What is $sin (arccos (\frac{5}{13}))$ $sin(arccos(5/13))$ ?

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What is $sin (arccos (\frac{5}{13}))$ $sin(arccos(5/13))$ ?

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Answer 1 · 2015-07-21T22:29:39

$\frac{12}{13}$ $12/13$

Explanation:

First consider that : $θ = arccos (\frac{5}{13})$ $theta=arccos(5/13)$

$θ$ $theta$ just represents an angle.

This means that we are looking for $sin (θ)!$ $color(red)sin(theta)!$

If $θ = arccos (\frac{5}{13})$ $theta=arccos(5/13)$ then,

$\Rightarrow cos (θ) = \frac{5}{13}$ $=>cos(theta)=5/13$

To find $sin (θ)$ $sin(theta)$ We use the identity : ${sin}^{2} (θ) = 1 - {cos}^{2} (θ)$ $sin^2(theta)=1-cos^2(theta)$

$\Rightarrow sin (θ) = \sqrt{1 - {cos}^{2} (θ)}$ $=>sin(theta)=sqrt(1-cos^2(theta)$

$\Rightarrow sin (θ) = \sqrt{1 - {(\frac{5}{13})}^{2}} = \sqrt{\frac{169 - 25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$ $=>sin(theta)=sqrt(1-(5/13)^2)=sqrt((169-25)/169)=sqrt(144/169)=color(blue)(12/13)$

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