Answer
See the step-by-step proof below.
Work Step by Step
Apply the identity given in the problem:
$ sin(A+B)=cos(\frac{\pi}{2}-A-B)=cos((\frac{\pi}{2}-A)-B)$
Here we use the formula:
$ cos(A'-B)=cosA'cosB+sinA'sinB $
We let $ A'=\frac{\pi}{2}-A $ and the above becomes:
$ sin(A+B)=cos((\frac{\pi}{2}-A)-B)=cos(\frac{\pi}{2}-A)cosB+sin(\frac{\pi}{2}-A)sinB=sinAcosB+cosAsinB $
Hence: $ sin(A+B)=sinAcosB+cosAsinB $
