Find the principal values of sin ^{-1}left(-f | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $y = \sin ^{-1}\left(-\frac{1}{2}\right)$.Then,$\sin y = -\frac{1}{2}$.We know that $\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$,so $-\sin

Vedclass · Accepted Answer

$-\frac{\pi}{6}$

Vedclass · Answer

(C) Let $y = \sin ^{-1}\left(-\frac{1}{2}\right)$.Then,$\sin y = -\frac{1}{2}$.We know that $\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$,so $-\sin \left(\frac{\pi}{6}\right) = \sin \left(-\frac{\pi}{6}\right)$.Thus,$\sin y = \sin \left(-\frac{\pi}{6}\right)$.The range of the principal value branch of $\sin ^{-1} x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.Since $-\frac{\pi}{6} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$,the principal value of $\sin ^{-1}\left(-\frac{1}{2}\right)$ is $-\frac{\pi}{6}$.

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

Similar Questions

${\left[ {\sin \left( {{{\tan }^{ - 1}}\frac{3}{4}} \right)} \right]^2} = $

If $f(x) = \sin^{-1}(\sin x)$; $x \in R$,then $f$ is

$\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)+\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)=$ . . . . . . .

If $\cos ^{-1} x = \sin ^{-1}(3x)$,then $x$ is equal to

Find the value of $\tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module