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Provide solution for RD Sharma maths class 12 Chapter Inverse trigromtery functions exercise 3.8 question 3 sub question (ii)

Answers (1)

$x=\pm \frac{1}{\sqrt{2}}$

Hint:

As we know that the value of $\operatorname{cos}\left(\cos ^{-1} x\right) \text { is } x \text { for } x \in[-1,1]$

Given:

We have to solve $\cos \left(2 \sin ^{-1}(-x)\right)=0$ to find the value of x.

Solution:

So, we convert $2\sin ^{-1}(x)$ into $\cos ^{-1}(x)$

$\cos \left(2 \sin ^{-1}(-x)\right)=\cos \left(-2 \sin ^{-1} x\right)$ [ $\because \sin ^{-1}(-x)=-\sin x$ ]

$=\cos \left(-2 \sin ^{-1} x\right)$ …(i) [ as, $\cos$ is even function, so $\cos(-x)= \cos x$ ]

Let’s suppose,

$\sin ^{-1} x =\alpha$ …(ii)

$\sin \alpha =x$ …(iii)

Given,

$\cos \left(2 \sin ^{-1}(-x)\right)=0$

$\cos \left(2 \sin ^{-1}(-x)\right)=0$ … [ from eqn(i) ]

$\cos 2 \alpha =0$ …[ using (ii) ]

$1-2 \sin ^2 \alpha =0$ [ $\cos 2 \alpha =1-2 \sin ^2 \alpha$ ]

$\begin{aligned} &1-2 x^{2}=0 \\ &2 x^{2}=1 \\ &x^{2}=\frac{1}{2} \\ &x=\pm \frac{1}{\sqrt{2}} \end{aligned}$

Therefore, the answer is $x=\pm \frac{1}{\sqrt{2}}$

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