How to prove the identity $\sin2x + \sin2y = 2\sin(x + y)\cos(x - y)$ I've been trying to prove the identity $$\sin2x + \sin2y = 2\sin(x + y)\cos(x - y).$$
So far I've used the identities based off of the compound angle formulas. I'm not quite sure if those identities would work with proving the above identity.
Thank you in advance.
 A: \begin{align}
   2\sin(x + y)\cos(x - y)
   &= 2(\sin x \cos y + \cos x \sin y)\cdot
       (\cos x \cos y + \sin x \sin y) \\
   &= 2\sin x \cos x(\cos^2 y + \sin^2 y) + 
      2\sin y \cos y(\cos^2 x + \sin^2 x) \\
   &= 2\sin x \cos x + 2\sin y \cos y \\
   &= \sin 2x + \sin 2y
\end{align}
A: Hint:
Use the linearisation formula
$$2\sin a\cos b=\sin(a+b)+\sin(a-b).$$
A: Hint:
If $\sin(A+B),\sin(A-B)$ can be used 
set $2x=A+B,2y=A-B\implies A= x+y, B=x-y$
and then expand $\sin(A\pm B)$
A: Add the two compound angle formulas below,
$$\sin(a+b) = \sin a \cos b + \cos a\sin b$$
$$\sin(a-b) = \sin a \cos b - \cos a\sin b$$
to get,
$$\sin(a+b)+ \sin(a-b) = 2\sin a \cos b$$
Then, let $a=x+y$ and $b=x-y$ to obtain
$$\sin2x + \sin2y = 2\sin(x + y)\cos(x - y)$$
A: Hint:
Use factor formula, or also known as sum-to-product formula. Note that the formula works both ways i.e. can be interpreted as product-to-sum as well when read from right to left.
A: Here is a proof of this identity using complex numbers.
Let $x, y \in \mathbb{R}^2$. By expanding 
$$ e^{i(x+y)}e^{i(x-y)} = \big( \cos(x+y) + i\sin(x+y) \big) \big( \cos(x-y) + i\sin(x-y) \big),$$
you obtain: 
$$ \Im\big( e^{i(x+y)}e^{i(x-y)} \big) = \cos(x+y)\sin(x-y) + \cos(x-y)\sin(x+y), $$
where $\Im(z)$ denotes the imaginary part of $z \in \mathbb{C}$. Therefore:
$$ \Im\big( e^{2ix} \big) = \sin(2x) = \cos(x+y)\sin(x-y) + \cos(x-y)\sin(x+y). \quad (\star) $$
By exchanging the role of $x$ and $y$ in $(\star)$, you also get:
$$ \Im\big( e^{2iy} \big) = \sin(2y) = -\cos(x+y) \sin(x-y) + \cos(x-y)\sin(x+y). \quad (\star\star)$$
Putting $(\star)$ and $(\star\star)$ together, you eventually get:

$$ \forall x, y \in \mathbb{R}^2, \; \sin(2x) + \sin(2y) = 2\cos(x-y)\sin(x+y). $$

