What is $cos (arcsin (\frac{3}{5}))$ $cos (Arcsin (3/5))$ ?

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What is $cos (arcsin (\frac{3}{5}))$ $cos (Arcsin (3/5))$ ?

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Answer 1 · 2015-07-21T22:45:58

$\frac{4}{5}$ $4/5$

Explanation:

First consider that : $θ = arcsin (\frac{3}{5})$ $theta=arcsin(3/5)$

$θ$ $theta$ just represents an angle.

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

If $θ = arcsin (\frac{3}{5})$ $theta=arcsin(3/5)$ then,

$\Rightarrow sin (θ) = \frac{3}{5}$ $=>sin(theta)=3/5$

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

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

$\Rightarrow cos (θ) = \sqrt{1 - {(\frac{3}{5})}^{2}} = \sqrt{\frac{25 - 9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$ $=>cos(theta)=sqrt(1-(3/5)^2)=sqrt((25-9)/25)=sqrt(16/25)=color(blue)(4/5)$

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What is $cos (arcsin (\frac{3}{5}))$ $cos (Arcsin (3/5))$ ?

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Explanation:

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