How to prove $\sin (x)+ \sin(y) = 2 \sin \left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right)$ using addition theorems? I am stuck at one task, it's to proof the following equation using addition theorems.
From a draft I understood that the equation is correct or true. But that's it.
What's a good way to explain it mathematically?
$$\sin (x)+ \sin(y) = 2 \sin \left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right)$$
As usual, any help is upvoted immediately.

The relevant addition formulae tell you that:
$$
\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)\\
\cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B)
$$
Applying these to $2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$ yields:
$$
\begin{align*}
2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right) & = 2\left[\sin(x/2)\cos(y/2) + \cos(x/2)\sin(y/2)\right]\left[\cos(x/2)\cos(y/2) + \sin(x/2)\sin(y/2)\right]\\
& = 2\sin(x/2)\cos(x/2)\cos^2(y/2) + 2\sin^2(x/2)\sin(y/2)\cos(y/2)\\
& \hspace{20pt}+ 2\cos^2(x/2)\sin(y/2)\cos(y/2) + 2\sin(x/2)\cos(x/2)\sin^2(y/2)\\
& = 2\sin(x/2)\cos(x/2)\left[\sin^2(y/2) + \cos^2(y/2)\right]\\
& \hspace{20pt}+ 2\sin(y/2)\cos(y/2)\left[\sin^2(x/2) + \cos^2(x/2)\right]\\
& = 2\sin(x/2)\cos(x/2) + 2\sin(y/2)\cos(y/2)
\end{align*}
$$
From here, can you make the final step?
