If pi leq x leq 2 pi,then cos^{-1}(cos x) is | vedclass

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

Question

(D) We know that the range of the principal value branch of $\cos^{-1} x$ is $[0, \pi]$.Given the interval $\pi \leq x \leq 2 \pi$.We use the property

Vedclass · Accepted Answer

$2 \pi - x$

Vedclass · Answer

(D) We know that the range of the principal value branch of $\cos^{-1} x$ is $[0, \pi]$.Given the interval $\pi \leq x \leq 2 \pi$.We use the property $\cos(2 \pi - x) = \cos x$.Since $\pi \leq x \leq 2 \pi$,we have $-\pi \geq -x \geq -2 \pi$.Adding $2 \pi$ to all sides,we get $2 \pi - \pi \geq 2 \pi - x \geq 2 \pi - 2 \pi$,which simplifies to $\pi \geq 2 \pi - x \geq 0$.Thus,$0 \leq 2 \pi - x \leq \pi$,which lies within the principal range $[0, \pi]$.Therefore,$\cos^{-1}(\cos x) = \cos^{-1}(\cos(2 \pi - x)) = 2 \pi - x$.

If $\pi \leq x \leq 2 \pi$,then $\cos^{-1}(\cos x)$ is equal to :-

Similar Questions

Evaluate: ${\tan ^{ - 1}}\left[ {\frac{{\cos x}}{{1 + \sin x}}} \right]$

If $0 < x < 1$,then $\cot ^{-1}\left( \frac{2x^2 - 1}{2x\sqrt{1 - x^2}} \right)$ is equal to

${\sin ^{ - 1}}\left[ {x\sqrt {1 - x} - \sqrt x \sqrt {1 - {x^2}} } \right] = $

$2 \tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)$ is equal to

$\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{8}$ has the value

Vedclass Test Series

Exam Paper Generator

Online Exam Module