Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Inverse Trigonometric Functions

Question:

The value of $sin^{-1}\left(sin(-600°)\right) is $

Options:

$\frac{\pi}{3}$

$\frac{-\pi}{3}$

$\frac{2\pi}{3}$

$-\frac{2\pi}{3}$

Correct Answer:

$\frac{\pi}{3}$

Explanation:

$sin^{-1}\left(sin(-600°)\right) =sin^{-1}\begin{Bmatrix}sin\left(-600×\frac{\pi}{180}\right)\end{Bmatrix}$

$sin^{-1}\begin{Bmatrix}sin\left(-\frac{10\pi}{3}\right)\end{Bmatrix}= -sin^{-1}-\left(-sin\frac{10\pi}{3}\right)$

$sin^{-1}\begin{Bmatrix}sin\left(3\pi +\frac{\pi}{3}\right)\end{Bmatrix}= sin^{-1}\begin{Bmatrix}-\left(-sin\frac{\pi}{3}\right)\end{Bmatrix}$

$= sin^{-1}\left(sin\frac{\pi}{3}\right) =\frac{\pi}{3}$