Try working from the left-hand side:

. . .sin(2x) + sin(2y) = 2sin(x)cos(x) + 2sin(y)cos(y)

. . . . .= 2[sin(x)cos(x) + sin(y)cos(y)]

Now work from the right-hand side, ignoring the "2" for the time being:

. . . . .sin(x + y) = sin(x)cos(y) + cos(x)sin(y)

. . . . .cos(x - y) = cos(x)cos(y) + sin(x)sin(y)

Multiplying:

. . . . .sin(x)cos(x)cos^2(y) + cos^2(x)sin(y)cos(y)

. . . . . . . .+ sin^2(x)sin(y)cos(y) + sin(x)sin^2(y)cos(x)

Rearranging a bit:

. . . . .sin(x)cos(x)[cos^2(y) + sin^(y)]

. . . . . . . .+ sin(y)cos(y)[cos^(x) + sin^2(x)]

This simplifies as:

. . . . .sin(x)cos(x) + sin(y)cos(y)

Throw that "2" back in, and you've got equality.

Of course, the above is not a proper "proof". For that, you'll need to start at one side, work down the where "they meet in the middle", and then work back up to the other side.