Target Exam

CUET

Subject

Section B1

Chapter

Inverse Trigonometric Functions

Question:

Find the principal value of $\sin^{-1} \left( \frac{1}{\sqrt{2}} \right)$.

Options:

$\frac{\pi}{6}$

$\frac{\pi}{3}$

$\frac{\pi}{4}$

$\frac{\pi}{2}$

Correct Answer:

$\frac{\pi}{4}$

Explanation:

The correct answer is Option (3) → $\frac{\pi}{4}$ ##

Let $\sin^{-1} \left( \frac{1}{\sqrt{2}} \right) = y$. Then, $\sin y = \frac{1}{\sqrt{2}}$.

We know that the range of the principal value branch of $\sin^{-1}$ is $\left( \frac{-\pi}{2}, \frac{\pi}{2} \right)$ and $\sin \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}}$.

Therefore, principal value of $\sin^{-1} \left( \frac{1}{\sqrt{2}} \right)$ is $\frac{\pi}{4}$.