Prove.
$\displaystyle
sin(2x) + sin(2y) = 2sin(x+y)cos(x-y)
$
sneaky, sneaky! this is actually one of the sum to product formulas in disguise!
anyway, start with the right side and expand using the addition formulas for sine and cosine
$\displaystyle \sin (A + B) = \sin A \cos B + \sin B \cos A$
$\displaystyle \cos (A - B) = \cos A \cos B + \sin A \sin B$
and try to simplify to get the left side, which, if you know your double angle formulas, you'd realize is $\displaystyle 2 \sin x \cos x + 2 \sin y \cos y$
Another way is to use Euler's formula
$\displaystyle \cos \theta = \frac{e^{i \theta} + e^{-i \theta}}{2},\;\;\;\sin \theta = \frac{e^{i \theta} - e^{-i \theta}}{2}$
Replace the $\displaystyle \theta $ with $\displaystyle (x+y)\;\;\;(x-y)$ where appropriate on the RHS, expand and use these expressions again, to get the LHS.
you can start with either side here and get the other pretty easily.
say we start with the left: multiply by $\displaystyle \frac {1 + \sin 2x}{1 + \sin 2x}$
it should be easy for you to clean up from there in a step or two (think "difference of two squares")
P.S. you should post new questions in a new thread