Question

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

Answer 1

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

Explanation:

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`