Target Exam

CUET

Subject

Section B1

Chapter

Inverse Trigonometric Functions

Question:

Find the value of $\sin^{-1} \left[ \sin \left( -\frac{17\pi}{8} \right) \right]$.

Options:

$\frac{\pi}{8}$

$-\frac{\pi}{8}$

$-\frac{17\pi}{8}$

$\frac{7\pi}{8}$

Correct Answer:

$-\frac{\pi}{8}$

Explanation:

The correct answer is Option (2) → $-\frac{\pi}{8}$ ##

$\sin^{-1} \left[ \sin \left( -\frac{17\pi}{8} \right) \right] = -\sin^{-1} \left[ \sin \left( 2\pi + \frac{\pi}{8} \right) \right]$

$= -\frac{\pi}{8}$