Precalculus, Chapter 5, 5.4, Section 5.4, Problem 61

You need to use the formula cos(a-b) = cos a*cos b + sin a*sin b and sin(a+b) = sin a*cos b + sin b*cos a, such that:
cos (pi - theta) = cos pi*cos theta + sin pi*sin theta
Since cos pi = -1 and sin pi = 0 , yields:
cos (pi - theta) = - cos theta
sin(pi/2 + theta) = sin(pi/2)*cos theta + sin theta*cos(pi/2)
Since sin(pi/2) = 1 and cos(pi/2) = 0, yields:
sin(pi/2 + theta) = cos theta
Replacing - cos theta for cos (pi - theta) and cos theta for in (pi/2 + theta) yields:
cos (pi - theta) + sin(pi/2 + theta) = - cos theta + cos theta = 0
Hence, checking if the identity is valid yields that cos (pi - theta) + sin(pi/2 + theta) = 0 holds.