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.
\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}
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 $\sin(A\pm B)$
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
$$\sin2x + \sin2y = 2\sin(x + y)\cos(x - y)$$
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.
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:
$$ \forall x, y \in \mathbb{R}^2, \; \sin(2x) + \sin(2y) = 2\cos(x-y)\sin(x+y). $$
