Find the principal value of $\sin^{-1} \left( \frac{1}{\sqrt{2}} \right)$. |
$\frac{\pi}{6}$ $\frac{\pi}{3}$ $\frac{\pi}{4}$ $\frac{3\pi}{4}$ |
$\frac{\pi}{4}$ |
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 $\frac{\pi}{4}$. |