How do you find the exact value of $cos (arcsin (\frac{5}{13}))$ $cos(arcsin(5/13))$ ?

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Answer 1 · 2016-12-26T11:04:24

The answer is $= \frac{12}{13}$ $=12/13$

We need

${cos}^{2} θ + {sin}^{2} θ = 1$ $cos^2theta+sin^2theta=1$

Let $y = arcsin (\frac{5}{13})$ $y=arcsin(5/13)$

So,

$sin y = \frac{5}{13}$ $siny=5/13$

${cos}^{2} y = 1 - {sin}^{2} y = 1 - \frac{25}{169} = \frac{144}{169}$ $cos^2y=1-sin^2y=1-25/169=144/169$

$cos y = \frac{12}{13}$ $cosy=12/13$

Therefore,

$cos (arcsin (\frac{5}{13})) = \frac{12}{13}$ $cos(arcsin(5/13))=12/13$

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