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  • provide solution for RD Sharma maths class 12 chapter Inverse Trignometric Functions exercise 3.7 question 1 sub question (ii)

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

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Answer: -\frac{\pi }{6}

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

Given:   \sin ^{-1}\left(\sin \frac{7 \pi}{6}\right)

Explanation:

First, we solve \left(\sin \frac{7 \pi}{6}\right)

            \operatorname{Sin} \frac{7 \pi}{6}=\sin \left(\pi+\frac{\pi}{6}\right)

As we know \sin \left(\frac{\pi}{6}\right)=\frac{1}{2}

           \sin \left(\pi+\frac{\pi}{6}\right)=\sin \left(-\frac{\pi}{6}\right) \quad \because \sin [\pi+\theta]=\sin (-\theta)

           \sin ^{-1}\left(\sin \frac{7 \pi}{6}\right)=\sin ^{-1}\left(-\frac{1}{2}\right)

Now,    \text { let } y=\sin ^{-1}\left(-\frac{1}{2}\right)

            \begin{aligned} &-\sin y=\frac{1}{2} \\ &-\sin \left(\frac{\pi}{6}\right)=\frac{1}{2} \\ &-\sin \left(\frac{\pi}{6}\right)=\sin \left(-\frac{\pi}{6}\right) \end{aligned}

The range of principal value of \sin ^{-1}\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \text { and } \sin \left(-\frac{\pi}{6}\right)=-\frac{1}{2}

            \therefore \sin ^{-1}\left(\sin \frac{\pi}{6}\right)=\frac{\pi}{6} \quad\left\{\text { As } \operatorname{Sin} \frac{7 \pi}{6}=\sin \left(\pi+\frac{\pi}{6}\right)\right\}

As          \begin{aligned} &\left[\sin ^{-1}(\sin x)\right]=x \end{aligned}

 

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