Introduction to Trigonometry and Basic Ratios (Right Triangle Trigonometry)

Basic Terms Related to Trigonometry (Angles, Sides of a Right Triangle)

Trigonometry is a branch of mathematics that studies the relationships between the sides and angles of triangles. The word "trigonometry" comes from the Greek words 'trigonon' (meaning triangle) and 'metron' (meaning measure).

Historically, the study of trigonometry began by focusing on the relationships within right-angled triangles.

Angles

An angle is formed by the rotation of a ray about its fixed endpoint. The initial position of the ray is called the initial side, and the final position after rotation is called the terminal side. The fixed endpoint is called the vertex.

Angles can be measured in various units, most commonly in degrees ($^\circ$) and radians (rad). We will discuss angle measurement systems in detail later.

For basic right-triangle trigonometry, we primarily deal with acute angles (angles between $0^\circ$ and $90^\circ$) and the right angle ($90^\circ$).

Right-Angled Triangle

A right-angled triangle (or right triangle) is a triangle where one of the interior angles measures exactly $90^\circ$. The side opposite the right angle is always the longest side and is called the hypotenuse. The sum of the other two angles in a right-angled triangle is $90^\circ$, meaning they are both acute angles (angles less than $90^\circ$).

In the triangle ABC shown above, $\angle B = 90^\circ$. Therefore, triangle ABC is a right-angled triangle. The angles $\angle A$ and $\angle C$ are acute angles, and $\angle A + \angle C = 90^\circ$.

Sides of a Right-Angled Triangle (Relative to an Acute Angle)

When we consider a right-angled triangle in the context of trigonometry, the names of the sides (other than the hypotenuse) are defined relative to one of the acute angles within the triangle. Let's consider a right-angled triangle ABC, with the right angle at B, and focus on an acute angle, say $\angle C$.

• Hypotenuse (H): This is the side opposite the $90^\circ$ angle. It is always the longest side in a right-angled triangle. In $\triangle$ABC, AC is the hypotenuse.

• Opposite Side (O) or Perpendicular (P): This is the side that is directly across from the acute angle you are considering. If we consider angle C, the opposite side is AB.

• Adjacent Side (A) or Base (B): This is the side that is next to the acute angle you are considering, but it is not the hypotenuse. If we consider angle C, the adjacent side is BC.

Important Note: The identification of the Opposite and Adjacent sides depends entirely on which acute angle you choose in the right triangle. The Hypotenuse, however, remains the same regardless of which acute angle is considered.

Let's illustrate this using the same triangle ABC, right-angled at B:

• With respect to angle C ($\theta$):

  • Hypotenuse: AC

  • Opposite Side (to C): AB

  • Adjacent Side (to C): BC

• With respect to angle A ($\phi$):

  • Hypotenuse: AC

  • Opposite Side (to A): BC

  • Adjacent Side (to A): AB

Understanding these terms and their dependence on the chosen angle is crucial for defining and working with trigonometric ratios.

Example 1: Identifying Sides in a Right Triangle

Answer:

First, let's visualise the triangle PQR, with the right angle at Q.

a) With respect to angle R:

• The Hypotenuse is the side opposite the right angle (at Q), which is PR. Length = 5 cm.

• The Opposite Side to angle R is the side across from angle R, which is PQ. Length = 3 cm.

• The Adjacent Side to angle R is the side next to angle R that is not the hypotenuse, which is QR. Length = 4 cm.

So, for $\angle R$: Hypotenuse = PR, Opposite = PQ, Adjacent = QR.

b) With respect to angle P:

• The Hypotenuse is still the side opposite the right angle (at Q), which is PR. Length = 5 cm.

• The Opposite Side to angle P is the side across from angle P, which is QR. Length = 4 cm.

• The Adjacent Side to angle P is the side next to angle P that is not the hypotenuse, which is PQ. Length = 3 cm.

So, for $\angle P$: Hypotenuse = PR, Opposite = QR, Adjacent = PQ.

Notice how the Opposite and Adjacent sides swap their roles when we change the reference acute angle from R to P.

Trigonometrical Ratios (Sine, Cosine, Tangent and their Reciprocals)

Trigonometric ratios are ratios of the lengths of two sides of a right-angled triangle, defined with respect to one of its acute angles. These ratios establish a relationship between the acute angles of a right triangle and the lengths of its sides.

Consider a right-angled triangle ABC, right-angled at B, with an acute angle $\theta$ (let's take $\theta = \angle C$).

The six main trigonometric ratios are defined as follows:

1. Sine ($\sin \theta$): This is the ratio of the length of the side opposite angle $\theta$ to the length of the hypotenuse.

   $ \sin \theta = \frac{\text{Length of Opposite Side}}{\text{Length of Hypotenuse}} = \frac{O}{H} = \frac{\text{Side AB}}{\text{Side AC}} $

2. Cosine ($\cos \theta$): This is the ratio of the length of the side adjacent to angle $\theta$ to the length of the hypotenuse.

   $ \cos \theta = \frac{\text{Length of Adjacent Side}}{\text{Length of Hypotenuse}} = \frac{A}{H} = \frac{\text{Side BC}}{\text{Side AC}} $

3. Tangent ($\tan \theta$): This is the ratio of the length of the side opposite angle $\theta$ to the length of the side adjacent to angle $\theta$.

   $ \tan \theta = \frac{\text{Length of Opposite Side}}{\text{Length of Adjacent Side}} = \frac{O}{A} = \frac{\text{Side AB}}{\text{Side BC}} $

4. Cosecant ($\text{cosec} \, \theta$ or $\csc \theta$): This is the ratio of the length of the hypotenuse to the length of the side opposite angle $\theta$. It is the reciprocal of sine.

   $ \text{cosec} \, \theta = \frac{\text{Length of Hypotenuse}}{\text{Length of Opposite Side}} = \frac{H}{O} = \frac{\text{Side AC}}{\text{Side AB}} $

5. Secant ($\sec \theta$): This is the ratio of the length of the hypotenuse to the length of the side adjacent to angle $\theta$. It is the reciprocal of cosine.

   $ \sec \theta = \frac{\text{Length of Hypotenuse}}{\text{Length of Adjacent Side}} = \frac{H}{A} = \frac{\text{Side AC}}{\text{Side BC}} $

6. Cotangent ($\cot \theta$): This is the ratio of the length of the side adjacent to angle $\theta$ to the length of the side opposite angle $\theta$. It is the reciprocal of tangent.

   $ \cot \theta = \frac{\text{Length of Adjacent Side}}{\text{Length of Opposite Side}} = \frac{A}{O} = \frac{\text{Side BC}}{\text{Side AB}} $

Mnemonic for Primary Ratios

Remembering these ratios can be made easier with mnemonics.

A common mnemonic is SOH CAH TOA:

• SOH: Sine = Opposite / Hypotenuse

• CAH: Cosine = Adjacent / Hypotenuse

• TOA: Tangent = Opposite / Adjacent

Another popular mnemonic in India uses "Pandit Badri Prasad, Har Har Bole, Sona Chandi Tole":

• $\frac{P}{H} \rightarrow \text{S} \rightarrow \sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}}$

• $\frac{B}{H} \rightarrow \text{C} \rightarrow \cos \theta = \frac{\text{Base}}{\text{Hypotenuse}}$

• $\frac{P}{B} \rightarrow \text{T} \rightarrow \tan \theta = \frac{\text{Perpendicular}}{\text{Base}}$

Here, 'Perpendicular' corresponds to the 'Opposite' side, and 'Base' corresponds to the 'Adjacent' side when referring to the acute angle.

The reciprocal ratios ($\text{cosec} \, \theta$, $\sec \theta$, $\cot \theta$) are simply the flips of $\sin \theta$, $\cos \theta$, and $\tan \theta$, respectively.

Note: For similar right-angled triangles, the ratio of corresponding sides is constant. This implies that the value of a trigonometric ratio for a given angle depends only on the angle itself and not on the size of the triangle.

Example 1: Calculating Trigonometric Ratios

Answer:

Relationships between Trigonometric Ratios

The six trigonometric ratios are closely related to each other through various identities. These relationships are fundamental and are used extensively in simplifying trigonometric expressions and solving trigonometric equations.

1. Reciprocal Relationships

These relationships arise directly from the definitions of the trigonometric ratios. Since cosecant is the ratio of Hypotenuse to Opposite and sine is the ratio of Opposite to Hypotenuse, they are reciprocals of each other. Similar relationships hold for cosine/secant and tangent/cotangent.

• Cosecant and Sine:

  $ \text{cosec} \, \theta = \frac{H}{O} $ and $ \sin \theta = \frac{O}{H} $

  Thus, $ \text{cosec} \, \theta = \frac{1}{O/H} = \frac{1}{\sin \theta} $. This gives us the identity:

  $\mathbf{\text{cosec} \, \theta = \frac{1}{\sin \theta}}$

  From this, we also get:

  $\mathbf{\sin \theta = \frac{1}{\text{cosec} \, \theta}}$ and $\mathbf{\sin \theta \cdot \text{cosec} \, \theta = 1}$

• Secant and Cosine:

  $ \sec \theta = \frac{H}{A} $ and $ \cos \theta = \frac{A}{H} $

  Thus, $ \sec \theta = \frac{1}{A/H} = \frac{1}{\cos \theta} $. This gives us the identity:

  $\mathbf{\sec \theta = \frac{1}{\cos \theta}}$

  From this, we also get:

  $\mathbf{\cos \theta = \frac{1}{\sec \theta}}$ and $\mathbf{\cos \theta \cdot \sec \theta = 1}$

• Cotangent and Tangent:

  $ \cot \theta = \frac{A}{O} $ and $ \tan \theta = \frac{O}{A} $

  Thus, $ \cot \theta = \frac{1}{O/A} = \frac{1}{\tan \theta} $. This gives us the identity:

  $\mathbf{\cot \theta = \frac{1}{\tan \theta}}$

  From this, we also get:

  $\mathbf{\tan \theta = \frac{1}{\cot \theta}}$ and $\mathbf{\tan \theta \cdot \cot \theta = 1}$

2. Quotient Relationships

These identities express tangent and cotangent in terms of sine and cosine.

• Tangent in terms of Sine and Cosine:

  Let's consider the ratio $\frac{\sin \theta}{\cos \theta}$. Using the definitions in a right triangle:

  $ \frac{\sin \theta}{\cos \theta} = \frac{O/H}{A/H} $

  To simplify, we can multiply the numerator by the reciprocal of the denominator:

  $ \frac{O}{H} \times \frac{H}{A} = \frac{O \times H}{H \times A} $

  Cancelling H from numerator and denominator:

  $ = \frac{O}{A} $

  From the definition of $\tan \theta$, we know that $\tan \theta = \frac{O}{A}$.

  Therefore, we have the identity:

  $\mathbf{\tan \theta = \frac{\sin \theta}{\cos \theta}}$

• Cotangent in terms of Sine and Cosine:

  Similarly, let's consider the ratio $\frac{\cos \theta}{\sin \theta}$:

  $ \frac{\cos \theta}{\sin \theta} = \frac{A/H}{O/H} $

  Multiplying by the reciprocal of the denominator:

  $ \frac{A}{H} \times \frac{H}{O} = \frac{A \times H}{H \times O} $

  Cancelling H:

  $ = \frac{A}{O} $

  From the definition of $\cot \theta$, we know that $\cot \theta = \frac{A}{O}$.

  Therefore, we have the identity:

  $\mathbf{\cot \theta = \frac{\cos \theta}{\sin \theta}}$

Note that this quotient identity for $\cot \theta$ is consistent with the reciprocal identity $\cot \theta = \frac{1}{\tan \theta} = \frac{1}{(\sin \theta / \cos \theta)} = \frac{\cos \theta}{\sin \theta}$.

3. Pythagorean Identities (Fundamental Identities)

These are among the most important identities in trigonometry. They are derived directly from the Pythagorean Theorem ($(\text{Opposite})^2 + (\text{Adjacent})^2 = (\text{Hypotenuse})^2$, or $O^2 + A^2 = H^2$) applied to a right triangle with acute angle $\theta$.

Derivation of $\sin^2 \theta + \cos^2 \theta = 1$

Start with the Pythagorean Theorem for the sides relative to angle $\theta$:

Divide both sides of the equation by the square of the Hypotenuse, $H^2$ (assuming $H \ne 0$, which is true for any triangle):

$ \frac{O^2}{H^2} + \frac{A^2}{H^2} = \frac{H^2}{H^2} $

This can be written as:

$ \left(\frac{O}{H}\right)^2 + \left(\frac{A}{H}\right)^2 = \left(\frac{H}{H}\right)^2 $

Recall the definitions: $\sin \theta = \frac{O}{H}$, $\cos \theta = \frac{A}{H}$, and $\frac{H}{H} = 1$. Substitute these into the equation:

$ (\sin \theta)^2 + (\cos \theta)^2 = (1)^2 $

Using the standard notation $\sin^2 \theta$ for $(\sin \theta)^2$ and $\cos^2 \theta$ for $(\cos \theta)^2$, we get the first Pythagorean identity:

$\mathbf{\sin^2 \theta + \cos^2 \theta = 1}$

This identity is valid for all values of $\theta$.

Derivation of $1 + \tan^2 \theta = \sec^2 \theta$

Start again with the Pythagorean Theorem:

This time, divide both sides of the equation by the square of the Adjacent side, $A^2$ (assuming $A \ne 0$, which means $\cos \theta \ne 0$, so $\theta \ne 90^\circ, 270^\circ, ...$):

$ \frac{O^2}{A^2} + \frac{A^2}{A^2} = \frac{H^2}{A^2} $

This can be written as:

$ \left(\frac{O}{A}\right)^2 + \left(\frac{A}{A}\right)^2 = \left(\frac{H}{A}\right)^2 $

Recall the definitions: $\tan \theta = \frac{O}{A}$, $\frac{A}{A} = 1$, and $\sec \theta = \frac{H}{A}$. Substitute these into the equation:

$ (\tan \theta)^2 + (1)^2 = (\sec \theta)^2 $

Using standard notation, we get the second Pythagorean identity:

$\mathbf{\tan^2 \theta + 1 = \sec^2 \theta}$ or $\mathbf{1 + \tan^2 \theta = \sec^2 \theta}$

This identity is valid for all values of $\theta$ for which $\sec \theta$ and $\tan \theta$ are defined (i.e., where $\cos \theta \ne 0$).

Derivation of $1 + \cot^2 \theta = \text{cosec}^2 \theta$

Start with the Pythagorean Theorem one last time:

This time, divide both sides of the equation by the square of the Opposite side, $O^2$ (assuming $O \ne 0$, which means $\sin \theta \ne 0$, so $\theta \ne 0^\circ, 180^\circ, 360^\circ, ...$):

$ \frac{O^2}{O^2} + \frac{A^2}{O^2} = \frac{H^2}{O^2} $

This can be written as:

$ \left(\frac{O}{O}\right)^2 + \left(\frac{A}{O}\right)^2 = \left(\frac{H}{O}\right)^2 $

Recall the definitions: $\frac{O}{O} = 1$, $\cot \theta = \frac{A}{O}$, and $\text{cosec} \, \theta = \frac{H}{O}$. Substitute these into the equation:

$ (1)^2 + (\cot \theta)^2 = (\text{cosec} \, \theta)^2 $

Using standard notation, we get the third Pythagorean identity:

$\mathbf{1 + \cot^2 \theta = \text{cosec}^2 \theta}$

This identity is valid for all values of $\theta$ for which $\text{cosec} \, \theta$ and $\cot \theta$ are defined (i.e., where $\sin \theta \ne 0$).

These three Pythagorean identities are fundamental and you should memorise them. They are used extensively to prove other identities and simplify trigonometric expressions.

Example 1: Using Relationships

Answer:

We are given $\sin \theta = \frac{3}{5}$ and $\theta$ is acute (so $\cos \theta$ will be positive).

We can use the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$ to find $\cos \theta$.

Substitute the value of $\sin \theta$:

$ \left(\frac{3}{5}\right)^2 + \cos^2 \theta = 1 $

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

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

$ \cos^2 \theta = 1 - \frac{9}{25} $

$ \cos^2 \theta = \frac{25}{25} - \frac{9}{25} $

$ \cos^2 \theta = \frac{25 - 9}{25} = \frac{16}{25} $

Take the square root of both sides:

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

Since $\theta$ is an acute angle, $\cos \theta$ must be positive.

So, $ \cos \theta = \frac{4}{5} $

Now, we can find $\tan \theta$ using the quotient identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$:

$ \tan \theta = \frac{3/5}{4/5} $

To divide fractions, multiply the numerator by the reciprocal of the denominator:

$ \tan \theta = \frac{3}{\cancel{5}} \times \frac{\cancel{5}}{4} = \frac{3}{4} $

Thus, for an acute angle $\theta$ with $\sin \theta = \frac{3}{5}$, we have $\cos \theta = \frac{4}{5}$ and $\tan \theta = \frac{3}{4}$.

Alternative Solution for Example 1: Using a Right Triangle

Answer:

Since $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{3}{5}$, we can consider a right triangle where the Opposite side is 3 units and the Hypotenuse is 5 units (we can take these as ratios, e.g., $3k$ and $5k$, but for finding ratios, using 3 and 5 directly is sufficient).

Let the Adjacent side be A. By the Pythagorean theorem:

$ (\text{Opposite})^2 + (\text{Adjacent})^2 = (\text{Hypotenuse})^2 $

$ 3^2 + A^2 = 5^2 $

$ 9 + A^2 = 25 $

$ A^2 = 25 - 9 $

$ A^2 = 16 $

$ A = \sqrt{16} = 4 $ (Since length must be positive)

So, the Adjacent side is 4 units.

Now, we can find $\cos \theta$ and $\tan \theta$ using the side lengths:

$ \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{4}{5} $

$ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{3}{4} $

This method using a right triangle gives the same results and provides a visual understanding of the ratios.