How do you evaluate $sin(arccos (3/5))$ ?

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Answer 1 · 2018-04-06T07:47:43

$sin(arccos(3/5))=4/5$

Let $arccos(3/5)=A$ , then $cosA=3/5$

and $sinA=sqrt(1-cos^2A)=sqrt(1-(3/5)^2)=sqrt(1-9/25)=4/5$

an as $A=arccos(3/5)$

$sin(arccos(3/5))=4/5$

How do you evaluate sin(arccos (3/5)) sin(arccos (3/5)) ?