How do you evaluate $sin ({cos}^{- 1} (\frac{3}{5}))$ $sin(cos^-1(3/5))$ ?

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How do you evaluate $sin ({cos}^{- 1} (\frac{3}{5}))$ $sin(cos^-1(3/5))$ ?

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Answer 1 · 2016-06-30T05:42:53

$sin ({cos}^{- 1} (\frac{3}{5})) = \pm \frac{4}{5}$ $sin(cos^(-1)(3/5))=+-4/5$

Explanation:

For evaluating $sin ({cos}^{- 1} (\frac{3}{5}))$ $sin(cos^(-1)(3/5))$ , let us assume

$cos x = \frac{3}{5}$ $cosx=3/5$ hence $x = {cos}^{- 1} (\frac{3}{5})$ $x=cos^(-1)(3/5)$ and

$sin x = \pm \sqrt{1 - {(\frac{3}{5})}^{2}} = \pm \sqrt{1 - \frac{9}{25}} = \pm \sqrt{\frac{16}{25}} = \pm \frac{4}{5}$ $sinx=+-sqrt(1-(3/5)^2)=+-sqrt(1-9/25)=+-sqrt(16/25)=+-4/5$

Hence $sin ({cos}^{- 1} (\frac{3}{5})) = sin x = \pm \frac{4}{5}$ $sin(cos^(-1)(3/5))=sinx=+-4/5$

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How do you evaluate $sin ({cos}^{- 1} (\frac{3}{5}))$ $sin(cos^-1(3/5))$ ?

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