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

$\operatorname{sech}^{-1}\left(\frac{1}{\sqrt{2}}\right)+\operatorname{cosech}^{-1}(-1)=$

Write the function in the simplest form: $\tan ^{-1} \frac{x}{\sqrt{a^{2}-x^{2}}}, |x| < a$

$\frac{\tan ^{-1}(\sqrt{3})-\sec ^{-1}(-2)}{\operatorname{cosec}^{-1}(-\sqrt{2})+\cos ^{-1}\left(-\frac{1}{2}\right)}=$

$\cos^{-1}\left(x^2 + \frac{1}{x^2} - 1\right) + \sin^{-1}\left(x^2 - \frac{1}{x^2}\right) + \tan^{-1}(x^2)$ is equal to (where $x \in R - \{0\}$)

The value of $\tan ^{-1}(-\sqrt{3})-\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)+\cos ^{-1}\left(\frac{-1}{2}\right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module