Given that a+b = pi/2, prove that sin 2a+sin 2b=2cos(a-b)?

We'll transform the sum sin 2a+sin 2b into a
product:

sin 2a+sin 2b = 2sin [(2a+2b)/2]*cos
[(2a-2b)/2]

We'll factorize by 2 inside the
brackets:

sin 2a+sin 2b = 2sin [2(a+b)/2]*cos
[2(a-b)/2]

We'll simplify and we'll
get:

sin 2a+sin 2b = 2sin [(a+b)]*cos
[(a-b)]

But, from enunciation, a+b =
pi/2.

sin 2a+sin 2b = 2sin [(pi/2)]*cos
[(a-b)]

But sin pi/2 = 1

sin
2a+sin 2b = 2cos [(a-b)] q.e.d.