Double Angle Formulas

Double Angle Formulas: Your Key to JEE Trigonometry

Hello JEE aspirants! Welcome to a crucial topic in Trigonometry: Double Angle Formulas. These formulas are not just theoretical tools; they are powerful shortcuts that can significantly simplify complex problems in JEE Main. Mastering them will give you a distinct advantage in tackling a wide range of questions, from trigonometry itself to calculus and coordinate geometry.

Understanding the Concept

Double angle formulas, as the name suggests, express trigonometric functions of an angle $2A$ in terms of trigonometric functions of the angle $A$. Think of them as special cases derived from the compound angle formulas. Instead of having two different angles, we have the same angle doubled. Let's dive into each formula.

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

Derivation: Recall the compound angle formula for sine: $\sin(A + B) = \sin A \cos B + \cos A \sin B$. Now, let $B = A$. We get:

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

Example: Suppose you need to find $\sin 2\theta$ and you know that $\sin \theta = \frac{3}{5}$ and $\theta$ is in the first quadrant. First, find $\cos \theta$ using the Pythagorean identity: $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - (\frac{3}{5})^2} = \frac{4}{5}$. Then, $\sin 2\theta = 2 \times \frac{3}{5} \times \frac{4}{5} = \frac{24}{25}$.

2. The Cosine Double Angle Formulas: $\cos 2A$

Cosine has three different forms for its double angle formula, each useful in different situations.

Derivation: Start with the cosine compound angle formula: $\cos(A + B) = \cos A \cos B - \sin A \sin B$. Let $B = A$.

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

Now, using the Pythagorean identity $\cos^2 A + \sin^2 A = 1$, we can derive the other two forms:

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

Example: If $\cos A = \frac{5}{13}$, find $\cos 2A$. We can use any of the three forms. Let's use $\cos 2A = 2\cos^2 A - 1 = 2 \times (\frac{5}{13})^2 - 1 = 2 \times \frac{25}{169} - 1 = \frac{50}{169} - 1 = -\frac{119}{169}$.

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

Derivation: Using the tangent compound angle formula: $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$. Let $B = A$.

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

Example: If $\tan A = \frac{3}{4}$, find $\tan 2A$. $\tan 2A = \frac{2 \times \frac{3}{4}}{1 - (\frac{3}{4})^2} = \frac{\frac{3}{2}}{1 - \frac{9}{16}} = \frac{\frac{3}{2}}{\frac{7}{16}} = \frac{3}{2} \times \frac{16}{7} = \frac{24}{7}$.

4. Expressing $\sin A$ in terms of $\tan(A/2)$

This is a handy formula when you have information about $\tan(A/2)$ and need to find $\sin A$. It's particularly useful in problems involving the substitution $t = \tan(A/2)$ to solve trigonometric equations.

Derivation: Let $A = 2 \times (A/2)$. Using the double angle formula for sine, $\sin A = \sin(2 \times (A/2)) = 2 \sin(A/2) \cos(A/2)$.

Now, divide and multiply by $\cos(A/2)$: $\sin A = 2 \frac{\sin(A/2)}{\cos(A/2)} \cos^2(A/2) = 2 \tan(A/2) \cos^2(A/2)$.

Since $\cos^2(A/2) = \frac{1}{\sec^2(A/2)} = \frac{1}{1 + \tan^2(A/2)}$, we get:

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

Example: If $\tan(A/2) = \frac{1}{2}$, then $\sin A = \frac{2 \times \frac{1}{2}}{1 + (\frac{1}{2})^2} = \frac{1}{1 + \frac{1}{4}} = \frac{1}{\frac{5}{4}} = \frac{4}{5}$.

Applications in Simplification

Double angle formulas are extremely useful in simplifying trigonometric expressions. Often, a complex expression can be reduced to a simpler form by recognizing and applying these formulas.

Example: Simplify $\frac{\sin 2x}{1 + \cos 2x}$.

Using the formulas $\sin 2x = 2 \sin x \cos x$ and $\cos 2x = 2\cos^2 x - 1$, we have:

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

JEE-Specific Tricks

In JEE, you might encounter problems where you need to cleverly apply double angle formulas in combination with other trigonometric identities.

Example: Find the value of $\cos 20^\circ \cos 40^\circ \cos 80^\circ$.

Multiply and divide by $\sin 20^\circ$:

$\frac{\sin 20^\circ \cos 20^\circ \cos 40^\circ \cos 80^\circ}{\sin 20^\circ} = \frac{\frac{1}{2} \sin 40^\circ \cos 40^\circ \cos 80^\circ}{\sin 20^\circ} = \frac{\frac{1}{4} \sin 80^\circ \cos 80^\circ}{\sin 20^\circ} = \frac{\frac{1}{8} \sin 160^\circ}{\sin 20^\circ}$

Since $\sin 160^\circ = \sin (180^\circ - 20^\circ) = \sin 20^\circ$, the expression simplifies to $\frac{1}{8}$.

Mastering double angle formulas is essential for your JEE preparation. Practice lots of problems, and you'll be well on your way to acing trigonometry!