How do you get the exact value of $cos [2 {sin}^{- 1} (\frac{\sqrt{2}}{2})]$ $cos[2 sin^-1(sqrt2 / 2)]$ ?

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How do you get the exact value of $cos [2 {sin}^{- 1} (\frac{\sqrt{2}}{2})]$ $cos[2 sin^-1(sqrt2 / 2)]$ ?

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Answer 1 · 2018-03-06T05:21:58

= 0

Explanation:

$sin x = \frac{\sqrt{2}}{2}$ $sin x = sqrt2/2$
$arcsin x = \frac{π}{4}$ $arcsin x = pi/4$
$cos ((2) (\frac{π}{4})) = cos (\frac{π}{2}) = 0$ $cos ((2)(pi/4)) = cos (pi/2) = 0$

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How do you get the exact value of $cos [2 {sin}^{- 1} (\frac{\sqrt{2}}{2})]$ $cos[2 sin^-1(sqrt2 / 2)]$ ?

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Explanation:

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How do you get the exact value of $cos [2 {sin}^{- 1} (\frac{\sqrt{2}}{2})]$ $cos[2 sin^-1(sqrt2 / 2)]$ ?