Inverse Circular Functions Ques 8
- Find the value of $\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)$ at $x=\frac{1}{5}$, where $0 \leq \cos ^{-1} x \leq \pi$ and $-\pi / 2 \leq \sin ^{-1} x \leq \pi / 2$.
(1981, 2M)
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Answer:
Correct Answer: 8.$\frac{-2 \sqrt{6}}{5}$
Solution:
- Let $f(x)=\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)$
$ \begin{aligned} & =\cos \cos ^{-1} x+\frac{\pi}{2} \quad \because \cos ^{-1} x+\sin ^{-1} x=\frac{\pi}{2} \\ & =-\sin \left(\cos ^{-1} x\right) \\ \Rightarrow \quad f(x) & =-\sin \left(\sin ^{-1} \sqrt{1-x^{2}}\right) \\ \Rightarrow \quad f \quad (\frac{1}{5}) & =-\sin \sin ^{-1} \sqrt{1-\frac{1}{5^{2}}} \\ & =-\sin (\sin ^{-1} \frac{2 \sqrt{6}}{5})=-\frac{2 \sqrt{6}}{5} \end{aligned} $
BETA
