Question

The minimum value of \( (\sin^{-1} x)^2 + (\cos^{-1} x)^2 \) in \( x \in \left[ \frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}} \right] \) is \( \frac{a \pi^2}{b} \), then \( a + b \) is:

Hint:

For any inverse trigonometric functions, use the identity \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \) to simplify expressions and find minimum or maximum values of the related functions.

Answer 1

Correct Answer: 9

Step 1: Use the identity for inverse trigonometric functions.
We are given the expression \( (\sin^{-1} x)^2 + (\cos^{-1} x)^2 \), and we need to minimize it in the given interval for \( x \). We use the identity: \[ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \] which holds for all \( x \in [0, 1] \). This means that: \[ (\sin^{-1} x)^2 + (\cos^{-1} x)^2 = \left( \frac{\pi}{2} - \sin^{-1} x \right)^2 + (\sin^{-1} x)^2 \]
Step 2: Simplify the expression.
Expanding the terms: \[ = \left( \frac{\pi}{2} \right)^2 - 2 \times \frac{\pi}{2} \times \sin^{-1} x + (\sin^{-1} x)^2 + (\sin^{-1} x)^2 \] \[ = \frac{\pi^2}{4} - \pi \sin^{-1} x + 2 (\sin^{-1} x)^2 \] Now, to minimize this expression, we need to consider the values of \( x \) in the given interval.
Step 3: Calculate the value at the boundary.
We are given the interval \( x \in \left[ \frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}} \right] \).
- For \( x = \frac{\sqrt{3}}{2} \), we calculate \( \sin^{-1} x = \frac{\pi}{3} \).
- For \( x = \frac{1}{\sqrt{2}} \), we calculate \( \sin^{-1} x = \frac{\pi}{4} \).
Thus, the minimum value of the expression occurs when \( x = \frac{1}{\sqrt{2}} \), yielding the value of the expression as: \[ (\sin^{-1} x)^2 + (\cos^{-1} x)^2 = \frac{a \pi^2}{b} = \frac{9 \pi^2}{4} \] Therefore, the minimum value of \( a + b \) is 9.