Show that- sin -1 2x- 1- x 2 -2cos -1 x - 1 - 2 - x- 1

Let cos ^ 1 x =θ⇒cosθ #160; =x LHS sin ^ 1 ( 2sinθ #160; cosθ #160; #160; ) #160; =sin ^ 1 ( sin 2θ #160; #160; ) #160; #1 ...

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Sufficient Condition for an Extrema

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Question

Show that: ${sin}^{- 1} (2 x \sqrt{1 - x^{2}}) = 2 {cos}^{- 1} x, \frac{1}{\sqrt{2}} \leq x \leq 1$

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Solution

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{sin}^{- 1} (2 x \sqrt{1 - x^{2}}) = 2 {cos}^{- 1} x, \frac{1}{\sqrt{2}} \leq x \leq 1

{cos}^{- 1} (2 x \sqrt{1 - x^{2}}) = (\frac{π}{2}) - 2 {cos}^{- 1} x, \frac{1}{\sqrt{2}} \leq x \leq 1

2 c o s^{- 1} x = s i n^{- 1} (2 x \sqrt{1 - x^{2}})

0 \leq x \leq 1

tan {\frac{1}{2} {sin}^{- 1} \frac{2 x}{1 + x^{2}} + \frac{1}{2} {cos}^{- 1} \frac{2 x}{1 + x^{2}}}

{sin}^{- 1} (2 x \sqrt{1 - x^{2}}) = 2 {sin}^{- 1} x

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