Answer
The identity is verified.
$sin(\frac{\pi}{6}+x)=\frac{1}{2}(cos~x+\sqrt 3~sin~x)$
Work Step by Step
$sin(u+v)=sin~u~cos~v+cos~u~sin~v$
$sin(\frac{\pi}{6}+x)=sin~\frac{\pi}{6}~cos~x+cos~\frac{\pi}{6}~sin~x=\frac{1}{2}~cos~x+\frac{\sqrt 3}{2}~sin~x=\frac{1}{2}(cos~x+\sqrt 3~sin~x)$
