# 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.

• Hint: Use $\cos x=2\cos^2(x/2)-1=1-2\sin^2(x/2)$ to continue.
– Feng
Commented Mar 11, 2020 at 1:21

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

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.

• Oh my god. How was I supposed to come up with this!? Thank you. Commented Mar 11, 2020 at 1:25
• @tomashauser mathematics is often deceiving. Commented Mar 11, 2020 at 1:35
• This proof is one of the most elegant ones I've come come across this site. Commented Mar 11, 2020 at 1:58

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

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

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

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

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