Find the principal value of cos ^{-1}left(-fr | vedclass

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

Question

(C) Let $y = \cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)$.Then $\cos y = -\frac{1}{\sqrt{2}}$.Since $\cos \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $y = \cos ^{-1}\left(-\frac{1}{\sqrt{2}}ight)$.Then $\cos y = -\frac{1}{\sqrt{2}}$.Since $\cos \left(\frac{\pi}{4}ight) = \frac{1}{\sqrt{2}}$,we have $\cos y = -\cos \left(\frac{\pi}{4}ight)$.Using the identity $\cos (\pi - 	heta) = -\cos 	heta$,we get $\cos y = \cos \left(\pi - \frac{\pi}{4}ight) = \cos \left(\frac{3 \pi}{4}ight)$.The range of the principal value branch of $\cos ^{-1}$ is $[0, \pi]$.Since $\frac{3 \pi}{4} \in [0, \pi]$,the principal value of $\cos ^{-1}\left(-\frac{1}{\sqrt{2}}ight)$ is $\frac{3 \pi}{4}$.

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

Similar Questions

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

If $\operatorname{Sinh}^{-1}(2)+\operatorname{Sinh}^{-1}(3)=\alpha$,then $\sinh \alpha=$

If $\sin^{-1} \theta = \sin^{-1}(\sin 5)$,then $\theta$ is

$\sin^{-1} (\sin \frac{3\pi}{5}) = \dots \dots \dots$

$\tan ^{-1}\left(\tan \frac{31 \pi}{6}\right)=$ . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module