How do you evaluate $cos ({sin}^{- 1} (\frac{1}{2}))$ $cos(sin^-1(1/2))$ ?

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How do you evaluate $cos ({sin}^{- 1} (\frac{1}{2}))$ $cos(sin^-1(1/2))$ ?

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Answer 1 · 2016-08-08T01:40:50

$cos ({sin}^{- 1} (\frac{1}{2})) = cos ⎛ ⎝ {cos}^{- 1} \sqrt{1 - {(\frac{1}{2})}^{2}} ⎞ ⎠ = \sqrt{1 - {(\frac{1}{2})}^{2}} = \frac{\sqrt{3}}{2} .$ $cos(sin^(-1) (1/2))=cos(cos^(-1) sqrt(1-(1/2)^2) )=sqrt(1-(1/2)^2)=sqrt 3/2.$

Explanation:

Use ${cos}^{- 1} x = {sin}^{- 1} (\pm \sqrt{1 - x^{2}})$ $cos^(-1) x = sin^(-1)(+- sqrt(1-x^2))$ .

Here the angle is in Q1. So, + sign is taken.

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How do you evaluate $cos ({sin}^{- 1} (\frac{1}{2}))$ $cos(sin^-1(1/2))$ ?

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