Answer
The identity is verified.
$sin(\frac{\pi}{2}+x)=cos~x$
Work Step by Step
We know that:
$sin\frac{\pi}{2}=1$ and $cos\frac{\pi}{2}=0$
$sin(u+v)=sin~u~cos~v+cos~u~sin~v$
$sin(\frac{\pi}{2}+x)=sin~\frac{\pi}{2}~cos~x+cos~\frac{\pi}{2}~sin~x=1~cos~x+0~sin~x=cos~x$
