How do you evaluate $sin (arccos (\frac{3}{5}))$ $sin(arccos (3/5))$ ?

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How do you evaluate $sin (arccos (\frac{3}{5}))$ $sin(arccos (3/5))$ ?

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Answer 1 · 2018-04-06T07:47:43

$sin (arccos (\frac{3}{5})) = \frac{4}{5}$ $sin(arccos(3/5))=4/5$

Explanation:

Let $arccos (\frac{3}{5}) = A$ $arccos(3/5)=A$ , then $cos A = \frac{3}{5}$ $cosA=3/5$

and $sin A = \sqrt{1 - {cos}^{2} A} = \sqrt{1 - {(\frac{3}{5})}^{2}} = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$ $sinA=sqrt(1-cos^2A)=sqrt(1-(3/5)^2)=sqrt(1-9/25)=4/5$

an as $A = arccos (\frac{3}{5})$ $A=arccos(3/5)$

$sin (arccos (\frac{3}{5})) = \frac{4}{5}$ $sin(arccos(3/5))=4/5$

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How do you evaluate $sin (arccos (\frac{3}{5}))$ $sin(arccos (3/5))$ ?

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