cos(2\pi + x)$: The angle $(2\pi + x)$ is coterminal with $x$. We know that $\cos(2n\pi + \theta) = \cos \theta$ for any integer $n$. So, $\cos(2\pi + x) = \cos x$.

$\cot\left(\frac{3\pi}{2} - x\right)$: The angle $\left(\frac{3\pi}{2} - x\right)$ is in the third quadrant. In the third quadrant, cotangent is positive. When the angle is $\frac{3\pi}{2} - \theta$, cotangent changes to tangent. So, $\cot\left(\frac{3\pi}{2} - x\right) = \tan x$.

$\cot(2\pi + x)$: The angle $(2\pi + x)$ is coterminal with $x$. We know that $\cot(2n\pi + \theta) = \cot \theta$ for any integer $n$. So, $\cot(2\pi + x) = \cot x$.

Now substitute these simplified terms back into the LHS expression:

Rewrite $\tan x$ and $\cot x$ in terms of $\sin x$ and $\cos x$:

$\tan x = \frac{\sin x}{\cos x}$

$\cot x = \frac{\cos x}{\sin x}$

Substitute these into the expression inside the brackets:

Combine the terms inside the brackets by finding a common denominator ($\cos x \sin x$):

Use the fundamental identity $\sin^2 x + \cos^2 x = 1$:

Substitute this back into the LHS expression:

Multiply the terms. The term $\sin x \cos x$ cancels out with the denominator (assuming $\sin x \neq 0$ and $\cos x \neq 0$).

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven (provided that the denominators are not zero). This requires $\sin x \neq 0$, $\cos x \neq 0$, $\cot(\frac{3\pi}{2}-x)$ is defined, and $\cot(2\pi+x)$ is defined.

Hence Proved.

To Prove

$\sin (n + 1)x \sin (n + 2)x + \cos (n + 1)x \cos (n + 2)x = \cos x$

Proof

We will start with the Left Hand Side (LHS) of the equation and simplify it.

LHS $= \sin (n + 1)x \sin (n + 2)x + \cos (n + 1)x \cos (n + 2)x$

Rearrange the terms in the LHS:

LHS $= \cos (n + 1)x \cos (n + 2)x + \sin (n + 1)x \sin (n + 2)x$

This expression has the form $\cos A \cos B + \sin A \sin B$, which is the expansion of $\cos(A - B)$.

Let $A = (n + 2)x$ and $B = (n + 1)x$.

Then the LHS is equal to $\cos(A - B)$.

Simplify the angle inside the cosine function:

$(n + 2)x - (n + 1)x = nx + 2x - (nx + x) = nx + 2x - nx - x = (nx - nx) + (2x - x) = 0 + x = x$.

So, the angle is $x$.

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven.

Alternatively, let $A = (n + 1)x$ and $B = (n + 2)x$.

Then the LHS is equal to $\cos(A - B)$:

Simplify the angle:

$(n + 1)x - (n + 2)x = nx + x - (nx + 2x) = nx + x - nx - 2x = (nx - nx) + (x - 2x) = 0 - x = -x$.

So, the angle is $-x$.

Use the property that $\cos(-\theta) = \cos \theta$:

This is equal to the RHS.

Since LHS = RHS, the identity is proven.

Hence Proved.

To Prove

$\cos \left( \frac{3\pi}{4} +x\right) - \cos \left( \frac{3\pi}{4} -x\right)=-\sqrt{2} \;\sin \;x$

Proof

We will start with the Left Hand Side (LHS) of the equation and simplify it.

LHS $= \cos \left( \frac{3\pi}{4} +x\right) - \cos \left( \frac{3\pi}{4} -x\right)$

We can use the difference of cosines formula (product-to-sum in reverse):

Let $A = \frac{3\pi}{4} + x$ and $B = \frac{3\pi}{4} - x$.

Calculate $\frac{A + B}{2}$:

$A + B = \left(\frac{3\pi}{4} + x\right) + \left(\frac{3\pi}{4} - x\right) = \frac{3\pi}{4} + \frac{3\pi}{4} + x - x = \frac{6\pi}{4} = \frac{3\pi}{2}$.

Calculate $\frac{A - B}{2}$:

$A - B = \left(\frac{3\pi}{4} + x\right) - \left(\frac{3\pi}{4} - x\right) = \frac{3\pi}{4} + x - \frac{3\pi}{4} + x = x + x = 2x$.

Substitute these into the difference of cosines formula:

Now, find the value of $\sin \left(\frac{3\pi}{4}\right)$. The angle $\frac{3\pi}{4}$ is in the second quadrant ($\frac{\pi}{2} < \frac{3\pi}{4} < \pi$). We can write $\frac{3\pi}{4}$ as $\pi - \frac{\pi}{4}$.

Using the reduction formula $\sin(\pi - \theta) = \sin \theta$:

We know that $\sin \left(\frac{\pi}{4}\right) = \sin 45^\circ = \frac{1}{\sqrt{2}}$.

So, $\sin \left(\frac{3\pi}{4}\right) = \frac{1}{\sqrt{2}}$.

Substitute this value back into the LHS expression:

Simplify the expression. We know that $2 = \sqrt{2} \times \sqrt{2}$.

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven.

Alternatively, using the sum/difference formulas for cosine:

$\cos(A + B) = \cos A \cos B - \sin A \sin B$

$\cos(A - B) = \cos A \cos B + \sin A \sin B$

LHS $= \cos \left( \frac{3\pi}{4} +x\right) - \cos \left( \frac{3\pi}{4} -x\right)$

LHS $= (\cos \frac{3\pi}{4} \cos x - \sin \frac{3\pi}{4} \sin x) - (\cos \frac{3\pi}{4} \cos x + \sin \frac{3\pi}{4} \sin x)$

LHS $= \cos \frac{3\pi}{4} \cos x - \sin \frac{3\pi}{4} \sin x - \cos \frac{3\pi}{4} \cos x - \sin \frac{3\pi}{4} \sin x$

Combine like terms. The terms $\cos \frac{3\pi}{4} \cos x$ and $-\cos \frac{3\pi}{4} \cos x$ cancel out:

LHS $= -2 \sin \frac{3\pi}{4} \sin x$.

As calculated before, $\sin \frac{3\pi}{4} = \frac{1}{\sqrt{2}}$.

This is equal to the RHS.

Since LHS = RHS, the identity is proven.

Hence Proved.

To Prove

$\sin^2 6x – \sin^2 4x = \sin 2x \sin 10x$

Proof

We will start with the Left Hand Side (LHS) of the equation and simplify it.

LHS $= \sin^2 6x – \sin^2 4x$

We can use the difference of squares formula: $a^2 - b^2 = (a - b)(a + b)$.

Here, $a = \sin 6x$ and $b = \sin 4x$.

Now, use the sum-to-product and difference-to-product formulas:

$\sin A - \sin B = 2 \cos \left(\frac{A + B}{2}\right) \sin \left(\frac{A - B}{2}\right)$

$\sin A + \sin B = 2 \sin \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right)$

Apply the difference formula to $(\sin 6x - \sin 4x)$ with $A=6x$ and $B=4x$:

$\frac{A + B}{2} = \frac{6x + 4x}{2} = \frac{10x}{2} = 5x$

$\frac{A - B}{2} = \frac{6x - 4x}{2} = \frac{2x}{2} = x$

So, $\sin 6x - \sin 4x = 2 \cos (5x) \sin (x)$.

Apply the sum formula to $(\sin 6x + \sin 4x)$ with $A=6x$ and $B=4x$:

$\frac{A + B}{2} = \frac{6x + 4x}{2} = 5x$

$\frac{A - B}{2} = \frac{6x - 4x}{2} = x$

So, $\sin 6x + \sin 4x = 2 \sin (5x) \cos (x)$.

Substitute these expressions back into the LHS:

Rearrange and group the terms:

LHS $= (2 \sin x \cos x)(2 \sin 5x \cos 5x)$

Use the double angle formula for sine: $\sin 2\theta = 2 \sin \theta \cos \theta$.

For the first group, $\theta = x$: $2 \sin x \cos x = \sin 2x$.

For the second group, $\theta = 5x$: $2 \sin 5x \cos 5x = \sin 2(5x) = \sin 10x$.

Substitute these back into the LHS expression:

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven.

Hence Proved.

To Prove

$\cos^2 2x – \cos^2 6x = \sin 4x \sin 8x$

Proof

We will start with the Left Hand Side (LHS) of the equation and simplify it.

LHS $= \cos^2 2x – \cos^2 6x$

We can use the difference of squares formula: $a^2 - b^2 = (a - b)(a + b)$.

Here, $a = \cos 2x$ and $b = \cos 6x$.

Now, use the sum-to-product and difference-to-product formulas:

$\cos A - \cos B = -2 \sin \left(\frac{A + B}{2}\right) \sin \left(\frac{A - B}{2}\right)$

$\cos A + \cos B = 2 \cos \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right)$

Apply the difference formula to $(\cos 2x - \cos 6x)$ with $A=2x$ and $B=6x$:

$\frac{A + B}{2} = \frac{2x + 6x}{2} = \frac{8x}{2} = 4x$

$\frac{A - B}{2} = \frac{2x - 6x}{2} = \frac{-4x}{2} = -2x$

So, $\cos 2x - \cos 6x = -2 \sin (4x) \sin (-2x)$.

Using $\sin (-\theta) = -\sin \theta$, we have $\sin (-2x) = -\sin 2x$.

So, $\cos 2x - \cos 6x = -2 \sin (4x) (-\sin 2x) = 2 \sin 4x \sin 2x$.

Apply the sum formula to $(\cos 2x + \cos 6x)$ with $A=2x$ and $B=6x$:

$\frac{A + B}{2} = \frac{2x + 6x}{2} = 4x$

$\frac{A - B}{2} = \frac{2x - 6x}{2} = -2x$

So, $\cos 2x + \cos 6x = 2 \cos (4x) \cos (-2x)$.

Using $\cos (-\theta) = \cos \theta$, we have $\cos (-2x) = \cos 2x$.

So, $\cos 2x + \cos 6x = 2 \cos 4x \cos 2x$.

Substitute these expressions back into the LHS:

Rearrange and group the terms:

LHS $= (2 \sin 4x \cos 4x)(2 \sin 2x \cos 2x)$

Use the double angle formula for sine: $\sin 2\theta = 2 \sin \theta \cos \theta$.

For the first group, $\theta = 4x$: $2 \sin 4x \cos 4x = \sin 2(4x) = \sin 8x$.

For the second group, $\theta = 2x$: $2 \sin 2x \cos 2x = \sin 2(2x) = \sin 4x$.

Substitute these back into the LHS expression:

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven.

Alternatively, using the identity $\cos^2 A - \cos^2 B = -\sin(A-B)\sin(A+B)$:

LHS $= \cos^2 2x – \cos^2 6x$

Let $A = 2x$ and $B = 6x$.

$A - B = 2x - 6x = -4x$

$A + B = 2x + 6x = 8x$

Using $\sin(-\theta) = -\sin \theta$, we have $\sin(-4x) = -\sin 4x$.

This is equal to the RHS.

Since LHS = RHS, the identity is proven.

Hence Proved.

To Prove

$\sin 2x + 2 \sin 4x + \sin 6x = 4 \cos^2 x \sin 4x$

Proof

We will start with the Left Hand Side (LHS) of the equation and simplify it.

LHS $= \sin 2x + 2 \sin 4x + \sin 6x$

Rearrange the terms to group $\sin 2x$ and $\sin 6x$:

LHS $= (\sin 6x + \sin 2x) + 2 \sin 4x$

Apply the sum-to-product formula for $\sin A + \sin B = 2 \sin \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right)$ to the term $(\sin 6x + \sin 2x)$:

Let $A = 6x$ and $B = 2x$.

$\frac{A + B}{2} = \frac{6x + 2x}{2} = \frac{8x}{2} = 4x$

$\frac{A - B}{2} = \frac{6x - 2x}{2} = \frac{4x}{2} = 2x$

So, $\sin 6x + \sin 2x = 2 \sin (4x) \cos (2x)$.

Substitute this back into the LHS expression:

Factor out the common term $2 \sin 4x$ from the LHS:

Use the double angle identity for cosine: $\cos 2x = 2 \cos^2 x - 1$.

From this, we have $\cos 2x + 1 = 2 \cos^2 x$.

Substitute this into the expression for LHS:

Multiply the terms:

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven.

Hence Proved.

To Prove

$\cot 4x (\sin 5x + \sin 3x) = \cot x (\sin 5x – \sin 3x)$

Proof

We will simplify both the Left Hand Side (LHS) and the Right Hand Side (RHS) separately and show that they are equal.

Consider the LHS:

LHS $= \cot 4x (\sin 5x + \sin 3x)$

Use the sum-to-product formula for $\sin A + \sin B = 2 \sin \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right)$ to the term $(\sin 5x + \sin 3x)$:

Let $A = 5x$ and $B = 3x$.

$\frac{A + B}{2} = \frac{5x + 3x}{2} = \frac{8x}{2} = 4x$

$\frac{A - B}{2} = \frac{5x - 3x}{2} = \frac{2x}{2} = x$

So, $\sin 5x + \sin 3x = 2 \sin (4x) \cos (x)$.

Substitute this back into the LHS expression:

Rewrite $\cot 4x$ as $\frac{\cos 4x}{\sin 4x}$:

Cancel the common term $\sin 4x$ (assuming $\sin 4x \neq 0$):

Consider the RHS:

RHS $= \cot x (\sin 5x – \sin 3x)$

Use the difference-to-product formula for $\sin A - \sin B = 2 \cos \left(\frac{A + B}{2}\right) \sin \left(\frac{A - B}{2}\right)$ to the term $(\sin 5x - \sin 3x)$:

Let $A = 5x$ and $B = 3x$.

$\frac{A + B}{2} = \frac{5x + 3x}{2} = 4x$

$\frac{A - B}{2} = \frac{5x - 3x}{2} = x$

So, $\sin 5x - \sin 3x = 2 \cos (4x) \sin (x)$.

Substitute this back into the RHS expression:

Rewrite $\cot x$ as $\frac{\cos x}{\sin x}$:

Cancel the common term $\sin x$ (assuming $\sin x \neq 0$):

Rearrange the terms:

We see that LHS $= 2 \cos 4x \cos x$ and RHS $= 2 \cos 4x \cos x$.

Since LHS = RHS, the identity is proven (provided that $\sin 4x \neq 0$ and $\sin x \neq 0$).

Hence Proved.

To Prove

$\frac{\cos \;9x \;-\; \cos \;5x}{\sin \;17x \;-\; \sin \;3x} = -\frac{\sin \;2x}{\cos \;10x}$

Proof

We will start with the Left Hand Side (LHS) of the equation and simplify it using sum/difference-to-product formulas.

LHS $= \frac{\cos \;9x \;-\; \cos \;5x}{\sin \;17x \;-\; \sin \;3x}$

Apply the difference-to-product formula for the numerator ($\cos A - \cos B = -2 \sin \left(\frac{A + B}{2}\right) \sin \left(\frac{A - B}{2}\right)$):

Let $A = 9x$ and $B = 5x$.

$\frac{A + B}{2} = \frac{9x + 5x}{2} = \frac{14x}{2} = 7x$

$\frac{A - B}{2} = \frac{9x - 5x}{2} = \frac{4x}{2} = 2x$

Numerator: $\cos 9x - \cos 5x = -2 \sin (7x) \sin (2x)$.

Apply the difference-to-product formula for the denominator ($\sin A - \sin B = 2 \cos \left(\frac{A + B}{2}\right) \sin \left(\frac{A - B}{2}\right)$):

Let $A = 17x$ and $B = 3x$.

$\frac{A + B}{2} = \frac{17x + 3x}{2} = \frac{20x}{2} = 10x$

$\frac{A - B}{2} = \frac{17x - 3x}{2} = \frac{14x}{2} = 7x$

Denominator: $\sin 17x - \sin 3x = 2 \cos (10x) \sin (7x)$.

Substitute the simplified expressions for the numerator and denominator back into the LHS:

Cancel the common terms ($2$ and $\sin (7x)$) (assuming $\sin 7x \neq 0$ and $\cos 10x \neq 0$):

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven (provided that the denominators are not zero, i.e., $\sin 7x \neq 0$ and $\cos 10x \neq 0$).

Hence Proved.

To Prove

$\frac{\sin \;5x \;+\; \sin \;3x}{\cos \;5x \;+\; \cos \;3x} = \tan 4x$

Proof

We will start with the Left Hand Side (LHS) of the equation and simplify it using sum-to-product formulas.

LHS $= \frac{\sin \;5x \;+\; \sin \;3x}{\cos \;5x \;+\; \cos \;3x}$

Apply the sum-to-product formula for the numerator ($\sin A + \sin B = 2 \sin \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right)$):

Let $A = 5x$ and $B = 3x$.

$\frac{A + B}{2} = \frac{5x + 3x}{2} = \frac{8x}{2} = 4x$

$\frac{A - B}{2} = \frac{5x - 3x}{2} = \frac{2x}{2} = x$

Numerator: $\sin 5x + \sin 3x = 2 \sin (4x) \cos (x)$.

Apply the sum-to-product formula for the denominator ($\cos A + \cos B = 2 \cos \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right)$):

Let $A = 5x$ and $B = 3x$.

$\frac{A + B}{2} = \frac{5x + 3x}{2} = 4x$

$\frac{A - B}{2} = \frac{5x - 3x}{2} = x$

Denominator: $\cos 5x + \cos 3x = 2 \cos (4x) \cos (x)$.

Substitute the simplified expressions for the numerator and denominator back into the LHS:

Cancel the common terms ($2$ and $\cos (x)$) (assuming $\cos 4x \neq 0$ and $\cos x \neq 0$):

Recall the definition of $\tan \theta = \frac{\sin \theta}{\cos \theta}$.

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven (provided that the denominator is not zero, i.e., $\cos 5x + \cos 3x \neq 0$, which means $\cos 4x \neq 0$ and $\cos x \neq 0$).

Hence Proved.

To Prove

$\frac{\sin \;x \;-\; \sin \;y}{\cos \;x \;+\; \cos \;y} = \tan \frac{x \;-\; y}{2}$

Proof

We will start with the Left Hand Side (LHS) of the equation and simplify it using difference/sum-to-product formulas.

LHS $= \frac{\sin \;x \;-\; \sin \;y}{\cos \;x \;+\; \cos \;y}$

Apply the difference-to-product formula for the numerator ($\sin A - \sin B = 2 \cos \left(\frac{A + B}{2}\right) \sin \left(\frac{A - B}{2}\right)$):

Let $A = x$ and $B = y$.

Numerator: $\sin x - \sin y = 2 \cos \left(\frac{x + y}{2}\right) \sin \left(\frac{x - y}{2}\right)$.

Apply the sum-to-product formula for the denominator ($\cos A + \cos B = 2 \cos \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right)$):

Let $A = x$ and $B = y$.

Denominator: $\cos x + \cos y = 2 \cos \left(\frac{x + y}{2}\right) \cos \left(\frac{x - y}{2}\right)$.

Substitute the simplified expressions for the numerator and denominator back into the LHS:

Cancel the common terms ($2$ and $\cos \left(\frac{x + y}{2}\right)$) (assuming $\cos \left(\frac{x + y}{2}\right) \neq 0$ and $\cos \left(\frac{x - y}{2}\right) \neq 0$):

Recall the definition of $\tan \theta = \frac{\sin \theta}{\cos \theta}$.

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven (provided that the denominator is not zero, i.e., $\cos x + \cos y \neq 0$).

Hence Proved.

To Prove

$\frac{\sin \;x \;+\; \sin \;3x}{\cos \;x \;+\; \cos \;3x} = \tan \; 2x$

Proof

We will start with the Left Hand Side (LHS) of the equation and simplify it using sum-to-product formulas.

LHS $= \frac{\sin \;x \;+\; \sin \;3x}{\cos \;x \;+\; \cos \;3x}$

Apply the sum-to-product formula for the numerator ($\sin A + \sin B = 2 \sin \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right)$):

Let $A = 3x$ and $B = x$.

$\frac{A + B}{2} = \frac{3x + x}{2} = \frac{4x}{2} = 2x$

$\frac{A - B}{2} = \frac{3x - x}{2} = \frac{2x}{2} = x$

Numerator: $\sin 3x + \sin x = 2 \sin (2x) \cos (x)$.

Apply the sum-to-product formula for the denominator ($\cos A + \cos B = 2 \cos \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right)$):

Let $A = 3x$ and $B = x$.

$\frac{A + B}{2} = \frac{3x + x}{2} = 2x$

$\frac{A - B}{2} = \frac{3x - x}{2} = x$

Denominator: $\cos 3x + \cos x = 2 \cos (2x) \cos (x)$.

Substitute the simplified expressions for the numerator and denominator back into the LHS:

Cancel the common terms ($2$ and $\cos (x)$) (assuming $\cos x \neq 0$ and $\cos 2x \neq 0$):

Recall the definition of $\tan \theta = \frac{\sin \theta}{\cos \theta}$.

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven (provided that the denominator is not zero, i.e., $\cos x + \cos 3x \neq 0$, which means $\cos x \neq 0$ and $\cos 2x \neq 0$).

Hence Proved.

To Prove

$\frac{\sin \;x\;-\;\sin\;3x}{\sin^{2}\;x\;-\;\cos^{2}\;x}=2\sin \;x$

Proof

We will start with the Left Hand Side (LHS) of the equation and simplify it.

LHS $= \frac{\sin \;x\;-\;\sin\;3x}{\sin^{2}\;x\;-\;\cos^{2}\;x}$

Apply the difference-to-product formula for the numerator ($\sin A - \sin B = 2 \cos \left(\frac{A + B}{2}\right) \sin \left(\frac{A - B}{2}\right)$):

Let $A = x$ and $B = 3x$.

$\frac{A + B}{2} = \frac{x + 3x}{2} = \frac{4x}{2} = 2x$

$\frac{A - B}{2} = \frac{x - 3x}{2} = \frac{-2x}{2} = -x$

Numerator: $\sin x - \sin 3x = 2 \cos (2x) \sin (-x)$.

Using the property $\sin (-\theta) = -\sin \theta$, we have $\sin (-x) = -\sin x$.

So, Numerator $= 2 \cos (2x) (-\sin x) = -2 \cos (2x) \sin x$.

Consider the denominator: $\sin^2 x - \cos^2 x$.

Recall the identity $\cos 2x = \cos^2 x - \sin^2 x$.

So, $\sin^2 x - \cos^2 x = -(\cos^2 x - \sin^2 x) = -\cos 2x$.

Substitute the simplified expressions for the numerator and denominator back into the LHS:

Cancel the negative signs and the common term $\cos 2x$ (assuming $\cos 2x \neq 0$):

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven (provided that the denominator is not zero, i.e., $\sin^2 x - \cos^2 x \neq 0$, which means $\cos 2x \neq 0$).

Hence Proved.

To Prove

$\frac{\cos \;4x \;+\; \cos \;3x \;+\; \cos \;2x}{\sin \;4x \;+\; \sin \;3x \;+\; \sin \;2x} = \cot \;3x$

Proof

We will start with the Left Hand Side (LHS) of the equation and simplify it.

LHS $= \frac{\cos \;4x \;+\; \cos \;3x \;+\; \cos \;2x}{\sin \;4x \;+\; \sin \;3x \;+\; \sin \;2x}$

Rearrange the terms in the numerator and denominator to group $\cos 4x$ with $\cos 2x$, and $\sin 4x$ with $\sin 2x$. The term with $3x$ is left alone for now.

Numerator $= (\cos 4x + \cos 2x) + \cos 3x$

Denominator $= (\sin 4x + \sin 2x) + \sin 3x$

Apply the sum-to-product formula for $\cos A + \cos B = 2 \cos \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right)$ to $(\cos 4x + \cos 2x)$:

Let $A = 4x$ and $B = 2x$.

$\frac{A + B}{2} = \frac{4x + 2x}{2} = \frac{6x}{2} = 3x$

$\frac{A - B}{2} = \frac{4x - 2x}{2} = \frac{2x}{2} = x$

So, $\cos 4x + \cos 2x = 2 \cos (3x) \cos (x)$.

Apply the sum-to-product formula for $\sin A + \sin B = 2 \sin \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right)$ to $(\sin 4x + \sin 2x)$:

Let $A = 4x$ and $B = 2x$.

$\frac{A + B}{2} = \frac{4x + 2x}{2} = 3x$

$\frac{A - B}{2} = \frac{4x - 2x}{2} = x$

So, $\sin 4x + \sin 2x = 2 \sin (3x) \cos (x)$.

Substitute these expressions back into the numerator and denominator:

Numerator $= (2 \cos 3x \cos x) + \cos 3x$

Denominator $= (2 \sin 3x \cos x) + \sin 3x$

Factor out the common term $\cos 3x$ from the numerator:

Factor out the common term $\sin 3x$ from the denominator:

Substitute the factored expressions back into the LHS:

Cancel the common term $(2 \cos x + 1)$ from the numerator and the denominator (assuming $2 \cos x + 1 \neq 0$ and $\sin 3x \neq 0$):

Recall the definition of $\cot \theta = \frac{\cos \theta}{\sin \theta}$.

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven (provided that the denominator is not zero, i.e., $\sin 4x + \sin 3x + \sin 2x \neq 0$).

Hence Proved.

To Prove

$\cot x \cot 2x – \cot 2x \cot 3x – \cot 3x \cot x = 1$

Proof

We can relate the angles $x$, $2x$, and $3x$ by the equation $3x = 2x + x$.

Take the cotangent of both sides:

Use the sum of angles formula for cotangent: $\cot(A + B) = \frac{\cot A \cot B - 1}{\cot B + \cot A}$.

Let $A = 2x$ and $B = x$.

Multiply both sides of the equation by $(\cot x + \cot 2x)$:

Distribute $\cot 3x$ on the left side:

Rewrite the terms:

Rearrange the terms to match the desired equation. Move the term $-1$ to the left side and move the terms $\cot x \cot 3x$ and $\cot 2x \cot 3x$ to the right side.

This is the same as the equation we were asked to prove, just with the sides swapped and terms reordered on the right side.

Thus, we have proven the identity (provided that $\cot(2x+x)$ is defined, which means $x$, $2x$, $3x$, $x+2x$ are not integer multiples of $\pi$).

Hence Proved.

To Prove

$\tan \;4x = \frac{4 \;\tan \;x (1\;-\;\tan^{2}\;x)}{1\;-\;6\;\tan^{2}\;x\;+\;\tan^{4}\;x}$

Proof

We will start with the Left Hand Side (LHS) and express $\tan 4x$ in terms of $\tan x$ using the double angle formula for tangent repeatedly.

We know that $\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}$.

First, express $\tan 4x$ as $\tan(2 \times 2x)$. Use the double angle formula with $\theta = 2x$:

Now, we need to express $\tan 2x$ in terms of $\tan x$. Use the double angle formula with $\theta = x$:

Substitute the expression for $\tan 2x$ from equation (ii) into equation (i):

Simplify the numerator and the denominator separately.

Numerator $= 2 \left(\frac{2 \tan x}{1 - \tan^2 x}\right) = \frac{4 \tan x}{1 - \tan^2 x}$.

Denominator $= 1 - \left(\frac{2 \tan x}{1 - \tan^2 x}\right)^2 = 1 - \frac{(2 \tan x)^2}{(1 - \tan^2 x)^2} = 1 - \frac{4 \tan^2 x}{(1 - \tan^2 x)^2}$.

Find a common denominator for the terms in the denominator:

Denominator $= \frac{(1 - \tan^2 x)^2}{(1 - \tan^2 x)^2} - \frac{4 \tan^2 x}{(1 - \tan^2 x)^2} = \frac{(1 - \tan^2 x)^2 - 4 \tan^2 x}{(1 - \tan^2 x)^2}$.

Expand the term $(1 - \tan^2 x)^2$ in the numerator of the denominator using $(a-b)^2 = a^2 - 2ab + b^2$:

$(1 - \tan^2 x)^2 = 1^2 - 2(1)(\tan^2 x) + (\tan^2 x)^2 = 1 - 2 \tan^2 x + \tan^4 x$.

So, Numerator of the Denominator $= (1 - 2 \tan^2 x + \tan^4 x) - 4 \tan^2 x = 1 - 2 \tan^2 x - 4 \tan^2 x + \tan^4 x = 1 - 6 \tan^2 x + \tan^4 x$.

Denominator $= \frac{1 - 6 \tan^2 x + \tan^4 x}{(1 - \tan^2 x)^2}$.

Now substitute the simplified numerator and denominator back into the expression for $\tan 4x$:

To simplify the complex fraction, multiply the numerator by the reciprocal of the denominator:

Cancel one factor of $(1 - \tan^2 x)$ from the numerator and denominator (assuming $1 - \tan^2 x \neq 0$):

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven (provided that the denominators are not zero).

Hence Proved.

To Prove

$\cos 4x = 1 – 8\sin^{2} x \cos^{2} x$

Proof

We will start with the Left Hand Side (LHS) and express $\cos 4x$ in terms of $\sin x$ and $\cos x$ using the double angle formula.

We know that $\cos 2\theta = 1 - 2\sin^2 \theta$.

Express $\cos 4x$ as $\cos(2 \times 2x)$. Use the double angle formula with $\theta = 2x$:

Now, use the double angle formula for sine: $\sin 2x = 2 \sin x \cos x$.

Square this expression: $\sin^2 2x = (2 \sin x \cos x)^2 = 2^2 \sin^2 x \cos^2 x = 4 \sin^2 x \cos^2 x$.

Substitute the expression for $\sin^2 2x$ back into the LHS:

Multiply the terms:

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven.

Alternatively, starting from the RHS:

RHS $= 1 – 8\sin^{2} x \cos^{2} x$

Rewrite the term $8\sin^2 x \cos^2 x$ as $2 \times (4 \sin^2 x \cos^2 x)$.

Recognize that $4 \sin^2 x \cos^2 x = (2 \sin x \cos x)^2 = \sin^2 2x$.

Recall the double angle formula $\cos 2\theta = 1 - 2\sin^2 \theta$.

Let $\theta = 2x$. Then $1 - 2\sin^2 2x = \cos 2(2x) = \cos 4x$.

This is equal to the LHS.

Since RHS = LHS, the identity is proven.

Hence Proved.

To Prove

$\cos 6x = 32 \cos^{6} x – 48\cos^{4} x + 18 \cos^{2} x – 1$

Proof

We will start with the Left Hand Side (LHS) and express $\cos 6x$ in terms of $\cos x$ using multiple angle formulas.

We can write $\cos 6x$ as $\cos(2 \times 3x)$ or $\cos(3 \times 2x)$. Let's use $\cos(2 \times 3x)$.

Use the double angle formula for cosine: $\cos 2\theta = 2 \cos^2 \theta - 1$.

Let $\theta = 3x$.

Now, we need to express $\cos 3x$ in terms of $\cos x$. Use the triple angle formula for cosine:

Let $\phi = x$. So, $\cos 3x = 4 \cos^3 x - 3 \cos x$.

Substitute this expression for $\cos 3x$ into equation (i):

Expand the squared term using the formula $(a - b)^2 = a^2 - 2ab + b^2$. Let $a = 4 \cos^3 x$ and $b = 3 \cos x$.

$(4 \cos^3 x - 3 \cos x)^2 = (4 \cos^3 x)^2 - 2(4 \cos^3 x)(3 \cos x) + (3 \cos x)^2$

$= 16 \cos^6 x - 24 \cos^4 x + 9 \cos^2 x$.

Substitute this expanded expression back into the LHS:

Distribute the 2 into the bracket:

LHS $= 2 \times 16 \cos^6 x - 2 \times 24 \cos^4 x + 2 \times 9 \cos^2 x - 1$

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven.

Hence Proved.

Given

$\sin x = \frac{3}{5}$, $x$ lies in the second quadrant.

$\cos y = -\frac{12}{13}$, $y$ lies in the second quadrant.

To Find

The value of $\sin (x + y)$.

Solution

We need to find the value of $\sin (x + y)$. The sum formula for sine is:

We are given $\sin x = \frac{3}{5}$ and $\cos y = -\frac{12}{13}$. We need to find $\cos x$ and $\sin y$.

Since $x$ lies in the second quadrant, $\cos x$ is negative.

Use the identity $\sin^2 x + \cos^2 x = 1$:

$\frac{9}{25} + \cos^2 x = 1$

$\cos^2 x = 1 - \frac{9}{25} = \frac{25 - 9}{25} = \frac{16}{25}$.

$\cos x = \pm\sqrt{\frac{16}{25}} = \pm\frac{4}{5}$.

Since $x$ is in the second quadrant, $\cos x = -\frac{4}{5}$.

Since $y$ lies in the second quadrant, $\sin y$ is positive.

Use the identity $\sin^2 y + \cos^2 y = 1$:

$\sin^2 y + \frac{144}{169} = 1$

$\sin^2 y = 1 - \frac{144}{169} = \frac{169 - 144}{169} = \frac{25}{169}$.

$\sin y = \pm\sqrt{\frac{25}{169}} = \pm\frac{5}{13}$.

Since $y$ is in the second quadrant, $\sin y = \frac{5}{13}$.

Now, substitute the values of $\sin x$, $\cos y$, $\cos x$, and $\sin y$ into formula (i):

Perform the multiplications:

$\sin(x + y) = \frac{3 \times (-12)}{5 \times 13} + \frac{(-4) \times 5}{5 \times 13}$

$\sin(x + y) = \frac{-36}{65} + \frac{-20}{65}$

Add the fractions:

The value of $\sin (x + y)$ is $-\frac{56}{65}$.

The final answer is:

$\sin (x + y) = -\frac{56}{65}$.

To Prove

$\cos \;2x \;\cos \frac{x}{2} - \cos \;3x \;\cos \frac{9x}{2} = \sin \;5x \;\sin \frac{5x}{2}$

Proof

We will start with the Left Hand Side (LHS) of the equation and simplify it using product-to-sum formulas.

LHS $= \cos \;2x \;\cos \frac{x}{2} - \cos \;3x \;\cos \frac{9x}{2}$

Use the product-to-sum formula $2 \cos A \cos B = \cos(A + B) + \cos(A - B)$. This means $\cos A \cos B = \frac{1}{2} [\cos(A + B) + \cos(A - B)]$.

Apply the formula to the first term $\cos 2x \cos \frac{x}{2}$. Let $A = 2x$ and $B = \frac{x}{2}$.

$A + B = 2x + \frac{x}{2} = \frac{4x + x}{2} = \frac{5x}{2}$.

$A - B = 2x - \frac{x}{2} = \frac{4x - x}{2} = \frac{3x}{2}$.

So, $\cos 2x \cos \frac{x}{2} = \frac{1}{2} \left[ \cos \left(\frac{5x}{2}\right) + \cos \left(\frac{3x}{2}\right) \right]$.

Apply the formula to the second term $\cos 3x \cos \frac{9x}{2}$. Let $A = \frac{9x}{2}$ and $B = 3x$ (or $A = 3x$ and $B = \frac{9x}{2}$). It's usually easier if A is the larger angle when finding the difference, but the formula is symmetric in A and B for the sum part. Let's use $A = \frac{9x}{2}$ and $B = 3x = \frac{6x}{2}$.

$A + B = \frac{9x}{2} + \frac{6x}{2} = \frac{15x}{2}$.

$A - B = \frac{9x}{2} - \frac{6x}{2} = \frac{3x}{2}$.

So, $\cos \frac{9x}{2} \cos 3x = \frac{1}{2} \left[ \cos \left(\frac{15x}{2}\right) + \cos \left(\frac{3x}{2}\right) \right]$.

Substitute these expressions back into the LHS:

Factor out $\frac{1}{2}$ and remove the inner brackets:

Combine like terms. Notice that the terms $\cos \left(\frac{3x}{2}\right)$ and $-\cos \left(\frac{3x}{2}\right)$ cancel out:

Apply the difference-to-product formula for $\cos A - \cos B = -2 \sin \left(\frac{A + B}{2}\right) \sin \left(\frac{A - B}{2}\right)$ to the term inside the bracket.

Let $A = \frac{5x}{2}$ and $B = \frac{15x}{2}$.

$\frac{A + B}{2} = \frac{\frac{5x}{2} + \frac{15x}{2}}{2} = \frac{\frac{20x}{2}}{2} = \frac{10x}{2} = 5x$.

$\frac{A - B}{2} = \frac{\frac{5x}{2} - \frac{15x}{2}}{2} = \frac{\frac{-10x}{2}}{2} = \frac{-5x}{2}$.

So, $\cos \left(\frac{5x}{2}\right) - \cos \left(\frac{15x}{2}\right) = -2 \sin (5x) \sin \left(\frac{-5x}{2}\right)$.

Using the property $\sin (-\theta) = -\sin \theta$, we have $\sin \left(\frac{-5x}{2}\right) = -\sin \left(\frac{5x}{2}\right)$.

So, $\cos \left(\frac{5x}{2}\right) - \cos \left(\frac{15x}{2}\right) = -2 \sin (5x) \left(-\sin \left(\frac{5x}{2}\right)\right) = 2 \sin 5x \sin \left(\frac{5x}{2}\right)$.

Substitute this back into the LHS expression:

Multiply the terms:

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven.

Hence Proved.

Given

The expression is $\tan \frac{\pi}{8}$.

To Find

The value of $\tan \frac{\pi}{8}$.

Solution

We need to find the value of $\tan \frac{\pi}{8}$. The angle $\frac{\pi}{8}$ is half of $\frac{\pi}{4}$. We can use the half-angle formula for tangent.

The half-angle formula for tangent is $\tan \frac{\theta}{2} = \frac{1 - \cos \theta}{\sin \theta}$ or $\tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta}$.

Let $\frac{\theta}{2} = \frac{\pi}{8}$. Then $\theta = 2 \times \frac{\pi}{8} = \frac{\pi}{4}$.

We know the values of $\sin \frac{\pi}{4}$ and $\cos \frac{\pi}{4}$.

$\sin \frac{\pi}{4} = \sin 45^\circ = \frac{1}{\sqrt{2}}$.

$\cos \frac{\pi}{4} = \cos 45^\circ = \frac{1}{\sqrt{2}}$.

Using the formula $\tan \frac{\theta}{2} = \frac{1 - \cos \theta}{\sin \theta}$ with $\theta = \frac{\pi}{4}$:

Substitute the values of $\sin \frac{\pi}{4}$ and $\cos \frac{\pi}{4}$:

Simplify the numerator:

$1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}}$.

Substitute back into the expression for $\tan \frac{\pi}{8}$:

Simplify the complex fraction by multiplying the numerator by the reciprocal of the denominator:

So, $\tan \frac{\pi}{8} = \sqrt{2} - 1$.

Alternatively, using the formula $\tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta}$ with $\theta = \frac{\pi}{4}$:

Substitute the values:

Simplify the denominator:

$1 + \frac{1}{\sqrt{2}} = \frac{\sqrt{2} + 1}{\sqrt{2}}$.

Substitute back into the expression for $\tan \frac{\pi}{8}$:

Simplify the complex fraction:

Rationalize the denominator by multiplying the numerator and denominator by the conjugate, $\sqrt{2} - 1$:

So, $\tan \frac{\pi}{8} = \sqrt{2} - 1$.

The final answer is:

$\tan \frac{\pi}{8} = \sqrt{2} - 1$.

Given

$\tan x = \frac{3}{4}$.

$\pi < x < \frac{3\pi}{2}$ (x lies in the third quadrant).

To Find

The values of $\sin \frac{x}{2}$, $\cos \frac{x}{2}$, and $\tan \frac{x}{2}$.

Solution

We are given that $\tan x = \frac{3}{4}$ and $\pi < x < \frac{3\pi}{2}$.

The condition $\pi < x < \frac{3\pi}{2}$ means $x$ is in the third quadrant. In the third quadrant, $\sin x$ and $\cos x$ are negative.

First, we need to find the values of $\sin x$ and $\cos x$. We can use the identity $\sec^2 x = 1 + \tan^2 x$.

$\sec x = \pm\sqrt{\frac{25}{16}} = \pm\frac{5}{4}$.

Since $x$ is in the third quadrant, $\sec x$ is negative.

Now, find $\cos x$ using the reciprocal identity $\cos x = \frac{1}{\sec x}$.

Find $\sin x$ using the identity $\tan x = \frac{\sin x}{\cos x}$, which means $\sin x = \tan x \times \cos x$.

Now, consider the quadrant for $\frac{x}{2}$. The given condition is $\pi < x < \frac{3\pi}{2}$.

Divide the inequality by 2:

This means $\frac{x}{2}$ lies in the second quadrant.

In the second quadrant, $\sin \frac{x}{2}$ is positive, $\cos \frac{x}{2}$ is negative, and $\tan \frac{x}{2}$ is negative.

Use the half-angle formulas:

$\sin^2 \frac{x}{2} = \frac{1 - \cos x}{2}$

$\cos^2 \frac{x}{2} = \frac{1 + \cos x}{2}$

$\tan \frac{x}{2} = \frac{1 - \cos x}{\sin x}$ (or $\tan \frac{x}{2} = \frac{\sin x}{1 + \cos x}$)

Substitute the value $\cos x = -\frac{4}{5}$ into the formulas for $\sin^2 \frac{x}{2}$ and $\cos^2 \frac{x}{2}$:

For $\sin^2 \frac{x}{2}$:

$\sin \frac{x}{2} = \pm\sqrt{\frac{9}{10}} = \pm\frac{3}{\sqrt{10}}$.

Since $\frac{x}{2}$ is in the second quadrant, $\sin \frac{x}{2}$ is positive.

Rationalize the denominator: $\sin \frac{x}{2} = \frac{3\sqrt{10}}{10}$.

For $\cos^2 \frac{x}{2}$:

$\cos \frac{x}{2} = \pm\sqrt{\frac{1}{10}} = \pm\frac{1}{\sqrt{10}}$.

Since $\frac{x}{2}$ is in the second quadrant, $\cos \frac{x}{2}$ is negative.

Rationalize the denominator: $\cos \frac{x}{2} = -\frac{\sqrt{10}}{10}$.

For $\tan \frac{x}{2}$, use the formula $\tan \frac{x}{2} = \frac{1 - \cos x}{\sin x}$. Substitute $\sin x = -\frac{3}{5}$ and $\cos x = -\frac{4}{5}$.

Simplify the complex fraction:

$\tan \frac{x}{2} = \frac{9}{\cancel{5}} \times \left(-\frac{\cancel{5}}{3}\right) = -\frac{9}{3} = -3$.

The values are:

$\sin \frac{x}{2} = \frac{3}{\sqrt{10}}$ (or $\frac{3\sqrt{10}}{10}$)

$\cos \frac{x}{2} = -\frac{1}{\sqrt{10}}$ (or $-\frac{\sqrt{10}}{10}$)

$\tan \frac{x}{2} = -3$

To Prove

$\cos^{2} x + \cos^{2} \left( x+\frac{\pi}{3} \right) + \cos^{2} \left( x-\frac{\pi}{3} \right) = \frac{3}{2}$

Proof

We will start with the Left Hand Side (LHS) of the equation and simplify it.

LHS $= \cos^{2} x + \cos^{2} \left( x+\frac{\pi}{3} \right) + \cos^{2} \left( x-\frac{\pi}{3} \right)$

Use the power reduction formula $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$.

Apply this formula to each term:

$\cos^2 x = \frac{1 + \cos 2x}{2}$

$\cos^2 \left( x+\frac{\pi}{3} \right) = \frac{1 + \cos \left( 2\left(x+\frac{\pi}{3}\right) \right)}{2} = \frac{1 + \cos \left( 2x+\frac{2\pi}{3} \right)}{2}$

$\cos^2 \left( x-\frac{\pi}{3} \right) = \frac{1 + \cos \left( 2\left(x-\frac{\pi}{3}\right) \right)}{2} = \frac{1 + \cos \left( 2x-\frac{2\pi}{3} \right)}{2}$

Substitute these expressions back into the LHS:

Combine the terms with the common denominator 2:

Consider the last two terms in the numerator: $\cos \left( 2x+\frac{2\pi}{3} \right) + \cos \left( 2x-\frac{2\pi}{3} \right)$.

Use the sum-to-product formula for $\cos A + \cos B = 2 \cos \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right)$.

Let $A = 2x+\frac{2\pi}{3}$ and $B = 2x-\frac{2\pi}{3}$.

$\frac{A + B}{2} = \frac{\left(2x+\frac{2\pi}{3}\right) + \left(2x-\frac{2\pi}{3}\right)}{2} = \frac{2x + 2x + \frac{2\pi}{3} - \frac{2\pi}{3}}{2} = \frac{4x}{2} = 2x$.

$\frac{A - B}{2} = \frac{\left(2x+\frac{2\pi}{3}\right) - \left(2x-\frac{2\pi}{3}\right)}{2} = \frac{2x + \frac{2\pi}{3} - 2x + \frac{2\pi}{3}}{2} = \frac{\frac{4\pi}{3}}{2} = \frac{4\pi}{6} = \frac{2\pi}{3}$.

So, $\cos \left( 2x+\frac{2\pi}{3} \right) + \cos \left( 2x-\frac{2\pi}{3} \right) = 2 \cos (2x) \cos \left(\frac{2\pi}{3}\right)$.

We need the value of $\cos \left(\frac{2\pi}{3}\right)$. The angle $\frac{2\pi}{3}$ is in the second quadrant ($\frac{\pi}{2} < \frac{2\pi}{3} < \pi$). We can write $\frac{2\pi}{3}$ as $\pi - \frac{\pi}{3}$.

Using the reduction formula $\cos(\pi - \theta) = -\cos \theta$:

We know that $\cos \left(\frac{\pi}{3}\right) = \cos 60^\circ = \frac{1}{2}$.

So, $\cos \left(\frac{2\pi}{3}\right) = -\frac{1}{2}$.

Substitute this value back into the sum-to-product expression:

$\cos \left( 2x+\frac{2\pi}{3} \right) + \cos \left( 2x-\frac{2\pi}{3} \right) = 2 \cos (2x) \left(-\frac{1}{2}\right) = -\cos 2x$.

Now substitute this back into the expression for LHS:

Simplify the numerator. The terms $\cos 2x$ and $-\cos 2x$ cancel out:

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven.

Hence Proved.

To Prove

$2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13} + \cos \frac{3 \pi}{13} + \cos \frac{5 \pi}{13} = 0$

Proof

We will start with the Left Hand Side (LHS) of the equation and simplify it.

LHS $= 2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13} + \cos \frac{3 \pi}{13} + \cos \frac{5 \pi}{13}$

Apply the product-to-sum formula for the first term $2 \cos A \cos B = \cos(A + B) + \cos(A - B)$.

Let $A = \frac{9 \pi}{13}$ and $B = \frac{\pi}{13}$.

$A + B = \frac{9 \pi}{13} + \frac{\pi}{13} = \frac{10 \pi}{13}$.

$A - B = \frac{9 \pi}{13} - \frac{\pi}{13} = \frac{8 \pi}{13}$.

So, $2 \cos \frac{9 \pi}{13} \cos \frac{\pi}{13} = \cos \left(\frac{10 \pi}{13}\right) + \cos \left(\frac{8 \pi}{13}\right)$.

Note that $\cos \frac{\pi}{13} \cos \frac{9 \pi}{13} = \cos \frac{9 \pi}{13} \cos \frac{\pi}{13}$ due to commutativity of multiplication.

Substitute this back into the LHS expression:

We can use the property $\cos(\pi - \theta) = -\cos \theta$. This is equivalent to $\cos(\pi - \theta) + \cos \theta = 0$.

Consider pairs of terms whose angles sum to $\pi$ ($\frac{13 \pi}{13}$):

$\frac{10 \pi}{13} + \frac{3 \pi}{13} = \frac{13 \pi}{13} = \pi$. So, $\cos \frac{10 \pi}{13} = \cos \left(\pi - \frac{3 \pi}{13}\right) = -\cos \frac{3 \pi}{13}$.

This implies $\cos \frac{10 \pi}{13} + \cos \frac{3 \pi}{13} = 0$.

$\frac{8 \pi}{13} + \frac{5 \pi}{13} = \frac{13 \pi}{13} = \pi$. So, $\cos \frac{8 \pi}{13} = \cos \left(\pi - \frac{5 \pi}{13}\right) = -\cos \frac{5 \pi}{13}$.

This implies $\cos \frac{8 \pi}{13} + \cos \frac{5 \pi}{13} = 0$.

Substitute these back into the LHS expression:

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven.

Alternatively, using sum-to-product on the last two terms:

LHS $= 2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13} + (\cos \frac{5 \pi}{13} + \cos \frac{3 \pi}{13})$

Apply sum-to-product formula $\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$ to $\cos \frac{5 \pi}{13} + \cos \frac{3 \pi}{13}$:

$A = \frac{5 \pi}{13}$, $B = \frac{3 \pi}{13}$.

$\frac{A+B}{2} = \frac{\frac{5 \pi}{13} + \frac{3 \pi}{13}}{2} = \frac{\frac{8 \pi}{13}}{2} = \frac{4 \pi}{13}$.

$\frac{A-B}{2} = \frac{\frac{5 \pi}{13} - \frac{3 \pi}{13}}{2} = \frac{\frac{2 \pi}{13}}{2} = \frac{\pi}{13}$.

So, $\cos \frac{5 \pi}{13} + \cos \frac{3 \pi}{13} = 2 \cos \frac{4 \pi}{13} \cos \frac{\pi}{13}$.

Substitute back into LHS:

Factor out $2 \cos \frac{\pi}{13}$:

Apply sum-to-product formula to $\cos \frac{9 \pi}{13} + \cos \frac{4 \pi}{13}$:

$A = \frac{9 \pi}{13}$, $B = \frac{4 \pi}{13}$.

$\frac{A+B}{2} = \frac{\frac{9 \pi}{13} + \frac{4 \pi}{13}}{2} = \frac{\frac{13 \pi}{13}}{2} = \frac{\pi}{2}$.

$\frac{A-B}{2} = \frac{\frac{9 \pi}{13} - \frac{4 \pi}{13}}{2} = \frac{\frac{5 \pi}{13}}{2} = \frac{5 \pi}{26}$.

So, $\cos \frac{9 \pi}{13} + \cos \frac{4 \pi}{13} = 2 \cos \frac{\pi}{2} \cos \frac{5 \pi}{26}$.

We know $\cos \frac{\pi}{2} = \cos 90^\circ = 0$.

So, $\cos \frac{9 \pi}{13} + \cos \frac{4 \pi}{13} = 2 \times 0 \times \cos \frac{5 \pi}{26} = 0$.

Substitute back into LHS:

This is equal to the RHS.

Since LHS = RHS, the identity is proven.

Hence Proved.

To Prove

$(\sin 3x + \sin x) \sin x + (\cos 3x – \cos x) \cos x = 0$

Proof

We will start with the Left Hand Side (LHS) of the equation and simplify it.

LHS $= (\sin 3x + \sin x) \sin x + (\cos 3x – \cos x) \cos x$

Distribute $\sin x$ into the first bracket and $\cos x$ into the second bracket:

Rearrange the terms to group the product terms:

Consider the first group: $\cos 3x \cos x + \sin 3x \sin x$. This is the expansion of $\cos(A - B)$.

Let $A = 3x$ and $B = x$.

So, $\cos 3x \cos x + \sin 3x \sin x = \cos(3x - x) = \cos 2x$.

Consider the second group: $\sin^2 x – \cos^2 x$.

Recall the double angle identity $\cos 2x = \cos^2 x - \sin^2 x$.

So, $\sin^2 x – \cos^2 x = -(\cos^2 x - \sin^2 x) = -\cos 2x$.

Substitute these simplified expressions back into the LHS:

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven.

Alternatively, using sum/difference-to-product formulas first:

LHS $= (\sin 3x + \sin x) \sin x + (\cos 3x – \cos x) \cos x$

Apply sum-to-product to $\sin 3x + \sin x$: $2 \sin \left(\frac{3x+x}{2}\right) \cos \left(\frac{3x-x}{2}\right) = 2 \sin 2x \cos x$.

Apply difference-to-product to $\cos 3x - \cos x$: $-2 \sin \left(\frac{3x+x}{2}\right) \sin \left(\frac{3x-x}{2}\right) = -2 \sin 2x \sin x$.

Substitute these back into LHS:

LHS $= 2 \sin 2x \cos x \sin x - 2 \sin 2x \sin x \cos x$

Rearrange the terms in the second part:

LHS $= 2 \sin 2x \sin x \cos x - 2 \sin 2x \sin x \cos x$

The two terms are identical and have opposite signs, so they cancel out.

This is equal to the RHS.

Since LHS = RHS, the identity is proven.

Hence Proved.

To Prove

$(\cos x + \cos y)^2 + (\sin x – \sin y)^2 = 4 \cos^{2} \frac{x \;+\; y}{2}$

Proof

We will start with the Left Hand Side (LHS) of the equation and expand the squared terms.

LHS $= (\cos x + \cos y)^2 + (\sin x – \sin y)^2$

Expand using $(a+b)^2 = a^2 + 2ab + b^2$ and $(a-b)^2 = a^2 - 2ab + b^2$:

$(\cos x + \cos y)^2 = \cos^2 x + 2 \cos x \cos y + \cos^2 y$

$(\sin x – \sin y)^2 = \sin^2 x - 2 \sin x \sin y + \sin^2 y$

Substitute these expansions back into the LHS:

LHS $= (\cos^2 x + 2 \cos x \cos y + \cos^2 y) + (\sin^2 x - 2 \sin x \sin y + \sin^2 y)$

Rearrange the terms to group $\sin^2 \theta + \cos^2 \theta$:

LHS $= (\cos^2 x + \sin^2 x) + (\cos^2 y + \sin^2 y) + 2 \cos x \cos y - 2 \sin x \sin y$

Use the fundamental identity $\sin^2 \theta + \cos^2 \theta = 1$:

Consider the term inside the bracket: $\cos x \cos y - \sin x \sin y$. This is the expansion of $\cos(A + B)$.

Let $A = x$ and $B = y$.

So, $\cos x \cos y - \sin x \sin y = \cos(x + y)$.

Substitute this back into the LHS expression:

Factor out 2:

Use the half-angle identity for cosine in the form $1 + \cos 2\theta = 2 \cos^2 \theta$.

Let $2\theta = x + y$, so $\theta = \frac{x + y}{2}$.

Then $1 + \cos(x + y) = 2 \cos^2 \left(\frac{x + y}{2}\right)$.

Substitute this back into the LHS expression:

Multiply the terms:

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven.

Alternate Proof

We will start with the Left Hand Side (LHS) of the equation and use sum/difference-to-product formulas directly.

LHS $= (\cos x + \cos y)^2 + (\sin x – \sin y)^2$

Apply sum-to-product formula to $\cos x + \cos y$:

Apply difference-to-product formula to $\sin x - \sin y$:

Substitute these into the LHS:

Square the terms:

LHS $= 4 \cos^2 \frac{x+y}{2} \cos^2 \frac{x-y}{2} + 4 \cos^2 \frac{x+y}{2} \sin^2 \frac{x-y}{2}$

Factor out the common term $4 \cos^2 \frac{x+y}{2}$:

Use the identity $\cos^2 \theta + \sin^2 \theta = 1$ with $\theta = \frac{x-y}{2}$:

This is equal to the RHS.

Since LHS = RHS, the identity is proven.

Hence Proved.

To Prove

$(\cos x – \cos y)^2 + (\sin x – \sin y)^2 = 4 \sin^{2} \frac{x \;-\; y}{2}$

Proof

We will start with the Left Hand Side (LHS) of the equation and expand the squared terms.

LHS $= (\cos x – \cos y)^2 + (\sin x – \sin y)^2$

Expand using $(a-b)^2 = a^2 - 2ab + b^2$:

$(\cos x – \cos y)^2 = \cos^2 x - 2 \cos x \cos y + \cos^2 y$

$(\sin x – \sin y)^2 = \sin^2 x - 2 \sin x \sin y + \sin^2 y$

Substitute these expansions back into the LHS:

Rearrange the terms to group $\sin^2 \theta + \cos^2 \theta$:

Use the fundamental identity $\sin^2 \theta + \cos^2 \theta = 1$:

Consider the term inside the bracket: $\cos x \cos y + \sin x \sin y$. This is the expansion of $\cos(A - B)$.

Let $A = x$ and $B = y$.

So, $\cos x \cos y + \sin x \sin y = \cos(x - y)$.

Substitute this back into the LHS expression:

Factor out 2:

Use the half-angle identity for sine in the form $1 - \cos 2\theta = 2 \sin^2 \theta$.

Let $2\theta = x - y$, so $\theta = \frac{x - y}{2}$.

Then $1 - \cos(x - y) = 2 \sin^2 \left(\frac{x - y}{2}\right)$.

Substitute this back into the LHS expression:

Multiply the terms:

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven.

Alternate Proof

We will start with the Left Hand Side (LHS) of the equation and use difference-to-product formulas directly.

LHS $= (\cos x – \cos y)^2 + (\sin x – \sin y)^2$

Apply difference-to-product formula to $\cos x - \cos y$:

Apply difference-to-product formula to $\sin x - \sin y$:

Substitute these into the LHS:

Square the terms:

LHS $= (-2)^2 \sin^2 \frac{x+y}{2} \sin^2 \frac{x-y}{2} + 2^2 \cos^2 \frac{x+y}{2} \sin^2 \frac{x-y}{2}$

LHS $= 4 \sin^2 \frac{x+y}{2} \sin^2 \frac{x-y}{2} + 4 \cos^2 \frac{x+y}{2} \sin^2 \frac{x-y}{2}$

Factor out the common term $4 \sin^2 \frac{x-y}{2}$:

Use the identity $\sin^2 \theta + \cos^2 \theta = 1$ with $\theta = \frac{x+y}{2}$:

This is equal to the RHS.

Since LHS = RHS, the identity is proven.

Hence Proved.

To Prove

$\sin x + \sin 3x + \sin 5x + \sin 7x = 4 \cos x \cos 2x \sin 4x$

Proof

We will start with the Left Hand Side (LHS) of the equation and simplify it by grouping terms and using sum-to-product formulas.

LHS $= \sin x + \sin 3x + \sin 5x + \sin 7x$

Group the terms strategically, for instance, group $\sin x$ with $\sin 7x$ and $\sin 3x$ with $\sin 5x$:

Apply the sum-to-product formula for $\sin A + \sin B = 2 \sin \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right)$ to each group.

For the first group $(\sin 7x + \sin x)$: Let $A = 7x$ and $B = x$.

$\frac{A + B}{2} = \frac{7x + x}{2} = \frac{8x}{2} = 4x$

$\frac{A - B}{2} = \frac{7x - x}{2} = \frac{6x}{2} = 3x$

So, $\sin 7x + \sin x = 2 \sin (4x) \cos (3x)$.

For the second group $(\sin 5x + \sin 3x)$: Let $A = 5x$ and $B = 3x$.

$\frac{A + B}{2} = \frac{5x + 3x}{2} = \frac{8x}{2} = 4x$

$\frac{A - B}{2} = \frac{5x - 3x}{2} = \frac{2x}{2} = x$

So, $\sin 5x + \sin 3x = 2 \sin (4x) \cos (x)$.

Substitute these expressions back into the LHS:

Factor out the common term $2 \sin 4x$ from the LHS:

Apply the sum-to-product formula for $\cos A + \cos B = 2 \cos \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right)$ to the term $(\cos 3x + \cos x)$:

Let $A = 3x$ and $B = x$.

$\frac{A + B}{2} = \frac{3x + x}{2} = \frac{4x}{2} = 2x$

$\frac{A - B}{2} = \frac{3x - x}{2} = \frac{2x}{2} = x$

So, $\cos 3x + \cos x = 2 \cos (2x) \cos (x)$.

Substitute this back into the LHS expression:

Multiply the terms and rearrange to match the RHS:

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven.

Hence Proved.

To Prove

$\frac{(\sin \;7x \;+\; \sin \;5x) \;+\; (\sin\; 9x \;+\; \sin \;3x)}{(\cos \;7x \;+\; \cos \;5x) \;+\; (\cos \;9x \;+\; \cos \;3x)} = \tan\;6x$

Proof

We will start with the Left Hand Side (LHS) of the equation and simplify it using sum-to-product formulas.

LHS $= \frac{(\sin \;7x \;+\; \sin \;5x) \;+\; (\sin\; 9x \;+\; \sin \;3x)}{(\cos \;7x \;+\; \cos \;5x) \;+\; (\cos \;9x \;+\; \cos \;3x)}$

Apply the sum-to-product formula for $\sin A + \sin B = 2 \sin \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right)$ to the terms in the numerator.

For $(\sin 7x + \sin 5x)$: $A=7x$, $B=5x$. $\frac{A+B}{2} = \frac{12x}{2} = 6x$, $\frac{A-B}{2} = \frac{2x}{2} = x$. So, $\sin 7x + \sin 5x = 2 \sin 6x \cos x$.

For $(\sin 9x + \sin 3x)$: $A=9x$, $B=3x$. $\frac{A+B}{2} = \frac{12x}{2} = 6x$, $\frac{A-B}{2} = \frac{6x}{2} = 3x$. So, $\sin 9x + \sin 3x = 2 \sin 6x \cos 3x$.

Substitute these into the numerator:

Numerator $= (2 \sin 6x \cos x) + (2 \sin 6x \cos 3x)$

Factor out the common term $2 \sin 6x$:

Apply the sum-to-product formula for $\cos A + \cos B = 2 \cos \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right)$ to the terms in the denominator.

For $(\cos 7x + \cos 5x)$: $A=7x$, $B=5x$. $\frac{A+B}{2} = 6x$, $\frac{A-B}{2} = x$. So, $\cos 7x + \cos 5x = 2 \cos 6x \cos x$.

For $(\cos 9x + \cos 3x)$: $A=9x$, $B=3x$. $\frac{A+B}{2} = 6x$, $\frac{A-B}{2} = 3x$. So, $\cos 9x + \cos 3x = 2 \cos 6x \cos 3x$.

Substitute these into the denominator:

Denominator $= (2 \cos 6x \cos x) + (2 \cos 6x \cos 3x)$

Factor out the common term $2 \cos 6x$:

Substitute the factored expressions for the numerator and denominator back into the LHS:

Cancel the common terms ($2$ and $(\cos x + \cos 3x)$) (assuming $\cos x + \cos 3x \neq 0$ and $\cos 6x \neq 0$):

Recall the definition of $\tan \theta = \frac{\sin \theta}{\cos \theta}$.

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven (provided that the denominator is not zero).

Hence Proved.

To Prove

$\sin 3x + \sin 2x – \sin x = 4\sin x \cos \frac{x}{2} \cos \frac{3x}{2}$

Proof

We will start with the Left Hand Side (LHS) of the equation and simplify it by grouping terms and using sum/difference-to-product formulas.

LHS $= \sin 3x + \sin 2x – \sin x$

Group $\sin 3x$ and $\sin x$:

Apply the difference-to-product formula for $\sin A - \sin B = 2 \cos \left(\frac{A + B}{2}\right) \sin \left(\frac{A - B}{2}\right)$ to the term $(\sin 3x - \sin x)$.

Let $A = 3x$ and $B = x$.

$\frac{A + B}{2} = \frac{3x + x}{2} = \frac{4x}{2} = 2x$

$\frac{A - B}{2} = \frac{3x - x}{2} = \frac{2x}{2} = x$

So, $\sin 3x - \sin x = 2 \cos (2x) \sin (x)$.

Substitute this back into the LHS expression:

Use the double angle formula for sine: $\sin 2x = 2 \sin x \cos x$.

Factor out the common term $2 \sin x$:

Apply the sum-to-product formula for $\cos A + \cos B = 2 \cos \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right)$ to the term $(\cos 2x + \cos x)$:

Let $A = 2x$ and $B = x$.

$\frac{A + B}{2} = \frac{2x + x}{2} = \frac{3x}{2}$

$\frac{A - B}{2} = \frac{2x - x}{2} = \frac{x}{2}$

So, $\cos 2x + \cos x = 2 \cos \left(\frac{3x}{2}\right) \cos \left(\frac{x}{2}\right)$.

Substitute this back into the LHS expression:

Multiply the terms and rearrange to match the RHS:

This is equal to the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is proven.

Hence Proved.

Given

$\tan x = -\frac{4}{3}$.

$x$ is in quadrant II.

To Find

The values of $\sin \frac{x}{2}$, $\cos \frac{x}{2}$, and $\tan \frac{x}{2}$.

Solution

We are given $\tan x = -\frac{4}{3}$ and that $x$ is in the second quadrant.

In the second quadrant, $\sin x$ is positive, and $\cos x$ is negative.

We can find $\cos x$ using the identity $\sec^2 x = 1 + \tan^2 x$.

$\sec x = \pm\sqrt{\frac{25}{9}} = \pm\frac{5}{3}$.

Since $x$ is in the second quadrant, $\sec x$ is negative.

Now, find $\cos x$ using the reciprocal identity $\cos x = \frac{1}{\sec x}$.

Find $\sin x$ using the identity $\tan x = \frac{\sin x}{\cos x}$, which means $\sin x = \tan x \times \cos x$.

Now, consider the quadrant for $\frac{x}{2}$. The given condition is $x$ in quadrant II. This means $\frac{\pi}{2} < x < \pi$.

Divide the inequality by 2:

This means $\frac{x}{2}$ lies in the first quadrant.

In the first quadrant, $\sin \frac{x}{2}$, $\cos \frac{x}{2}$, and $\tan \frac{x}{2}$ are all positive.

Use the half-angle formulas:

$\sin^2 \frac{x}{2} = \frac{1 - \cos x}{2}$

$\cos^2 \frac{x}{2} = \frac{1 + \cos x}{2}$

$\tan \frac{x}{2} = \frac{1 - \cos x}{\sin x}$ (or $\tan \frac{x}{2} = \frac{\sin x}{1 + \cos x}$)

Substitute the value $\cos x = -\frac{3}{5}$ into the formulas for $\sin^2 \frac{x}{2}$ and $\cos^2 \frac{x}{2}$:

For $\sin^2 \frac{x}{2}$:

$\sin \frac{x}{2} = \pm\sqrt{\frac{4}{5}} = \pm\frac{2}{\sqrt{5}}$.

Since $\frac{x}{2}$ is in the first quadrant, $\sin \frac{x}{2}$ is positive.

Rationalize the denominator: $\sin \frac{x}{2} = \frac{2\sqrt{5}}{5}$.

For $\cos^2 \frac{x}{2}$:

$\cos \frac{x}{2} = \pm\sqrt{\frac{1}{5}} = \pm\frac{1}{\sqrt{5}}$.

Since $\frac{x}{2}$ is in the first quadrant, $\cos \frac{x}{2}$ is positive.

Rationalize the denominator: $\cos \frac{x}{2} = \frac{\sqrt{5}}{5}$.

For $\tan \frac{x}{2}$, use the formula $\tan \frac{x}{2} = \frac{1 - \cos x}{\sin x}$. Substitute $\sin x = \frac{4}{5}$ and $\cos x = -\frac{3}{5}$.

Simplify the complex fraction:

$\tan \frac{x}{2} = \frac{8}{\cancel{5}} \times \frac{\cancel{5}}{4} = \frac{8}{4} = 2$.

The values are:

$\sin \frac{x}{2} = \frac{2}{\sqrt{5}}$ (or $\frac{2\sqrt{5}}{5}$)

$\cos \frac{x}{2} = \frac{1}{\sqrt{5}}$ (or $\frac{\sqrt{5}}{5}$)

$\tan \frac{x}{2} = 2$

Given

$\cos x = -\frac{1}{3}$.

$x$ in quadrant III.

To Find

The values of $\sin \frac{x}{2}$, $\cos \frac{x}{2}$, and $\tan \frac{x}{2}$.

Solution

We are given $\cos x = -\frac{1}{3}$ and that $x$ is in the third quadrant.

The condition $x$ in quadrant III means $\pi < x < \frac{3\pi}{2}$.

Consider the quadrant for $\frac{x}{2}$. The given condition is $\pi < x < \frac{3\pi}{2}$.

Divide the inequality by 2:

This means $\frac{x}{2}$ lies in the second quadrant.

In the second quadrant, $\sin \frac{x}{2}$ is positive, $\cos \frac{x}{2}$ is negative, and $\tan \frac{x}{2}$ is negative.

We are given $\cos x = -\frac{1}{3}$. We can directly use the half-angle formulas:

$\sin^2 \frac{x}{2} = \frac{1 - \cos x}{2}$

$\cos^2 \frac{x}{2} = \frac{1 + \cos x}{2}$

$\tan \frac{x}{2} = \frac{1 - \cos x}{\sin x}$ (or $\tan \frac{x}{2} = \frac{\sin x}{1 + \cos x}$)

First, find $\sin x$. Since $x$ is in the third quadrant, $\sin x$ is negative.

Use the identity $\sin^2 x + \cos^2 x = 1$:

$\sin^2 x + \frac{1}{9} = 1$

$\sin^2 x = 1 - \frac{1}{9} = \frac{9 - 1}{9} = \frac{8}{9}$.

$\sin x = \pm\sqrt{\frac{8}{9}} = \pm\frac{\sqrt{8}}{3} = \pm\frac{2\sqrt{2}}{3}$.

Since $x$ is in the third quadrant, $\sin x = -\frac{2\sqrt{2}}{3}$.

Substitute the value $\cos x = -\frac{1}{3}$ into the formulas for $\sin^2 \frac{x}{2}$ and $\cos^2 \frac{x}{2}$:

For $\sin^2 \frac{x}{2}$:

$\sin \frac{x}{2} = \pm\sqrt{\frac{2}{3}} = \pm\frac{\sqrt{2}}{\sqrt{3}}$.

Since $\frac{x}{2}$ is in the second quadrant, $\sin \frac{x}{2}$ is positive.

Rationalize the denominator: $\sin \frac{x}{2} = \frac{\sqrt{2} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{\sqrt{6}}{3}$.

For $\cos^2 \frac{x}{2}$:

$\cos \frac{x}{2} = \pm\sqrt{\frac{1}{3}} = \pm\frac{1}{\sqrt{3}}$.

Since $\frac{x}{2}$ is in the second quadrant, $\cos \frac{x}{2}$ is negative.

Rationalize the denominator: $\cos \frac{x}{2} = -\frac{1 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = -\frac{\sqrt{3}}{3}$.

For $\tan \frac{x}{2}$, use the formula $\tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$.

Simplify the square root: $\sqrt{\frac{6}{3}} = \sqrt{2}$.

The values are:

$\sin \frac{x}{2} = \sqrt{\frac{2}{3}}$ (or $\frac{\sqrt{6}}{3}$)

$\cos \frac{x}{2} = -\sqrt{\frac{1}{3}}$ (or $-\frac{\sqrt{3}}{3}$)

$\tan \frac{x}{2} = -\sqrt{2}$

Given

$\sin x = \frac{1}{4}$.

$x$ in quadrant II.

To Find

The values of $\sin \frac{x}{2}$, $\cos \frac{x}{2}$, and $\tan \frac{x}{2}$.

Solution

We are given $\sin x = \frac{1}{4}$ and that $x$ is in the second quadrant.

The condition $x$ in quadrant II means $\frac{\pi}{2} < x < \pi$.

Consider the quadrant for $\frac{x}{2}$. The given condition is $\frac{\pi}{2} < x < \pi$.

Divide the inequality by 2:

This means $\frac{x}{2}$ lies in the first quadrant.

In the first quadrant, $\sin \frac{x}{2}$, $\cos \frac{x}{2}$, and $\tan \frac{x}{2}$ are all positive.

To use the half-angle formulas $\sin^2 \frac{x}{2} = \frac{1 - \cos x}{2}$ and $\cos^2 \frac{x}{2} = \frac{1 + \cos x}{2}$, we need to find $\cos x$.

Since $x$ is in the second quadrant, $\cos x$ is negative.

Use the identity $\sin^2 x + \cos^2 x = 1$:

$\frac{1}{16} + \cos^2 x = 1$

$\cos^2 x = 1 - \frac{1}{16} = \frac{16 - 1}{16} = \frac{15}{16}$.

$\cos x = \pm\sqrt{\frac{15}{16}} = \pm\frac{\sqrt{15}}{4}$.

Since $x$ is in the second quadrant, $\cos x = -\frac{\sqrt{15}}{4}$.

Substitute the value $\cos x = -\frac{\sqrt{15}}{4}$ into the formulas for $\sin^2 \frac{x}{2}$ and $\cos^2 \frac{x}{2}$:

For $\sin^2 \frac{x}{2}$:

$\sin \frac{x}{2} = \pm\sqrt{\frac{4 + \sqrt{15}}{8}}$.

Since $\frac{x}{2}$ is in the first quadrant, $\sin \frac{x}{2}$ is positive.

We can simplify this expression by multiplying the numerator and denominator under the square root by 2:

$\sqrt{\frac{2(4 + \sqrt{15})}{2 \times 8}} = \sqrt{\frac{8 + 2\sqrt{15}}{16}} = \frac{\sqrt{8 + 2\sqrt{15}}}{4}$.

Note that $8 + 2\sqrt{15}$ is in the form $a+b+2\sqrt{ab}$, where $a=5$ and $b=3$. So, $\sqrt{8 + 2\sqrt{15}} = \sqrt{(\sqrt{5}+\sqrt{3})^2} = \sqrt{5}+\sqrt{3}$.

So, $\sin \frac{x}{2} = \frac{\sqrt{5}+\sqrt{3}}{4}$.

For $\cos^2 \frac{x}{2}$:

$\cos \frac{x}{2} = \pm\sqrt{\frac{4 - \sqrt{15}}{8}}$.

Since $\frac{x}{2}$ is in the first quadrant, $\cos \frac{x}{2}$ is positive.

Similar to $\sin \frac{x}{2}$, we can simplify this:

$\sqrt{\frac{2(4 - \sqrt{15})}{2 \times 8}} = \sqrt{\frac{8 - 2\sqrt{15}}{16}} = \frac{\sqrt{8 - 2\sqrt{15}}}{4}$.

Note that $8 - 2\sqrt{15}$ is in the form $a+b-2\sqrt{ab}$, where $a=5$ and $b=3$. So, $\sqrt{8 - 2\sqrt{15}} = \sqrt{(\sqrt{5}-\sqrt{3})^2} = \sqrt{5}-\sqrt{3}$ (since $\sqrt{5} > \sqrt{3}$).

So, $\cos \frac{x}{2} = \frac{\sqrt{5}-\sqrt{3}}{4}$.

For $\tan \frac{x}{2}$, use the formula $\tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$.

Rationalize the denominator by multiplying by the conjugate, $\sqrt{5}+\sqrt{3}$:

$\tan \frac{x}{2} = \frac{(\sqrt{5}+\sqrt{3})(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})} = \frac{(\sqrt{5}+\sqrt{3})^2}{(\sqrt{5})^2 - (\sqrt{3})^2}$

$= \frac{(\sqrt{5})^2 + 2\sqrt{5}\sqrt{3} + (\sqrt{3})^2}{5 - 3} = \frac{5 + 2\sqrt{15} + 3}{2} = \frac{8 + 2\sqrt{15}}{2}$

Factor out 2 from the numerator:

The values are:

$\sin \frac{x}{2} = \frac{\sqrt{5}+\sqrt{3}}{4}$

$\cos \frac{x}{2} = \frac{\sqrt{5}-\sqrt{3}}{4}$

$\tan \frac{x}{2} = 4 + \sqrt{15}$