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:

$\frac{\pi}{6}$

$\frac{\pi}{3}$

$\frac{\pi}{4}$

$\frac{3\pi}{4}$

Correct Answer:

$\frac{\pi}{4}$

Explanation:

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

Step 1: Let

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

$\Rightarrow \sin y = \frac{1}{\sqrt{2}}$

Step 2: $\sin y = \frac{1}{\sqrt{2}}$

$\Rightarrow \sin y = \sin \frac{\pi}{4} \Rightarrow y = \frac{\pi}{4}$

Step 3: We know that the range of the principal value branch of $\sin^{-1}x$ is $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.

Therefore, $y = \frac{\pi}{4}$

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