Prove $\cos x +\cos y = 2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})$

Question

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.

Answer 1

$$\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b)$$ $$\cos(a-b) = \cos(a)\cos(b) + \sin(a)\sin(b)$$

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.

Answer 2

Add these equations:

$\cos x=\cos\dfrac{x+y}2\cos\dfrac{x-y}2-\sin\dfrac{x+y}2\sin\dfrac{x-y}2$

$\cos y=\cos\dfrac{x+y}2\cos\dfrac{x-y}2+\sin\dfrac{x+y}2\sin\dfrac{x-y}2$

Answer 3

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

Answer 4

This problem is also an exercise in McMullen's "Trig Identities". Here is the book’s approach ( with hints from the author). The start is similar to the OP solution.

Show:

$ \displaystyle \cos x + \cos y = 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\left [\cos \frac{x}{2}\cos \frac{y}{2} - \sin \frac{x}{2}\sin \frac{y}{2} \right]\left [\cos \frac{x}{2}\cos \frac{y}{2} + \sin \frac{x}{2}\sin \frac{y}{2} \right]$

Now multiply the two terms.

$ \displaystyle = 2\left [\cos^2 \frac{x}{2}\cos^2 \frac{y}{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} \cos \frac{x}{2}\cos \frac{y}{2} -\sin^2 \frac{x}{2}\sin^2 \frac{y}{2} \right]$

Simplify, by adding like terms

$ \displaystyle = 2\left [\cos^2 \frac{x}{2}\cos^2 \frac{y}{2} -\sin^2 \frac{x}{2}\sin^2 \frac{y}{2} \right]$

Key step : Distribute $2$ and expand

$\displaystyle = \cos^2 \frac{x}{2}\cos^2 \frac{y}{2} + \cos^2 \frac{x}{2}\cos^2 \frac{y}{2} -\sin^2 \frac{x}{2}\sin^2 \frac{y}{2} -\sin^2 \frac{x}{2}\sin^2 \frac{y}{2}$

Apply the pythagorean identities

$ \displaystyle = \left( 1 - \sin^2 \frac{x}{2} \right)\cos^2 \frac{y}{2} + \cos^2 \frac{x}{2}\left( 1 - \sin^2 \frac{y}{2} \right) -\left( 1 - \cos^2 \frac{x }{2} \right)\sin^2 \frac{y}{2} -\sin^2 \frac{x}{2}\left( 1 - \cos^2 \frac{y}{2} \right)$

$ \displaystyle = \cos^2 \frac{y}{2} - \sin^2 \frac{x}{2}\cos^2 \frac{y}{2} + \cos^2 \frac{x}{2} - \cos^2 \frac{x}{2}\sin^2 \frac{y}{2} -\sin^2 \frac{y}{2} + \cos^2 \frac{x}{2}\sin^2 \frac{y}{2} -\sin^2 \frac{x}{2} + \sin^2 \frac{x}{2}\cos^2 \frac{y}{2}$

Rearrange

$ \displaystyle\cos^2 \frac{y}{2} -\sin^2 \frac{y}{2} + \cos^2 \frac{x}{2} -\sin^2 \frac{x}{2} + \sin^2 \frac{x}{2}\cos^2 \frac{y}{2}- \sin^2 \frac{x}{2}\cos^2 \frac{y}{2} + \cos^2 \frac{x}{2}\sin^2 \frac{y}{2} - \cos^2 \frac{x}{2}\sin^2 \frac{y}{2} $

Simplify

$ \displaystyle \cos^2 \frac{y}{2} -\sin^2 \frac{y}{2} + \cos^2 \frac{x}{2} -\sin^2 \frac{x}{2} $

$ \displaystyle \cos(2u) = \cos^2 u - sin ^ 2 u$

Now let $u = \frac{y}{2}$ Hence: $2u = y$

Thus: $ \displaystyle\cos y = \cos^2 \frac{y}{2} - sin ^ 2 \frac{y}{2}$ Similarly apply for $ \displaystyle x$

Finally

$ \displaystyle= \cos x + \cos y$