<s>[INST] 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.
 [/INST] Hint: $ x \rightarrow (x+y)/2 $ and $y \rightarrow (x-y)/2$ in the second & third equations. Now add them together.
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