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

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(D) We know that the principal value branch of ${\cos ^{ - 1}}x$ is $[0, \pi]$.Given the interval $\pi \le x \le 2\pi$,the value of $x$ lies outside t

Vedclass · Accepted Answer

$2\pi - x$

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

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

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