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  • explain solution RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 1 sub question (iv) maths

explain solution RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 1 sub question (iv) maths

Answers (1)

Answer:  -\frac{\pi }{7}

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

Given:   \sin ^{-1}\left(\sin \frac{13 \pi}{7}\right)

Explanation:

First we solve  \left(\sin \frac{13 \pi}{7}\right)

               \frac{13 \pi}{7}=2 \pi-\frac{\pi}{7}

                \sin \left(2 \pi-\frac{\pi}{7}\right)=\sin \left(-\frac{\pi}{7}\right) \quad \because[\sin (2 \pi-\theta)=\sin (-\theta)]

By substituting these value in

                \sin ^{-1}\left(\sin \frac{13 \pi}{7}\right)=\sin ^{-1}\left(\sin -\frac{\pi}{7}\right)

As    \sin ^{-1}(\sin x)=x \text { with } x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

The range of principal value of \sin ^{-1} \text { is }\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

                \sin ^{-1}\left(\sin \frac{13 \pi}{7}\right)=-\frac{\pi}{7}

 

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