SATHEE: Inverse Circular Functions 2 Question 14

Inverse Circular Functions 2 Question 14

14. Find the value of $\cos (2 \cos^{- 1} x + \sin^{- 1} x)$ at $x = \frac{1}{5}$ , where $0 \leq \cos^{- 1} x \leq π$ and $- π / 2 \leq \sin^{- 1} x \leq π / 2$ .

(1981, 2M)

Integer Answer Type Question

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Answer:

Correct Answer: 14. $\frac{- 2 \sqrt{6}}{5}$

Solution:

Let $f (x) = \cos (2 \cos^{- 1} x + \sin^{- 1} x)$

$\begin{aligned} = \cos \cos^{- 1} x + \frac{π}{2} ∵ \cos^{- 1} x + \sin^{- 1} x = \frac{π}{2} \\ = - \sin (\cos^{- 1} x) \\ \Rightarrow f (x) & = - \sin (\sin^{- 1} \sqrt{1 - x^{2}}) \\ \Rightarrow f \frac{1}{5} & = - \sin \sin^{- 1} \sqrt{1 - \frac{1}{5^{2}}} \\ = - \sin \sin^{- 1} \frac{2 \sqrt{6}}{5} = - \frac{2 \sqrt{6}}{5} \end{aligned}$

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Inverse Circular Functions 2 Question 14

14. Find the value of cos⁡(2cos−1⁡x+sin−1⁡x) at x=15, where 0≤cos−1⁡x≤π and −π/2≤sin−1⁡x≤π/2.

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Answer:

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14. Find the value of $\cos (2 \cos^{- 1} x + \sin^{- 1} x)$ at $x = \frac{1}{5}$ , where $0 \leq \cos^{- 1} x \leq π$ and $- π / 2 \leq \sin^{- 1} x \leq π / 2$ .