Question:
If \( \sin^{-1}x + \pi = y \), then
If \( \sin^{-1}x + \pi = y \), then
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Tip 1: Always memorize the principal value branches of all six inverse trigonometric functions.
Tip 2: When an equation involves an inverse function, the bounds of that function dictate the bounds of the other side of the equation.
Tip 2: When an equation involves an inverse function, the bounds of that function dictate the bounds of the other side of the equation.
Updated On: Sep 10, 2026
- \( -\frac{\pi}{2} \leq y \leq \frac{\pi}{2} \)
- \( -\frac{3\pi}{2} \leq y \leq -\frac{\pi}{2} \)
- \( \frac{\pi}{2} \leq y \leq \frac{3\pi}{2} \)
- \( 0 \leq y \leq \pi \)
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The Correct Option is C
Solution and Explanation
Concept:
• The Range (Principal Value Branch) of \( f(x) = \sin^{-1}x \) is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).
• This means for any defined \( x \), \( -\frac{\pi}{2} \leq \sin^{-1}x \leq \frac{\pi}{2} \).
Step 1: Isolate the inverse trigonometric term
Given the equation: \[ \sin^{-1}x + \pi = y \] Subtract \( \pi \) from both sides to get: \[ \sin^{-1}x = y - \pi \]
Step 2: Apply the standard range constraint
We know the bounds for the arcsine function: \[ -\frac{\pi}{2} \leq \sin^{-1}x \leq \frac{\pi}{2} \]
Step 3: Substitute and solve the inequality for \( y \)
Replace \( \sin^{-1}x \) with \( y - \pi \): \[ -\frac{\pi}{2} \leq y - \pi \leq \frac{\pi}{2} \]
Step 4: Add \( \pi \) to all parts of the inequality
\[ \pi - \frac{\pi}{2} \leq y \leq \pi + \frac{\pi}{2} \] \[ \frac{\pi}{2} \leq y \leq \frac{3\pi}{2} \]
• The Range (Principal Value Branch) of \( f(x) = \sin^{-1}x \) is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).
• This means for any defined \( x \), \( -\frac{\pi}{2} \leq \sin^{-1}x \leq \frac{\pi}{2} \).
Step 1: Isolate the inverse trigonometric term
Given the equation: \[ \sin^{-1}x + \pi = y \] Subtract \( \pi \) from both sides to get: \[ \sin^{-1}x = y - \pi \]
Step 2: Apply the standard range constraint
We know the bounds for the arcsine function: \[ -\frac{\pi}{2} \leq \sin^{-1}x \leq \frac{\pi}{2} \]
Step 3: Substitute and solve the inequality for \( y \)
Replace \( \sin^{-1}x \) with \( y - \pi \): \[ -\frac{\pi}{2} \leq y - \pi \leq \frac{\pi}{2} \]
Step 4: Add \( \pi \) to all parts of the inequality
\[ \pi - \frac{\pi}{2} \leq y \leq \pi + \frac{\pi}{2} \] \[ \frac{\pi}{2} \leq y \leq \frac{3\pi}{2} \]
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