KCET · Maths · Inverse Trigonometric Functions
\(2 \cos ^{-1} x=\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)\) is valid for all values of ' \(x\) ' satisfying
- A \(0 \leq x \leq \frac{1}{\sqrt{2}}\)
- B \(-1 \leq x \leq 1\)
- C \(0 \leq x \leq 1\)
- D \(\frac{1}{\sqrt{2}} \leq x \leq 1\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{\sqrt{2}} \leq x \leq 1\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & 2 \cos ^{-1} x=\sin ^{-1}\left(2 x \sqrt{1-x^2}\right) \\ & x=\cos \theta \Rightarrow \theta=\cos ^{-1} x \\ & \sin ^{-1}(2 \cos \theta \sin \theta)=\sin ^{-1}(\sin 2 \theta)=2 \theta \in[0,2 \pi] \\ & =2 \cos ^{-1} x \text { when } \theta \in[0, \pi / 4] \\ & \Rightarrow 2 \cos ^{-1} x \text { when } \cos ^{-1} x \in[0, \pi / 4]\end{aligned}\)
\(\Rightarrow 2 \cos ^{-1} \mathrm{x}\) when \(\mathrm{x} \in\left[\frac{1}{\sqrt{2}}, 1\right]\)
\(\Rightarrow 2 \cos ^{-1} \mathrm{x}\) when \(\mathrm{x} \in\left[\frac{1}{\sqrt{2}}, 1\right]\)
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