Question

How do you evaluate `y=sin(x + (pi/2) )`?

Answer 1

`y=cos x`

Explanation:

`y=sin(x+(pi/2)) ?`

`sin(a+b)=sin a*cos b+cos a*sin b`

`y=sin x*cos (pi/2)+cos x*sin(pi/2)`

`"plug "sin (pi/2)=1" ; "cos(pi/2)=0`

`y=sin x*0 +cos x*1`

`y=0+cos x`

`y=cos x`