There exist real numbers $x$ and $y$ such that $\cos(x+y)=\cos x+\cos y$ I’m trying to either prove or disprove the statement that there exist real numbers x and y such that $\cos(x+y)=\cos x+\cos y$, though I quickly encountered a brick wall after expanding the LHS:
$\cos x\cos y - \sin x\sin y = \cos x + \cos y$
My question is, is there a different approach to solving this problem, or is what I started doing the right way? I couldn’t find the problem online, so I would really appreciate your responses
 A: You can restrict yourself to the case $y=-x$.

  Then you only have to find a value $x$ such that $\cos x = 1/2$.  As $\cos 0=1$ and $\cos x=\cos (-x)$, you are done.

A: Now others have given answers and hints, one technique to show there is a solution to this kind of equation is to use the intermediate value theorem - show that there are values where the LHS is less than the RHS (simply maximise RHS) and also where LHS is greater than RHS (minimise RHS). This is easy to do because the range of values on the RHS is greater than the range on the LHS and strictly contains it (use the formula for the sum of two cosines).
The fact that there are two variables means you need to take a little care to make it work, but it is a straightforward argument.
I mention this because the Intermediate Value Theorem should be in your toolkit, even if it is not always the first thing to try.
A: Alternative approach:
Since the posted problem provides the trigonometry tag, it is unclear if Real Analysis methods are intended by the problem composer.
Also, the specific tool that you attempted to use is the multiple angle formula for the cosine function.  Your posting made no mention of (for example) the intermediate value theorem.
This is interesting, because the title of your posting suggests that you regard the domain of the cosine function to be real numbers, rather than angles.  This seems somewhat contradictory.  That is, when I was in school, the domain of the cosine function became a Real Number, rather than an angle, only when I started Calculus.
Anyway, I will structure a response based on only trigonometry methods.
It is sufficient to find one example.  One such example that has not been mentioned yet is if
$\cos(x) = 0 \implies ~~(\text{for example})~~ x ~~\text{could equal}~~ \pi/2.$
Then, solving for $(y)$, you have that
$[0 \times \cos(y)] - [1 \times \sin(y)] = \cos(y)$.
This implies that
$$-\sin(y) = \cos(y) \implies \sin^2(y) = \cos^2(y).$$
Since you know that $\sin^2(y) + \cos^2(y) = 1,$ 
you then know that $\cos^2(y) = \dfrac{1}{2}.$
Then, one obvious solution is $(y) = -\pi/4.$
A: $$\begin{aligned}& \cos(x+y)=\cos(x)+\cos(y)\Rightarrow \\
& \cos(x) \cos(y)-\sin(x)\sin(y)=\cos(x)+\cos(y)\Rightarrow\\
& (1-\cos(x))\cos(y)+\sin(x)\sin(y)=-\cos(x)\end{aligned}$$
$$(\cos x=1 {\rm\ gives\ } 0=-1) \Rightarrow \cos(x)\neq 1 \Rightarrow \cos(x)<1$$
$$A=\sqrt{(1-\cos(x))^2+(\sin(x))^2}=\sqrt{2(1-\cos x)} > 0$$
$$\exists \phi\in\mathbb{R}:\sin(\phi)=\frac{1-\cos(x)}{A}, \cos(\phi)=\frac{\sin(x)}{A}$$ $$A\sin(\phi)\cos(y)+A\cos(\phi)\sin(y)=-\cos(x)\Rightarrow\sin(y+\phi)=-\frac{\cos(x)}{A}$$
$$\left|-\frac{\cos(x)}{A}\right|\leq 1 \Rightarrow \cos^2(x)\leq A^2\Rightarrow \cos^2(x)+2\cos(x)\leq 2$$
$$\cos(x)=t,t^2+2t\leq 2 \Rightarrow (t+1)^2\leq 3\Rightarrow -1-\sqrt{3}\leq t \leq \sqrt{3}-1$$
$$|\cos(x)|\leq 1, -1-\sqrt{3}\leq \cos(x) \leq \sqrt{3}-1 \Rightarrow -1\leq \cos(x) \leq \sqrt{3}-1$$
Then for any $x$ such that $-1\leq\cos(x)\leq \sqrt{3}-1$ exists $y$ such that $$\cos(x+y)=\cos(x)+\cos(y)$$
A: There are infinite solutions, beyond those of $x=y$. How to prove this?
Take your expansion of $\cos x\cos y - \sin x\sin y = \cos x + \cos y$ and apply the following tan-half-angle substitutions, $u = \tan \left( \tfrac{x}{2} \right)$ and $v = \tan \left( \tfrac{y}{2} \right)$
$$\begin{aligned}
  \cos x & = \frac{1-u^2}{1+u^2} \\
  \sin x & = \frac{2u}{1+u^2} \\
  \cos y & = \frac{1-v^2}{1+v^2} \\
  \sin y & = \frac{2v}{1+v^2} \\
\end{aligned}$$
which transforms the expansion (with some grouping/simplifications) into
$$ u^2 (3 v^2 -1)-4 u v - v^2 = 1$$
which is a hyperbolic curve, and for every value of $u$ there will be up to two solutions for $v$, and vice versa. Any pair of $(u,v)$ that satisfies the above, will also satisfy the original equation with the corresponding $(x,y)$.

(u,v) plot
