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explain solution RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 2 sub question (ii) maths

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Answer:   \frac{3\pi }{4}

Hint: The principal value branch of function  \cos ^{-1}  is  [0,\pi ]

Given:   \cos ^{-1}\left(\cos \frac{5 \pi}{4}\right)


First we solve \cos \frac{5\pi }{4}

                    \begin{aligned} &\cos \frac{5 \pi}{4}=\cos \left(\pi+\frac{\pi}{4}\right) \\ &\cos \left(\pi+\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}} \quad \because[\cos (-\theta)=\cos \theta] \end{aligned}

As we know that \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}

                    \cos \left(\frac{5 \pi}{4}\right)=-\frac{1}{\sqrt{2}}

By substituting these value in \cos ^{-1}\left\{\cos \left(\frac{5 \pi}{4}\right)\right\}

we get,

                   \cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)

Since   [\cos (-\theta)=\cos \theta]

Let            y=\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)

                \begin{aligned} &\cos y=-\frac{1}{\sqrt{2}} \\ &-\cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \\ &\cos \left(\pi-\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}} \\ &\cos \left(\frac{3 \pi}{4}\right)=-\frac{1}{\sqrt{2}} \end{aligned}

Hence, range of principal value of  \cos ^{-1} \text { is }[0, \pi] \text { and } \cos \left(\frac{3 \pi}{4}\right)=-\frac{1}{\sqrt{2}}

\cos ^{-1}\left\{\cos \left(\frac{5 \pi}{4}\right)\right\}=\frac{3 \pi}{4} \therefore \cos ^{-1}(\cos x)=x, x \in[0, \pi]

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