Prove $\cos x +\cos y = 2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})$ Prove that $$\cos(x) + \cos(y) = 2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})$$ holds true for any $x, y \in \mathbb{R}$.
Even though I managed to prove its brother $\sin(x) + \sin(y)$, I haven't been able to tackle this one.
Important identities needed for the proof:
$$\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$$
$$\cos(x-y)=\cos(x)\cos(y)+\sin(x)\sin(y)$$
Let's go:
$$2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2}) = 2\cos(\frac{x}{2}+ \frac{y}{2})\cos(\frac{x}{2} - \frac{y}{2}) = $$
$$ = 2(\cos(\frac{x}{2})\cos(\frac{y}{2}) - \sin(\frac{x}{2})\sin(\frac{y}{2}))(\cos(\frac{x}{2})\cos(\frac{y}{2}) + \sin(\frac{x}{2})\sin(\frac{y}{2})) = $$
$$ = 2(\cos^2(\frac{x}{2})\cos^2(\frac{y}{2}) - \sin^2(\frac{x}{2})\sin^2(\frac{y}{2})) = $$
$$ = 2\cos^2(\frac{x}{2})\cos^2(\frac{y}{2}) - 2\sin^2(\frac{x}{2})\sin^2(\frac{y}{2})$$
Now I tried, I believe, almost every possible replacement by deriving from $\cos^2(x) + \sin^2(x) = 1$ and sadly nothing worked.

$$\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b)$$
$$\cos(a-b) = \cos(a)\cos(b) + \sin(a)\sin(b)$$
Adding the two,
$$\cos(a+b)+\cos(a-b) = 2\cos(a)\cos(b)$$
Setting $a = \frac{x+y}{2}$ and $b = \frac{x-y}{2}$, we get
$$\cos(x) + \cos(y) = 2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$$
as desired.
