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 $\cos ^{-1}\left(-\frac{1}{2}\right) = y$.Then,$\cos y = -\frac{1}{2}$.Since $\cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$,we have $\cos y =

Vedclass · Accepted Answer

$\frac{2 \pi}{3}$

Vedclass · Answer

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

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

Similar Questions

The principal value of $\cos^{-1}[\cos(-680^{\circ})]$ is equal to: . . . . . . .

If $y = \tan^{-1} \left[ \frac{x - \sqrt{1 - x^2}}{x + \sqrt{1 - x^2}} \right]$, then $\frac{dy}{dx} =$

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

The principal value of $\cot^{-1}\left(\frac{-1}{\sqrt{3}}\right)$ is equal to . . . . . . .

If $2\cos^{-1}\sqrt{\frac{1+x}{2}} = \frac{\pi}{2},$ then $x = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module