Triple Angle and Half Angle Formulas

Triple and Half Angle Formulas: Your JEE Main Toolkit

Welcome, future engineers! In this lesson, we'll delve into the world of triple and half angle formulas. These trigonometric identities are super handy for simplifying complex expressions and solving problems quickly, especially in JEE Main. Mastering these formulas will not only boost your speed but also deepen your understanding of trigonometric relationships.

Understanding the Concepts

Triple and half angle formulas are derived from the fundamental compound angle formulas we've already learned. Instead of dealing with sums or differences of angles, we're now focusing on multiples (specifically, three times an angle) and fractions (half of an angle). Think of it as zooming in and out on the unit circle!

1. Triple Angle Formulas

These formulas express trigonometric functions of $3A$ in terms of trigonometric functions of $A$.

a. Sine of 3A: $\sin 3A = 3\sin A - 4\sin^3 A$

Derivation:

We can derive this using the compound angle formula for sine: $\sin(A + B) = \sin A \cos B + \cos A \sin B$.

Let's write $3A$ as $A + 2A$:

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

Now, use the double angle formulas: $\cos 2A = 1 - 2\sin^2 A$ and $\sin 2A = 2\sin A \cos A$.

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

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

Since $\cos^2 A = 1 - \sin^2 A$:

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

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

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

Example: If $\sin A = \frac{1}{2}$, then $\sin 3A = 3(\frac{1}{2}) - 4(\frac{1}{2})^3 = \frac{3}{2} - \frac{1}{2} = 1$.

b. Cosine of 3A: $\cos 3A = 4\cos^3 A - 3\cos A$

Derivation:

Similar to the sine derivation, we start with $\cos(A + B) = \cos A \cos B - \sin A \sin B$.

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

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

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

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

Since $\sin^2 A = 1 - \cos^2 A$:

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

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

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

Example: If $\cos A = \frac{1}{2}$, then $\cos 3A = 4(\frac{1}{2})^3 - 3(\frac{1}{2}) = \frac{1}{2} - \frac{3}{2} = -1$.

c. Tangent of 3A: $\tan 3A = \frac{3\tan A - \tan^3 A}{1 - 3\tan^2 A}$

Derivation:

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

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

Use the double angle formula: $\tan 2A = \frac{2\tan A}{1 - \tan^2 A}$.

$$\tan 3A = \frac{\tan A + \frac{2\tan A}{1 - \tan^2 A}}{1 - \tan A \cdot \frac{2\tan A}{1 - \tan^2 A}}$$

Multiply the numerator and denominator by $(1 - \tan^2 A)$:

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

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

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

Example: If $\tan A = \frac{1}{\sqrt{3}}$, then $\tan 3A = \frac{3(\frac{1}{\sqrt{3}}) - (\frac{1}{\sqrt{3}})^3}{1 - 3(\frac{1}{\sqrt{3}})^2} = \frac{\sqrt{3} - \frac{1}{3\sqrt{3}}}{1 - 1}$, which is undefined (as expected, since $A = 30^\circ$ and $3A = 90^\circ$).

2. Half Angle Formulas

These formulas express trigonometric functions of $\frac{A}{2}$ in terms of trigonometric functions of $A$.

a. Sine of A/2: $\sin\frac{A}{2} = \pm\sqrt{\frac{1 - \cos A}{2}}$

Derivation:

Start with the double angle formula: $\cos 2\theta = 1 - 2\sin^2 \theta$. Let $\theta = \frac{A}{2}$.

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

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

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

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

Important: The $\pm$ sign depends on the quadrant in which $\frac{A}{2}$ lies. If $\frac{A}{2}$ is in the first or second quadrant, $\sin\frac{A}{2}$ is positive; if it's in the third or fourth, it's negative.

Example: Let $A = 60^\circ$. Then $\cos A = \frac{1}{2}$, and $\sin 30^\circ = \sin\frac{60^\circ}{2} = \sqrt{\frac{1 - \frac{1}{2}}{2}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$ (positive root since $30^\circ$ is in the first quadrant).

b. Cosine of A/2: $\cos\frac{A}{2} = \pm\sqrt{\frac{1 + \cos A}{2}}$

Derivation:

Start with the double angle formula: $\cos 2\theta = 2\cos^2 \theta - 1$. Let $\theta = \frac{A}{2}$.

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

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

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

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

Important: Again, the $\pm$ sign depends on the quadrant of $\frac{A}{2}$. If $\frac{A}{2}$ is in the first or fourth quadrant, $\cos\frac{A}{2}$ is positive; if it's in the second or third, it's negative.

Example: Let $A = 60^\circ$. Then $\cos A = \frac{1}{2}$, and $\cos 30^\circ = \cos\frac{60^\circ}{2} = \sqrt{\frac{1 + \frac{1}{2}}{2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$ (positive root since $30^\circ$ is in the first quadrant).

3. Sub-Multiple Angle Formulas

The term "sub-multiple angles" simply refers to the use of half-angle formulas, but with a slight change in perspective. Instead of considering $\frac{A}{2}$ as half of $A$, you can think of $A$ as a sub-multiple of $2A$. Therefore, the same half-angle formulas are applied.

These are simply a restatement of the half-angle formulas. They are useful when you're given information about an angle and need to find the trigonometric functions of half that angle.

4. Converting Between Forms

Sometimes, you'll need to convert between different trigonometric forms to simplify an expression or solve an equation. This often involves using Pythagorean identities ($\sin^2 A + \cos^2 A = 1$), or the double angle formulas in reverse.

Example: Suppose you have an expression involving $\sin 3A$ and you need to express it in terms of $\cos A$. You would first expand $\sin 3A$ and then use the Pythagorean identity to convert the sine terms into cosine terms.

JEE-Specific Tricks

In JEE Main, you'll often encounter problems that combine these formulas with other trigonometric concepts. Look for opportunities to apply these formulas to simplify complex expressions within a larger problem.

That's it for this lesson on triple and half angle formulas! Practice these formulas with a variety of problems, and you'll be well-prepared to tackle any JEE Main question that comes your way. Good luck!