Composition and Conversion Formulas

Composition and Conversion Formulas in Inverse Trigonometry

Hello JEE aspirants! This lesson delves into the fascinating world of composition and conversion formulas involving inverse trigonometric functions. Mastering these formulas is crucial for simplifying complex expressions and efficiently solving problems in JEE Main. You'll often encounter these concepts directly or indirectly in various problems, so let's get started!

Understanding Composition of Trigonometric and Inverse Trigonometric Functions

Let's begin with understanding how trigonometric functions and their inverses interact. The key is to remember the domain and range of each function.

1. Functions of Inverse Functions: $\sin(\sin^{-1}x)$, $\cos(\cos^{-1}x)$, etc.

These are straightforward. When you apply a trigonometric function to the inverse of the same function, you essentially "undo" the operation. However, this holds true only within certain domains.

Explanation: The function $\sin^{-1}x$ gives you an angle whose sine is $x$. So, if you take the sine of that angle, you naturally get back $x$. The condition $|x| \leq 1$ arises because the domain of $\sin^{-1}x$ is $[-1, 1]$. If $x$ is outside this range, $\sin^{-1}x$ is not defined.

Example: $\sin(\sin^{-1}0.5) = 0.5$. But $\sin(\sin^{-1}2)$ is undefined.

Similarly:

• $\cos(\cos^{-1}x) = x, \quad |x| \leq 1$
• $\tan(\tan^{-1}x) = x, \quad x \in \mathbb{R}$ (all real numbers)
• $\csc(\csc^{-1}x) = x, \quad |x| \geq 1$
• $\sec(\sec^{-1}x) = x, \quad |x| \geq 1$
• $\cot(\cot^{-1}x) = x, \quad x \in \mathbb{R}$

2. Inverse Functions of Trigonometric Functions: $\sin^{-1}(\sin x)$, $\cos^{-1}(\cos x)$, etc.

This is where it gets a bit trickier. The inverse trigonometric functions return an angle within a specific range (the principal value branch). Therefore, $\sin^{-1}(\sin x)$ is not always equal to $x$.

Explanation: $\sin^{-1}(\sin x) = x$ only if $x$ lies within the principal value branch of $\sin^{-1}$, which is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Outside this interval, you need to adjust $x$ to find an angle within the principal value branch that has the same sine value.

Example: $\sin^{-1}(\sin \frac{\pi}{4}) = \frac{\pi}{4}$ (since $\frac{\pi}{4}$ is in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$). But $\sin^{-1}(\sin \frac{3\pi}{4}) = \sin^{-1}(\sin (\pi - \frac{\pi}{4})) = \sin^{-1}(\sin \frac{\pi}{4}) = \frac{\pi}{4}$. Note that $\sin^{-1}(\sin \frac{3\pi}{4})$ is NOT $\frac{3\pi}{4}$ because $\frac{3\pi}{4}$ is not in the principal value branch.

Similarly:

• $\cos^{-1}(\cos x) = x, \quad x \in [0, \pi]$
• $\tan^{-1}(\tan x) = x, \quad x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
• $\csc^{-1}(\csc x) = x, \quad x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}$
• $\sec^{-1}(\sec x) = x, \quad x \in [0, \pi] - \{\frac{\pi}{2}\}$
• $\cot^{-1}(\cot x) = x, \quad x \in (0, \pi)$

Converting Between Inverse Trigonometric Functions

Being able to convert between different inverse trigonometric functions is another valuable skill. These conversions often involve constructing a right-angled triangle.

Explanation: Let $\sin^{-1}x = \theta$. Then, $\sin\theta = x = \frac{x}{1}$. Consider a right-angled triangle where the opposite side is $x$ and the hypotenuse is $1$. The adjacent side would then be $\sqrt{1-x^2}$. Therefore, $\cos\theta = \frac{\sqrt{1-x^2}}{1}$, which implies $\theta = \cos^{-1}\sqrt{1-x^2}$. Hence, $\sin^{-1}x = \cos^{-1}\sqrt{1-x^2}$. The condition $x \geq 0$ arises because the principal value of $\cos^{-1}$ lies in $[0, \pi/2]$ when $x\geq 0$.

Explanation: Let $\tan^{-1}x = \theta$. Then, $\tan\theta = x = \frac{x}{1}$. Imagine a right-angled triangle with the opposite side as $x$ and the adjacent side as $1$. The hypotenuse will be $\sqrt{1+x^2}$. Thus, $\sin\theta = \frac{x}{\sqrt{1+x^2}}$, leading to $\theta = \sin^{-1}\frac{x}{\sqrt{1+x^2}}$. Therefore, $\tan^{-1}x = \sin^{-1}\frac{x}{\sqrt{1+x^2}}$.

Explanation: This one is direct. Since $\csc\theta = \frac{1}{\sin\theta}$, it follows that $\csc^{-1}x = \sin^{-1}\frac{1}{x}$. The condition $|x| \geq 1$ comes from the fact that the domain of $\csc^{-1}x$ is $(-\infty, -1] \cup [1, \infty)$, and equivalently, $|\frac{1}{x}| \leq 1$, which is the domain requirement of $\sin^{-1}\frac{1}{x}$.

Other useful conversions:

• $\cos^{-1}x = \sin^{-1}\sqrt{1-x^2}, \quad x \geq 0$
• $\cot^{-1}x = \tan^{-1}\frac{1}{x}, \quad x > 0$
• $\sec^{-1}x = \cos^{-1}\frac{1}{x}, \quad |x| \geq 1$

Simplifying Nested Expressions

Many JEE problems involve simplifying expressions where inverse trigonometric functions are nested within other trigonometric or inverse trigonometric functions.

Example: Simplify $\tan(\cos^{-1}x)$.

Solution: Let $\cos^{-1}x = \theta$. Then, $\cos\theta = x$. We need to find $\tan\theta$. We know $\sin^2\theta + \cos^2\theta = 1$, so $\sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - x^2}$. Therefore, $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\sqrt{1 - x^2}}{x}$. Hence, $\tan(\cos^{-1}x) = \frac{\sqrt{1 - x^2}}{x}$.

JEE-Specific Tricks

1. Know the Principal Value Branches: Always keep the principal value branches of inverse trigonometric functions in mind. This is crucial for correctly simplifying expressions.
2. Use Substitutions: In complex expressions, substitute inverse trigonometric functions with $\theta$, $\phi$, etc. This often simplifies the problem and makes it easier to visualize.
3. Right-Angled Triangle Approach: For conversions and simplifications, always visualize a right-angled triangle.
4. Check for Domain Restrictions: Always be mindful of domain restrictions on both trigonometric and inverse trigonometric functions.

By understanding these concepts, formulas, and tricks, you'll be well-equipped to tackle problems involving composition and conversion formulas in inverse trigonometry in your JEE Main exam. Keep practicing, and good luck!