Problem 68 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\left(\sin ^{-1} x\right)^{2}+\left(\cos ^{-1} x\right)^{2}=\frac{5 \pi^{2}}{8}\), then \(x\) is equal to (A) 1 (B) \(-1\) (C) \(\frac{1}{\sqrt{2}}\) (D) \(-\frac{1}{\sqrt{2}}\)

Step-by-Step Solution

Answer

The value of \(x\) is \(\frac{1}{\sqrt{2}}\).

Step 1: Understanding Inverse Trigonometric Identities

We know that for any angle \( \theta \), \( \sin^{-1}x = \theta \) implies that \( \sin(\theta) = x \). Similarly, \( \cos^{-1}x = \theta \) implies that \( \cos(\theta) = x \). One of the fundamental properties is that \( \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \).

Step 2: Explore the given Equation

The problem gives us the equation: \( (\sin^{-1}x)^2 + (\cos^{-1}x)^2 = \frac{5\pi^2}{8} \). To solve this, we should use the identity from Step 1: \( \cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x \). We substitute this relation into the given equation.

Step 3: Substituting and Simplifying

Substitute \( \cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x \) into the equation: \[ (\sin^{-1}x)^2 + \left( \frac{\pi}{2} - \sin^{-1}x \right)^2 = \frac{5\pi^2}{8} \]Expand \( \left( \frac{\pi}{2} - \sin^{-1}x \right)^2 \) and simplify the expression.

Step 4: Solving the Quadratic Equation

Expanding the squared terms gives: \[ (\sin^{-1}x)^2 + \left( \frac{\pi^2}{4} - \pi \sin^{-1}x + (\sin^{-1}x)^2 \right) = \frac{5\pi^2}{8} \]Combine like terms, and you'll obtain a quadratic equation in terms of \((\sin^{-1}x)\).

Step 5: Finding \(x\) from \(\sin^{-1}x\)

After simplification, this quadratic can be solved for \(\sin^{-1}x\). By solving the quadratic, we find \(\sin^{-1}x = \frac{\pi}{4} \) or \(\sin^{-1}x = \frac{3\pi}{4}\). Thus, \(x = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \) and \(x = \sin\left(\frac{3\pi}{4}\right) = \frac{1}{\sqrt{2}} \).

Key Concepts

Trigonometric Identities

Trigonometric identities are fundamental to solving problems involving inverse trigonometric functions. One essential identity is:

• \( \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \)

This identity shows the complementary nature of sine and cosine angles. For any real number \( x \) in the domain of these functions, the inverse sine and inverse cosine together always sum to \( \frac{\pi}{2} \).
When dealing with problems like the one stated, this identity helps us replace terms and make equations easier to handle.
Understanding how these identities interconnect simplifies solving complex equations like inverse trigonometric functions squared.

Quadratic Equation

A quadratic equation is an equation of the form:

• \( ax^2 + bx + c = 0 \)

It's called "quadratic" because "quad" means square.
In problems involving inverse trigonometric identities, quadratic forms often appear after expanding and combining terms.
Once you have a quadratic equation, you can solve it either by factoring, completing the square, or using the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

For our exercise, the expanded equations in terms of \( \sin^{-1}x \) led to a quadratic that upon solving gave possible angle solutions back in radians.

Mathematical Simplification

Mathematical simplification is about reducing expressions to their simplest form.
This step often involves applying algebraic rules and identities effectively. In the context of our exercise, we start by expanding terms:

• Express \( \left( \frac{\pi}{2} - \sin^{-1}x \right)^2 \)
• Combine all like terms

Simplification is vital because it turns a complex expression into a manageable one, making it easier to solve and understand.
The equation \( (rac{\pi^2}{4} - \pi\sin^{-1}x + {\sin^{-1}x}^2) \) simplifies into a neat quadratic form.
Through simplification, solutions become clearer, showing the true nature of the trigonometric relationships in play.