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need solution for RD Sharma maths class 12 chapter Inverse Trignometric Functions exercise 3.7 question 1 sub question (vii)

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Answer:   {\pi }-3

Hint: The principal value branch of function  \sin ^{-1}  is  \left [ -\frac{\pi }{2},\frac{\pi }{2} \right ]

Given:    \sin ^{-1}\left(\sin 3\right)


We know that,\sin ^{-1}(\sin x)=x \text { with } x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]   which is approximately equals to \left [ -1.57,1.57 \right ]

But here x=3 , which do not lie on the above range.

∴  We know that  \sin (\pi-x)=\sin (x)

        \begin{aligned} &\sin (\pi-3)=\sin (3) \text { also } \pi-3 \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \\ &\sin ^{-1}(\sin 3)=\sin ^{-1}(\sin \pi-3) \\ &\text { Then, } \sin ^{-1}(\sin \pi-3)=\pi-3 \end{aligned}


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