## Algebra and Trigonometry 10th Edition

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