Let A=2C, B=2D
sin2C + sin2D = 2sin(C+D)cos(C-D) =
LHS = 2sin(C)cos(C)+2sin(D)cos(D)
RHS = 2[sin(C)cos(D)+sin(D)cos(C)][cos(C)cos(D)+sin(C)sin(D)]
Crossmultiply RHS. 2[sin(C)cos(D)cos(C)cos(D)+sin(C)cos(D)sin(C)sin(D)+sin(D)cos(C)cos(C)cos(D)+sin(D)cos(C)sin(C)sin(D)] = 2[sin(C)cos(C)cos(D)cos(D)+sin(D)cos(D)sin(C)sin(C)+sin(D)cos(D)cos(C)cos(C)+sin(C)cos(C)sin(D)sin(D)] = LHS by pythagorean identity (sin^2(x)+cos^2(x) = 1)
Dirty, but it works.
Alternatively, re-express sin((A+B)/2) = cos(90 - (A+B)/2). Now let 90-(A+B)/2 = X and (A-B)/2 = Y. X-Y = 90-A, X+Y = 90-B. Look familiar?
(Hint: use cosAcosB = .5(cos(A-B)+cos(A+B))
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