Question

How do you evaluate `sin (cos^-1 (sqrt 2/2))`?

Answer 1

`sqrt2/2`

Explanation:

`cos(x)=sqrt2/2` at `pi/4` radians therefore `cos(pi/4)=sqrt2/2`

`cos^-1(cos(pi/4))=cos^-1(sqrt2/2) rarr pi/4=cos^-1(sqrt2/2)`

Sub this into original equation

`sin(cos^-1(sqrt2/2)) rarr sin(pi/4)`

`sin(pi/4)=sqrt2/2`