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Trigonometric Functions of Sum and Difference of Two Angles Double Angle Formulas Half Angle Formulas (Implicit based on multiple angles)
Triple Angle Formulas (Implicit based on sum/multiple angles)

Trigonometric Identities: Compound and Multiple Angles

Trigonometric Functions of Sum and Difference of Two Angles

The identities for the sum and difference of two angles, also known as compound angle formulas, are fundamental in trigonometry. They allow us to express trigonometric functions of angles like $A+B$ or $A-B$ in terms of the trigonometric functions of the individual angles A and B. These identities are crucial for simplifying expressions, solving equations, and deriving other important formulas.

Derivation of $\cos(A - B)$ using the Unit Circle

One common method to derive the compound angle formulas starts with the cosine of the difference of two angles using the unit circle and the distance formula.

Consider a unit circle centered at the origin O(0,0). Let angles A and B be in standard position, with their terminal sides intersecting the unit circle at points P and Q, respectively.

The coordinates of P are $(\cos A, \sin A)$ and the coordinates of Q are $(\cos B, \sin B)$.

Unit circle showing points P and Q for angles A and B used in deriving cos(A-B). Chord PQ is drawn.

The angle between the terminal sides OP and OQ is $\angle QOP$. If we assume $A > B > 0$, then $\angle QOP = A - B$.

Now, we calculate the square of the distance between points P and Q using the distance formula $ d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 $:

$ PQ^2 = (\cos A - \cos B)^2 + (\sin A - \sin B)^2 $

Expand the squared terms:

$ PQ^2 = (\cos^2 A - 2\cos A \cos B + \cos^2 B) + (\sin^2 A - 2\sin A \sin B + \sin^2 B) $

Rearrange the terms and group using the Pythagorean identity $ \sin^2 \theta + \cos^2 \theta = 1 $:

$ PQ^2 = (\cos^2 A + \sin^2 A) + (\cos^2 B + \sin^2 B) - 2(\cos A \cos B + \sin A \sin B) $

$\sin^2 A + \cos^2 A = 1$

[Pythagorean Identity]

$\sin^2 B + \cos^2 B = 1$

[Pythagorean Identity]

Substitute these into the expression for $PQ^2$:

$PQ^2 = 1 + 1 - 2(\cos A \cos B + \sin A \sin B) $

$ PQ^2 = 2 - 2(\cos A \cos B + \sin A \sin B) $

... (1)

Now, consider the same chord PQ, but rotate the entire figure so that the angle B's terminal side aligns with the positive x-axis. The point Q moves to Q'(1, 0). Since the relative angle between OP and OQ is $A-B$, the new point P' will correspond to the angle $A-B$ in standard position. Its coordinates will be $(\cos(A-B), \sin(A-B))$.

Unit circle showing points P' and Q' for angles A-B and 0 after rotation. Chord P'Q' is drawn.

The distance P'Q' is the same as PQ. Now calculate the square of the distance between P'$(\cos(A-B), \sin(A-B))$ and Q'(1, 0) using the distance formula:

$ P'Q'^2 = (\cos(A-B) - 1)^2 + (\sin(A-B) - 0)^2 $

Expand the squared terms:

$ P'Q'^2 = (\cos^2(A-B) - 2\cos(A-B) + 1) + \sin^2(A-B) $

Rearrange and group using the Pythagorean identity $ \sin^2 \theta + \cos^2 \theta = 1 $ (applied to angle $A-B$):

$ P'Q'^2 = (\cos^2(A-B) + \sin^2(A-B)) + 1 - 2\cos(A-B) $

$\sin^2(A-B) + \cos^2(A-B) = 1$

[Pythagorean Identity]

Substitute this into the expression for $P'Q'^2$:

$P'Q'^2 = 1 + 1 - 2\cos(A-B)$

$ P'Q'^2 = 2 - 2\cos(A-B) $

... (2)

Since $PQ^2 = P'Q'^2$, we equate the expressions from Equation (1) and Equation (2):

$2 - 2(\cos A \cos B + \sin A \sin B) = 2 - 2\cos(A-B) $

[Equating (1) and (2)]

Subtract 2 from both sides:

$ -2(\cos A \cos B + \sin A \sin B) = -2\cos(A-B) $

Divide both sides by -2:

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

Thus, we get the first compound angle identity:

$\mathbf{\cos(A - B) = \cos A \cos B + \sin A \sin B}$

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

... (7.1)

This derivation holds for any angles A and B.

Other Sum and Difference Formulas

Using the identity for $ \cos(A-B) $ and fundamental identities (like even/odd properties and complementary angle identities), we can derive the other formulas.

1. Formula for $\cos(A + B)$

Replace B with (-B) in the identity for $ \cos(A - B) $ (Equation 7.1):

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

Apply Equation (7.1) with B replaced by -B:

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

Recall that cosine is an even function ($ \cos(-B) = \cos B $) and sine is an odd function ($ \sin(-B) = -\sin B $). Substitute these:

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

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

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

... (7.2)

2. Formula for $\sin(A + B)$

Use the complementary angle identity $ \sin \theta = \cos(90^\circ - \theta) $ (or $ \sin \theta = \cos(\pi/2 - \theta) $ in radians). Let $ \theta = A + B $:

$ \sin(A + B) = \cos(90^\circ - (A + B)) = \cos((90^\circ - A) - B) $

Now, apply the $ \cos(X - Y) $ formula (Equation 7.1) where $ X = 90^\circ - A $ and $ Y = B $:

$ \sin(A + B) = \cos(90^\circ - A) \cos B + \sin(90^\circ - A) \sin B $

Recall the complementary angle identities: $ \cos(90^\circ - A) = \sin A $ and $ \sin(90^\circ - A) = \cos A $. Substitute these:

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

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

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

... (7.3)

3. Formula for $\sin(A - B)$

Replace B with (-B) in the identity for $ \sin(A + B) $ (Equation 7.3):

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

Apply Equation (7.3) with B replaced by -B:

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

Recall $ \cos(-B) = \cos B $ and $ \sin(-B) = -\sin B $. Substitute these:

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

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

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

... (7.4)

4. Formula for $\tan(A + B)$

Use the quotient identity $ \tan \theta = \frac{\sin \theta}{\cos \theta} $. Let $ \theta = A + B $:

$ \tan(A + B) = \frac{\sin(A + B)}{\cos(A + B)} $

Substitute the formulas for $ \sin(A + B) $ (Equation 7.3) and $ \cos(A + B) $ (Equation 7.2):

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

To express this in terms of $ \tan A $ and $ \tan B $, divide both the numerator and the denominator by $ \cos A \cos B $, assuming $ \cos A \ne 0 $ and $ \cos B \ne 0 $:

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

Simplify each term:

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

Recall the quotient identity $ \tan \theta = \frac{\sin \theta}{\cos \theta} $:

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

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

... (7.5)

5. Formula for $\tan(A - B)$

Replace B with (-B) in the identity for $ \tan(A + B) $ (Equation 7.5):

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

Recall that tangent is an odd function ($ \tan(-B) = -\tan B $). Substitute this:

$ \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} $

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

... (7.6)

6. Formula for $\cot(A + B)$

Use the reciprocal identity $ \cot \theta = \frac{1}{\tan \theta} $ and the formula for $ \tan(A + B) $ (Equation 7.5):

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

To express this in terms of $ \cot A $ and $ \cot B $, replace $ \tan A = \frac{1}{\cot A} $ and $ \tan B = \frac{1}{\cot B} $ (assuming $ \cot A \ne 0 $ and $ \cot B \ne 0 $):

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

Simplify the numerator and denominator by finding common denominators:

Numerator: $ 1 - \frac{1}{\cot A \cot B} = \frac{\cot A \cot B - 1}{\cot A \cot B} $

Denominator: $ \frac{1}{\cot A} + \frac{1}{\cot B} = \frac{\cot B + \cot A}{\cot A \cot B} $

Substitute back into the expression for $ \cot(A + B) $:

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

Multiply the numerator by the reciprocal of the denominator:

$ \cot(A + B) = \frac{\cot A \cot B - 1}{\cancel{\cot A \cot B}} \times \frac{\cancel{\cot A \cot B}}{\cot B + \cot A} $

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

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

... (7.7)

7. Formula for $\cot(A - B)$

Replace B with (-B) in the identity for $ \cot(A + B) $ (Equation 7.7):

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

Recall that cotangent is an odd function ($ \cot(-B) = -\cot B $). Substitute this:

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

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

Factor out -1 from the numerator and rearrange the denominator:

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

To make the denominator look nicer, factor out -1 from the denominator as well: $ \cot A - \cot B = -(\cot B - \cot A) $

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

Cancel the -1 terms:

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

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

... (7.8)

Summary of Sum and Difference Formulas

Here is a summary of the compound angle formulas:

Examples

Example 1. Find the value of $\sin 75^\circ$.

Answer:

Solution:

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

Use the formula for $\sin(A + B)$, with $A = 45^\circ$ and $B = 30^\circ$:

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

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

Substitute the known values from the table of specific angles:

$ \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} $.

$ \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) $

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

$ = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} $

Combine the fractions (they have a common denominator):

$ = \frac{\sqrt{3} + 1}{2\sqrt{2}} $

To rationalise the denominator, multiply the numerator and denominator by $\sqrt{2}$:

$ = \frac{(\sqrt{3} + 1) \times \sqrt{2}}{(2\sqrt{2}) \times \sqrt{2}} = \frac{\sqrt{3}\sqrt{2} + 1\sqrt{2}}{2 \times 2} = \frac{\sqrt{6} + \sqrt{2}}{4} $

Thus, $ \sin 75^\circ = \frac{\sqrt{6} + \sqrt{2}}{4} $.

This value can also be obtained using $75^\circ = 90^\circ - 15^\circ$ or other combinations, but $45^\circ+30^\circ$ is the most common approach.

Example 2. Find the value of $\cos 15^\circ$.

Answer:

Solution:

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

Use the formula for $\cos(A - B)$, with $A = 45^\circ$ and $B = 30^\circ$:

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

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

Substitute the known values:

$ \cos 45^\circ = \frac{1}{\sqrt{2}} $, $ \cos 30^\circ = \frac{\sqrt{3}}{2} $, $ \sin 45^\circ = \frac{1}{\sqrt{2}} $, $ \sin 30^\circ = \frac{1}{2} $.

$ \cos 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) $

$ = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} $

$ = \frac{\sqrt{3} + 1}{2\sqrt{2}} $

Rationalising the denominator:

$ = \frac{(\sqrt{3} + 1) \times \sqrt{2}}{(2\sqrt{2}) \times \sqrt{2}} = \frac{\sqrt{6} + \sqrt{2}}{4} $

Thus, $ \cos 15^\circ = \frac{\sqrt{6} + \sqrt{2}}{4} $.

Notice that $ \sin 75^\circ = \cos 15^\circ $. This is consistent with the complementary angle identity $ \sin \theta = \cos(90^\circ - \theta) $, since $ 75^\circ + 15^\circ = 90^\circ $.

Example 3. Prove that $\frac{\tan A + \tan B}{\cot A + \cot B} = \tan A \tan B$.

Answer:

To Prove:

$ \frac{\tan A + \tan B}{\cot A + \cot B} = \tan A \tan B $

Proof:

Start with the Left-Hand Side (LHS):

LHS = $ \frac{\tan A + \tan B}{\cot A + \cot B} $

Convert cotangent terms to tangent using the reciprocal identity $ \cot \theta = \frac{1}{\tan \theta} $:

LHS = $ \frac{\tan A + \tan B}{\frac{1}{\tan A} + \frac{1}{\tan B}} $

Simplify the denominator by finding a common denominator:

$ \frac{1}{\tan A} + \frac{1}{\tan B} = \frac{\tan B}{\tan A \tan B} + \frac{\tan A}{\tan A \tan B} = \frac{\tan B + \tan A}{\tan A \tan B} = \frac{\tan A + \tan B}{\tan A \tan B} $

Substitute this back into the LHS expression:

LHS = $ \frac{\tan A + \tan B}{\frac{\tan A + \tan B}{\tan A \tan B}} $

To divide by a fraction, multiply by its reciprocal:

LHS = $ (\tan A + \tan B) \times \frac{\tan A \tan B}{\tan A + \tan B} $

Assuming $ \tan A + \tan B \ne 0 $, cancel the common factor $ (\tan A + \tan B) $:

LHS = $ \cancel{(\tan A + \tan B)} \times \frac{\tan A \tan B}{\cancel{(\tan A + \tan B)}} $

LHS = $ \tan A \tan B $

This is equal to the Right-Hand Side (RHS).

Therefore, LHS = RHS. The identity is Proved.

Note for Competitive Exams

Compound angle formulas are fundamental for solving problems involving angles that are sums or differences of standard angles (like $15^\circ, 75^\circ, 105^\circ$, etc.) and for deriving many other identities, including the double angle and half-angle formulas. Memorise the formulas for $ \sin(A \pm B) $ and $ \cos(A \pm B) $. The formulas for $ \tan(A \pm B) $ and $ \cot(A \pm B) $ can be derived from these using the quotient and reciprocal identities if needed, but memorising them saves time. Practice using these formulas to expand expressions, evaluate values, and prove identities. Be careful with signs!

Double Angle Formulas

Double angle formulas are a special case of the sum formulas where the two angles are equal, i.e., $A = B$. These identities express the trigonometric functions of $2A$ in terms of the trigonometric functions of A. They are particularly useful for simplifying expressions and solving equations involving double angles.

1. Sine Double Angle Formula: $\sin(2A)$

Start with the sine sum formula (Equation 7.3):

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

Set $ B = A $:

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

Combine like terms:

$ \sin(2A) = 2 \sin A \cos A $

$\sin(2A) = 2 \sin A \cos A$

... (7.9)

Alternative form in terms of Tangent ($ \tan A $)

We can express $ \sin(2A) $ using $ \tan A $. Start with $ \sin(2A) = 2 \sin A \cos A $. Divide and multiply by $ \cos A $:

$ \sin(2A) = 2 \frac{\sin A}{\cos A} \cos^2 A $

Recognize $ \frac{\sin A}{\cos A} = \tan A $ and $ \cos^2 A = \frac{1}{\sec^2 A} $. Use the Pythagorean identity $ \sec^2 A = 1 + \tan^2 A $.

$ \sin(2A) = 2 \tan A \left(\frac{1}{1 + \tan^2 A}\right) $

$ \sin(2A) = \frac{2 \tan A}{1 + \tan^2 A} $

$\sin(2A) = \frac{2 \tan A}{1 + \tan^2 A}$

... (7.10)

This form is useful when working with expressions involving $ \tan A $.

2. Cosine Double Angle Formulas: $\cos(2A)$

Start with the cosine sum formula (Equation 7.2):

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

Set $ B = A $:

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

$ \cos(2A) = \cos^2 A - \sin^2 A $

$\cos(2A) = \cos^2 A - \sin^2 A$

... (7.11)

Alternative forms using $\sin^2 A + \cos^2 A = 1$

We can express $ \cos(2A) $ purely in terms of $ \cos A $ or $ \sin A $ using the Pythagorean identity $ \sin^2 A + \cos^2 A = 1 $.

Equations (7.12) and (7.13) are very useful, especially for expressing $ \sin^2 A $ and $ \cos^2 A $ in terms of $ \cos(2A) $:
From (7.12): $ 2 \cos^2 A = 1 + \cos(2A) \implies \cos^2 A = \frac{1 + \cos(2A)}{2} $
From (7.13): $ 2 \sin^2 A = 1 - \cos(2A) \implies \sin^2 A = \frac{1 - \cos(2A)}{2} $
These last two relations are often referred to as power reduction formulas.

Alternative form in terms of Tangent ($ \tan A $)

Start with $ \cos(2A) = \cos^2 A - \sin^2 A $ (Equation 7.11). Divide numerator and denominator by $ 1 = \cos^2 A + \sin^2 A $ (assuming $ \cos A \ne 0 $):

$ \cos(2A) = \frac{\cos^2 A - \sin^2 A}{\cos^2 A + \sin^2 A} $

Divide every term in the numerator and denominator by $ \cos^2 A $:

$ \cos(2A) = \frac{\frac{\cos^2 A}{\cos^2 A} - \frac{\sin^2 A}{\cos^2 A}}{\frac{\cos^2 A}{\cos^2 A} + \frac{\sin^2 A}{\cos^2 A}} = \frac{1 - (\frac{\sin A}{\cos A})^2}{1 + (\frac{\sin A}{\cos A})^2} $

Recall the quotient identity $ \tan A = \frac{\sin A}{\cos A} $:

$ \cos(2A) = \frac{1 - \tan^2 A}{1 + \tan^2 A} $

$\cos(2A) = \frac{1 - \tan^2 A}{1 + \tan^2 A}$

... (7.14)

3. Tangent Double Angle Formula: $\tan(2A)$

Start with the tangent sum formula (Equation 7.5):

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

Set $ B = A $ (assuming $ \tan A $ is defined and $ \tan A \ne \pm 1 $):

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

Combine like terms:

$ \tan(2A) = \frac{2 \tan A}{1 - \tan^2 A} $

$\tan(2A) = \frac{2 \tan A}{1 - \tan^2 A}$

... (7.15)

This formula is defined when $ \tan A $ is defined ($ A \ne \frac{\pi}{2} + n\pi $) and $ 1 - \tan^2 A \ne 0 $ ($ \tan^2 A \ne 1 $, which means $ \tan A \ne \pm 1 $, i.e., $ A \ne \frac{\pi}{4} + n\frac{\pi}{2} $).

Summary of Double Angle Formulas

Here is a summary of the double angle formulas:

Examples

Example 1. If $\sin A = \frac{3}{5}$ and A is an acute angle, find the values of $\sin 2A$, $\cos 2A$, and $\tan 2A$.

Answer:

Solution:

Given $ \sin A = \frac{3}{5} $ and A is acute ($0^\circ < A < 90^\circ$).

First, we need to find $ \cos A $ and $ \tan A $. Since A is acute, $\cos A$ will be positive. Use the identity $ \sin^2 A + \cos^2 A = 1 $:

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

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

$ \cos^2 A = 1 - \frac{9}{25} = \frac{25 - 9}{25} = \frac{16}{25} $

$ \cos A = \sqrt{\frac{16}{25}} = \frac{4}{5} $ (since A is acute, $\cos A > 0$)

Now find $ \tan A $ using the quotient identity:

$ \tan A = \frac{\sin A}{\cos A} = \frac{3/5}{4/5} = \frac{3}{\cancel{5}} \times \frac{\cancel{5}}{4} = \frac{3}{4} $

Now, use the double angle formulas:

a) Find $ \sin 2A $:

$ \sin 2A = 2 \sin A \cos A $

$ \sin 2A = 2 \times \left(\frac{3}{5}\right) \times \left(\frac{4}{5}\right) = 2 \times \frac{12}{25} = \frac{24}{25} $

So, $ \sin 2A = \frac{24}{25} $.

b) Find $ \cos 2A $:

We can use any of the $\cos 2A$ formulas. Let's use $ \cos 2A = \cos^2 A - \sin^2 A $:

$ \cos 2A = \left(\frac{4}{5}\right)^2 - \left(\frac{3}{5}\right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{16 - 9}{25} = \frac{7}{25} $

Alternatively, using $ \cos 2A = 2 \cos^2 A - 1 $:

$ \cos 2A = 2 \left(\frac{4}{5}\right)^2 - 1 = 2 \times \frac{16}{25} - 1 = \frac{32}{25} - 1 = \frac{32 - 25}{25} = \frac{7}{25} $

Alternatively, using $ \cos 2A = 1 - 2 \sin^2 A $:

$ \cos 2A = 1 - 2 \left(\frac{3}{5}\right)^2 = 1 - 2 \times \frac{9}{25} = 1 - \frac{18}{25} = \frac{25 - 18}{25} = \frac{7}{25} $

Alternatively, using $ \cos 2A = \frac{1 - \tan^2 A}{1 + \tan^2 A} $ with $ \tan A = \frac{3}{4} $:

$ \cos 2A = \frac{1 - (\frac{3}{4})^2}{1 + (\frac{3}{4})^2} = \frac{1 - \frac{9}{16}}{1 + \frac{9}{16}} = \frac{\frac{16-9}{16}}{\frac{16+9}{16}} = \frac{\frac{7}{16}}{\frac{25}{16}} = \frac{7}{\cancel{16}} \times \frac{\cancel{16}}{25} = \frac{7}{25} $

All formulas give $ \cos 2A = \frac{7}{25} $.

c) Find $ \tan 2A $:

We can use $ \tan 2A = \frac{\sin 2A}{\cos 2A} $ or the formula $ \tan 2A = \frac{2 \tan A}{1 - \tan^2 A} $.

Using $ \tan 2A = \frac{\sin 2A}{\cos 2A} $:

$ \tan 2A = \frac{24/25}{7/25} = \frac{24}{\cancel{25}} \times \frac{\cancel{25}}{7} = \frac{24}{7} $

Using $ \tan 2A = \frac{2 \tan A}{1 - \tan^2 A} $ with $ \tan A = \frac{3}{4} $:

$ \tan 2A = \frac{2 (\frac{3}{4})}{1 - (\frac{3}{4})^2} = \frac{\frac{6}{4}}{1 - \frac{9}{16}} = \frac{\frac{3}{2}}{\frac{16-9}{16}} = \frac{\frac{3}{2}}{\frac{7}{16}} $

$ = \frac{3}{\cancel{2}_1} \times \frac{\cancel{16}^8}{7} = \frac{3 \times 8}{7} = \frac{24}{7} $

Both methods give $ \tan 2A = \frac{24}{7} $.

Since $ \sin 2A = 24/25 $ and $ \cos 2A = 7/25 $, $ \sin 2A > 0 $ and $ \cos 2A > 0 $. This means $ 2A $ is in Quadrant I. Since A is acute, $0 < A < 90^\circ$, so $0 < 2A < 180^\circ$. Given $ \cos 2A = 7/25 > 0 $, $2A$ must be in Quadrant I, so $0 < 2A < 90^\circ$. The calculated values are consistent.

Example 2. Simplify the expression $\frac{1 - \cos 2x}{\sin 2x}$.

Answer:

Solution:

We use the double angle formulas for $ \cos 2x $ and $ \sin 2x $. The formula for $ \cos 2x $ involving $ 1 $ is $ \cos 2x = 1 - 2 \sin^2 x $. Rearranging this gives $ 1 - \cos 2x = 2 \sin^2 x $.

The formula for $ \sin 2x $ is $ \sin 2x = 2 \sin x \cos x $.

Substitute these into the expression:

$ \frac{1 - \cos 2x}{\sin 2x} = \frac{2 \sin^2 x}{2 \sin x \cos x} $

Assuming $ \sin x \ne 0 $ and $ \cos x \ne 0 $, we can cancel common factors:

$ = \frac{\cancel{2} \sin^{\cancel{2}} x}{\cancel{2} \cancel{\sin x} \cos x} = \frac{\sin x}{\cos x} $

Using the quotient identity, $ \frac{\sin x}{\cos x} = \tan x $.

Therefore, the simplified expression is $\tan x$.

Note for Competitive Exams

Double angle formulas are extremely important and frequently used. Memorise all forms, especially the three forms of $ \cos(2A) $, as they are useful in different contexts (e.g., simplifying expressions involving squares of sine or cosine, integrating $\sin^2 x$ or $\cos^2 x$). The formulas for $ \sin(2A) $ and $ \cos(2A) $ in terms of $ \tan A $ are also very useful for questions involving tangent substitutions or when only tangent is given. Practice manipulating expressions and solving equations using these identities.


Half Angle Formulas (Implicit based on multiple angles)

The Half-Angle formulas are trigonometric identities that express the trigonometric functions of half an angle, $A/2$, in terms of the trigonometric functions of the full angle, $A$. These identities are very useful for evaluating trigonometric values of angles that are half of standard angles (like $22.5^\circ$ or $15^\circ$) or for simplifying expressions and solving equations.

These formulas are derived directly from the Cosine Double Angle formulas.

Derivation from Cosine Double Angle Formulas

Recall the double angle formulas for cosine from the previous section:

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

... (7.13)

$\cos(2\theta) = 2 \cos^2 \theta - 1$

... (7.12)

We rearrange these formulas to isolate the squared terms, which are functions of the single angle $\theta$:

From (7.13): $ \cos(2\theta) = 1 - 2 \sin^2 \theta $

Add $ 2 \sin^2 \theta $ to both sides and subtract $ \cos(2\theta) $ from both sides:

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

From (7.12): $ \cos(2\theta) = 2 \cos^2 \theta - 1 $

Add 1 to both sides:

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

Now, to get formulas for half angles, let $ 2\theta = A $. This implies $ \theta = A/2 $. Substitute this into the rearranged formulas:

$ 2 \sin^2 (A/2) = 1 - \cos A $

$ 2 \cos^2 (A/2) = 1 + \cos A $

Half Angle Formulas

  1. Formula for $\sin(A/2)$:

    From $ 2 \sin^2 (A/2) = 1 - \cos A $, divide by 2:

    $\sin^2 (A/2) = \frac{1 - \cos A}{2}$

    Take the square root of both sides:

    $ \sin(A/2) = \pm \sqrt{\frac{1 - \cos A}{2}} $

    $\sin(A/2) = \pm \sqrt{\frac{1 - \cos A}{2}}$

    ... (7.16)

    The choice of the positive or negative sign depends on the quadrant in which the angle $A/2$ lies. For example, if $A/2$ is in Quadrant I or II, $ \sin(A/2) $ is positive, so we use the '+' sign. If $A/2$ is in Quadrant III or IV, $ \sin(A/2) $ is negative, so we use the '-' sign.

  2. Formula for $\cos(A/2)$:

    From $ 2 \cos^2 (A/2) = 1 + \cos A $, divide by 2:

    $\cos^2 (A/2) = \frac{1 + \cos A}{2}$

    Take the square root of both sides:

    $ \cos(A/2) = \pm \sqrt{\frac{1 + \cos A}{2}} $

    $\cos(A/2) = \pm \sqrt{\frac{1 + \cos A}{2}}$

    ... (7.17)

    The choice of the positive or negative sign depends on the quadrant in which the angle $A/2$ lies. If $A/2$ is in Quadrant I or IV, $ \cos(A/2) $ is positive, so we use the '+' sign. If $A/2$ is in Quadrant II or III, $ \cos(A/2) $ is negative, so we use the '-' sign.

  3. Formula for $\tan(A/2)$:

    Use the quotient identity $ \tan(A/2) = \frac{\sin(A/2)}{\cos(A/2)} $:

    $ \tan(A/2) = \frac{\pm \sqrt{\frac{1 - \cos A}{2}}}{\pm \sqrt{\frac{1 + \cos A}{2}}} $

    Under the assumption that the sign is chosen consistently based on the quadrant of $A/2$, we can combine the square roots:

    $ \tan(A/2) = \pm \sqrt{\frac{\frac{1 - \cos A}{\cancel{2}}}{\frac{1 + \cos A}{\cancel{2}}}} = \pm \sqrt{\frac{1 - \cos A}{1 + \cos A}} $

    $ \tan(A/2) = \pm \sqrt{\frac{1 - \cos A}{1 + \cos A}} $

    $\tan(A/2) = \pm \sqrt{\frac{1 - \cos A}{1 + \cos A}}$

    ... (7.18)

    The choice of the positive or negative sign depends on the quadrant in which the angle $A/2$ lies. The tangent is positive in Quadrants I and III and negative in Quadrants II and IV.

Alternative forms for $\tan(A/2)$ (without $\pm$)

There are two other useful forms for $ \tan(A/2) $ that do not require the $ \pm $ sign. These are often preferred.

Summary of Half Angle Formulas

Here is a summary of the half angle formulas:

Examples

Example 1. Find the value of $\sin 22.5^\circ$.

Answer:

Solution:

We want to find $ \sin 22.5^\circ $. Notice that $ 22.5^\circ $ is half of $ 45^\circ $. So, we can use the half-angle formula for sine with $ A = 45^\circ $. The angle $ A/2 = 22.5^\circ $ lies in Quadrant I, where sine is positive ($ \sin 22.5^\circ > 0 $), so we use the '+' sign in the formula $ \sin(A/2) = \pm \sqrt{\frac{1 - \cos A}{2}} $.

Using $ A = 45^\circ $:

$ \sin(45^\circ/2) = \sqrt{\frac{1 - \cos 45^\circ}{2}} $

From the table of specific angles, $ \cos 45^\circ = \frac{1}{\sqrt{2}} $.

$ \sin 22.5^\circ = \sqrt{\frac{1 - \frac{1}{\sqrt{2}}}{2}} $

Simplify the expression inside the square root:

$ \frac{1 - \frac{1}{\sqrt{2}}}{2} = \frac{\frac{\sqrt{2} - 1}{\sqrt{2}}}{2} = \frac{\sqrt{2} - 1}{\sqrt{2} \times 2} = \frac{\sqrt{2} - 1}{2\sqrt{2}} $

So,

$ \sin 22.5^\circ = \sqrt{\frac{\sqrt{2} - 1}{2\sqrt{2}}} $

This can be further simplified if desired, but it is often left in this form or rationalised by multiplying numerator and denominator under the square root by $2\sqrt{2}$ or $\sqrt{2}$. Let's rationalize the denominator inside the root first:

$ \frac{\sqrt{2} - 1}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{4} - \sqrt{2}}{2 \times 2} = \frac{2 - \sqrt{2}}{4} $

So,

$ \sin 22.5^\circ = \sqrt{\frac{2 - \sqrt{2}}{4}} = \frac{\sqrt{2 - \sqrt{2}}}{\sqrt{4}} = \frac{\sqrt{2 - \sqrt{2}}}{2} $

Therefore, $ \sin 22.5^\circ = \frac{\sqrt{2 - \sqrt{2}}}{2} $.

Example 2. Find the value of $\tan 15^\circ$.

Answer:

Solution:

We want to find $ \tan 15^\circ $. Notice that $ 15^\circ $ is half of $ 30^\circ $. So, we can use a half-angle formula for tangent with $ A = 30^\circ $. The angle $ A/2 = 15^\circ $ lies in Quadrant I, where tangent is positive ($ \tan 15^\circ > 0 $). We can use any of the tangent half-angle formulas. Let's use $ \tan(A/2) = \frac{1 - \cos A}{\sin A} $.

Using $ A = 30^\circ $:

$ \tan(30^\circ/2) = \frac{1 - \cos 30^\circ}{\sin 30^\circ} $

From the table of specific angles, $ \cos 30^\circ = \frac{\sqrt{3}}{2} $ and $ \sin 30^\circ = \frac{1}{2} $.

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

Simplify the numerator:

$ 1 - \frac{\sqrt{3}}{2} = \frac{2}{2} - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2} $

Substitute back into the expression for $ \tan 15^\circ $:

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

Therefore, $ \tan 15^\circ = 2 - \sqrt{3} $.

Alternative Solution using $ \tan(A/2) = \frac{\sin A}{1 + \cos A} $

Using $ A = 30^\circ $:

$ \tan(30^\circ/2) = \frac{\sin 30^\circ}{1 + \cos 30^\circ} $

Substitute the known values:

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

Multiply the numerator by the reciprocal of the denominator:

$ = \frac{1}{\cancel{2}} \times \frac{\cancel{2}}{2 + \sqrt{3}} = \frac{1}{2 + \sqrt{3}} $

Rationalise the denominator by multiplying by the conjugate $(2 - \sqrt{3})$:

$ = \frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{(2)^2 - (\sqrt{3})^2} = \frac{2 - \sqrt{3}}{4 - 3} = \frac{2 - \sqrt{3}}{1} = 2 - \sqrt{3} $

Both alternative formulas give the same result.

Note for Competitive Exams

Half-angle formulas are less frequently used than sum/difference or double-angle formulas but are essential when dealing with half-integer multiples of standard angles (like $15^\circ, 22.5^\circ, 67.5^\circ$, etc.). Remember that $\sin(A/2)$ and $\cos(A/2)$ formulas involve a $\pm$ sign, and determining the correct sign based on the quadrant of $A/2$ is crucial. The tangent half-angle formulas $ \frac{1 - \cos A}{\sin A} $ and $ \frac{\sin A}{1 + \cos A} $ are often preferred because they avoid the $\pm$ ambiguity. The tangent half-angle substitution $ t = \tan(A/2) $ (implying $ \sin A = \frac{2t}{1+t^2} $ and $ \cos A = \frac{1-t^2}{1+t^2} $) is extremely important in calculus for transforming rational functions of sine and cosine into rational functions of $t$.

Triple Angle Formulas (Implicit based on sum/multiple angles)

The Triple Angle formulas are identities that express the trigonometric functions of $3A$ in terms of trigonometric functions of $A$. These formulas are derived using the sum formulas and the double angle formulas. They are useful in solving certain types of trigonometric equations and simplifying expressions involving angles that are triple a known angle.

1. Sine Triple Angle Formula: $\sin(3A)$

To derive $ \sin(3A) $, we can write $ 3A = 2A + A $ and use the sine sum formula $ \sin(X+Y) = \sin X \cos Y + \cos X \sin Y $ with $ X = 2A $ and $ Y = A $.

$ \sin(3A) = \sin(2A + A) = \sin(2A) \cos A + \cos(2A) \sin A $

Now substitute the double angle formulas for $ \sin(2A) $ and $ \cos(2A) $ to express everything in terms of functions of A. We aim for a formula purely in terms of $ \sin A $. So, we use $ \sin(2A) = 2 \sin A \cos A $ and the form of $ \cos(2A) $ that involves only $ \sin A $: $ \cos(2A) = 1 - 2 \sin^2 A $ (Equation 7.13).

$ \sin(3A) = (2 \sin A \cos A) \cos A + (1 - 2 \sin^2 A) \sin A $

$ \sin(3A) = 2 \sin A \cos^2 A + \sin A - 2 \sin^3 A $

We still have a $ \cos^2 A $ term. Use the Pythagorean identity $ \cos^2 A = 1 - \sin^2 A $:

$ \sin(3A) = 2 \sin A (1 - \sin^2 A) + \sin A - 2 \sin^3 A $

Distribute $ 2 \sin A $:

$ \sin(3A) = 2 \sin A - 2 \sin^3 A + \sin A - 2 \sin^3 A $

Combine like terms ($ 2 \sin A + \sin A = 3 \sin A $ and $ -2 \sin^3 A - 2 \sin^3 A = -4 \sin^3 A $):

$ \sin(3A) = 3 \sin A - 4 \sin^3 A $

$\sin(3A) = 3 \sin A - 4 \sin^3 A$

... (7.21)

2. Cosine Triple Angle Formula: $\cos(3A)$

To derive $ \cos(3A) $, we write $ 3A = 2A + A $ and use the cosine sum formula $ \cos(X+Y) = \cos X \cos Y - \sin X \sin Y $ with $ X = 2A $ and $ Y = A $.

$ \cos(3A) = \cos(2A + A) = \cos(2A) \cos A - \sin(2A) \sin A $

Now substitute the double angle formulas. We aim for a formula purely in terms of $ \cos A $. So, we use the form of $ \cos(2A) $ that involves only $ \cos A $: $ \cos(2A) = 2 \cos^2 A - 1 $ (Equation 7.12), and $ \sin(2A) = 2 \sin A \cos A $.

$ \cos(3A) = (2 \cos^2 A - 1) \cos A - (2 \sin A \cos A) \sin A $

Distribute $ \cos A $ in the first term and simplify the second term:

$ \cos(3A) = 2 \cos^3 A - \cos A - 2 \sin^2 A \cos A $

We still have a $ \sin^2 A $ term. Use the Pythagorean identity $ \sin^2 A = 1 - \cos^2 A $:

$ \cos(3A) = 2 \cos^3 A - \cos A - 2 (1 - \cos^2 A) \cos A $

Distribute $ -2 \cos A $ in the bracket:

$ \cos(3A) = 2 \cos^3 A - \cos A - (2 \cos A - 2 \cos^3 A) $

$ \cos(3A) = 2 \cos^3 A - \cos A - 2 \cos A + 2 \cos^3 A $

Combine like terms ($ 2 \cos^3 A + 2 \cos^3 A = 4 \cos^3 A $ and $ -\cos A - 2 \cos A = -3 \cos A $):

$ \cos(3A) = 4 \cos^3 A - 3 \cos A $

$\cos(3A) = 4 \cos^3 A - 3 \cos A$

... (7.22)

3. Tangent Triple Angle Formula: $\tan(3A)$

To derive $ \tan(3A) $, we write $ 3A = 2A + A $ and use the tangent sum formula $ \tan(X+Y) = \frac{\tan X + \tan Y}{1 - \tan X \tan Y} $ with $ X = 2A $ and $ Y = A $.

$ \tan(3A) = \tan(2A + A) = \frac{\tan(2A) + \tan A}{1 - \tan(2A) \tan A} $

Substitute the double angle formula for tangent: $ \tan(2A) = \frac{2 \tan A}{1 - \tan^2 A} $ (Equation 7.15), assuming $ \tan A $ is defined and $ \tan(2A) $ is defined.

$ \tan(3A) = \frac{\frac{2 \tan A}{1 - \tan^2 A} + \tan A}{1 - \left(\frac{2 \tan A}{1 - \tan^2 A}\right) \tan A} $

Simplify the numerator and the denominator separately by finding a common denominator:

Numerator: $ \frac{2 \tan A}{1 - \tan^2 A} + \tan A = \frac{2 \tan A + \tan A (1 - \tan^2 A)}{1 - \tan^2 A} = \frac{2 \tan A + \tan A - \tan^3 A}{1 - \tan^2 A} = \frac{3 \tan A - \tan^3 A}{1 - \tan^2 A} $

Denominator: $ 1 - \left(\frac{2 \tan A}{1 - \tan^2 A}\right) \tan A = 1 - \frac{2 \tan^2 A}{1 - \tan^2 A} = \frac{(1 - \tan^2 A) \times 1}{1 - \tan^2 A} - \frac{2 \tan^2 A}{1 - \tan^2 A} = \frac{1 - \tan^2 A - 2 \tan^2 A}{1 - \tan^2 A} = \frac{1 - 3 \tan^2 A}{1 - \tan^2 A} $

Now, substitute the simplified numerator and denominator back into the expression for $ \tan(3A) $:

$ \tan(3A) = \frac{\frac{3 \tan A - \tan^3 A}{1 - \tan^2 A}}{\frac{1 - 3 \tan^2 A}{1 - \tan^2 A}} $

Assuming $ 1 - \tan^2 A \ne 0 $ and $ 1 - 3 \tan^2 A \ne 0 $, we can multiply the numerator by the reciprocal of the denominator:

$ \tan(3A) = \frac{3 \tan A - \tan^3 A}{\cancel{1 - \tan^2 A}} \times \frac{\cancel{1 - \tan^2 A}}{1 - 3 \tan^2 A} $

$ \tan(3A) = \frac{3 \tan A - \tan^3 A}{1 - 3 \tan^2 A} $

$\tan(3A) = \frac{3 \tan A - \tan^3 A}{1 - 3 \tan^2 A}$

... (7.23)

This formula is valid when $ \tan A $ is defined, $ \tan^2 A \ne 1 $, and $ 1 - 3 \tan^2 A \ne 0 $ (i.e., $ \tan^2 A \ne 1/3 $ or $ \tan A \ne \pm 1/\sqrt{3} $).

Summary of Triple Angle Formulas

Here is a summary of the triple angle formulas:

Examples

Example 1. Find the value of $\sin 90^\circ$ using the triple angle formula, taking $A = 30^\circ$.

Answer:

Solution:

We use the formula $ \sin(3A) = 3 \sin A - 4 \sin^3 A $ with $ A = 30^\circ $. Then $ 3A = 3 \times 30^\circ = 90^\circ $.

Substitute $ A = 30^\circ $ into the formula:

$ \sin(3 \times 30^\circ) = 3 \sin 30^\circ - 4 \sin^3 30^\circ $

$ \sin 90^\circ = 3 \sin 30^\circ - 4 (\sin 30^\circ)^3 $

From the table of specific angles, $ \sin 30^\circ = \frac{1}{2} $.

$ \sin 90^\circ = 3 \times \left(\frac{1}{2}\right) - 4 \times \left(\frac{1}{2}\right)^3 $

$ \sin 90^\circ = \frac{3}{2} - 4 \times \left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right) = \frac{3}{2} - 4 \times \frac{1}{8} $

$ \sin 90^\circ = \frac{3}{2} - \frac{\cancel{4}^1}{\cancel{8}_2} = \frac{3}{2} - \frac{1}{2} $

$ \sin 90^\circ = \frac{3 - 1}{2} = \frac{2}{2} = 1 $

This matches the known value $ \sin 90^\circ = 1 $.

Example 2. Prove that $ \cos 3A = 4 \cos^3 A - 3 \cos A $. (This is the derivation shown earlier, but presented in a proof format).

Answer:

To Prove:

$ \cos 3A = 4 \cos^3 A - 3 \cos A $

Proof:

Start with the Left-Hand Side (LHS):

LHS = $ \cos 3A $

Rewrite $ 3A $ as $ 2A + A $:

LHS = $ \cos(2A + A) $

Use the cosine sum formula $ \cos(X+Y) = \cos X \cos Y - \sin X \sin Y $ with $ X=2A $ and $ Y=A $:

LHS = $ \cos(2A) \cos A - \sin(2A) \sin A $

Substitute the double angle formulas: Use $ \cos(2A) = 2 \cos^2 A - 1 $ (Equation 7.12) and $ \sin(2A) = 2 \sin A \cos A $ (Equation 7.9).

LHS = $ (2 \cos^2 A - 1) \cos A - (2 \sin A \cos A) \sin A $

Multiply and simplify:

LHS = $ 2 \cos^3 A - \cos A - 2 \sin^2 A \cos A $

Use the Pythagorean identity $ \sin^2 A = 1 - \cos^2 A $ to express $ \sin^2 A $ in terms of $ \cos A $:

LHS = $ 2 \cos^3 A - \cos A - 2 (1 - \cos^2 A) \cos A $

Distribute $ -2 \cos A $:

LHS = $ 2 \cos^3 A - \cos A - (2 \cos A - 2 \cos^3 A) $

Remove the parenthesis and change signs:

LHS = $ 2 \cos^3 A - \cos A - 2 \cos A + 2 \cos^3 A $

Combine like terms:

LHS = $ (2 \cos^3 A + 2 \cos^3 A) + (-\cos A - 2 \cos A) $

LHS = $ 4 \cos^3 A - 3 \cos A $

This is equal to the Right-Hand Side (RHS).

Therefore, LHS = RHS. The identity is Proved.

Note for Competitive Exams

Triple angle formulas are less common than double angle ones but are occasionally required. Memorise the formulas for $ \sin(3A) $ and $ \cos(3A) $. The formula for $ \tan(3A) $ can be derived if needed, but it's better to memorise it for speed. These formulas are often used in problems involving powers of $\sin A$ or $\cos A$ (like $ \sin^3 A $ or $ \cos^3 A $), as they can be rearranged to express these powers in terms of single and triple angles (e.g., $ 4 \sin^3 A = 3 \sin A - \sin 3A $). They are also used in solving certain polynomial equations that can be transformed into trigonometric form.