Finding parametrized solutions to a trigonometric equation I want to find what pairs $(x,y)$ satisfy $\cos(x-y) = \cos(x) - \cos(y)$. To start, I defined a function $f(x,y) = \cos(x-y) - \cos(x) + \cos(y)$. I want to find functions $x(t), y(t)$ s.t. $f(x(t), y(t)) = 0$ I didn't really know where to go, so I tried putting the cosines in exponential form:
$$
\begin{align}
0 &= \cos(x-y) - \cos(x) - \cos(y)\\
0 &= \frac{1}{2}\big[e^{i(x-y)}+e^{i(y-x)}-e^{ix}-e^{-ix}+e^{iy}+e^{-iy}\big]\\
\end{align}
$$
From here I wasn't sure what to do. I tried multiplying through to get rid of negative exponents, rearranging into a neater format, factoring, etc. to no avail. I'm not even sure that's a useful first step, it just was the only one I could think of.
Any tips or ideas?
Also feel free to add or edit tags, wasn't quite sure what to put this under.
 A: I'm not sure if your method with exponentials allows you to conclude, but here's an overview of how I'd do it.
Although this isn't linear algebra, you can still get an idea of how to proceed by looking at your number of equations and unknowns. You have two unknowns for one equation, so if you add in one more unknown (parameter $t$) without any additional constraints, it'll only make things more complicated.
Instead you can just find an expression of $y$ as a function of $x$, that will give you a parametric expression of the form $x(t)=t$ and $y(t)$.
\begin{align*}
0=\cos(x-y)-\cos x+\cos y
&~\iff~ \cos x=\cos x\cos y+\sin x\sin y+\cos y \\
&~\iff~ \cos x=(\cos x+1) \cos y+\sin x\sin y
\end{align*}
Now there's somewhat of a trick to solve trigonometric equations of the form
$a\cos y+b\sin y=c$. Assuming that $a^2+b^2>0$, you can rewrite that as
$$
\frac a{\sqrt{a^2+b^2}}\cos y+\frac b{\sqrt{a^2+b^2}}\sin y=\frac c{\sqrt{a^2+b^2}}
~\iff~ A\cos y+B\sin y=C
$$
Notice that $A^2+B^2=1$, that means there is $\theta\in\mathbb R$ (unique if taken in $[0,2\pi)$) such that
$A\cos y+B\sin y=\cos\theta\cos y+\sin\theta\sin y=\cos(\theta-y)$.
From there, the equation admits solution(s) if and only if $\lvert C\rvert\le 1$.

This method can be applied to your equation
$\cos x= (\cos x+1)\cos y+\sin x\sin y$.
Just as a heads-up, if you apply the method I outlined in that exact manner,
you should obtain
$\sqrt{a^2+b^2}=2\lvert\cos\left(\frac x2\right)\rvert$, which is somewhat not nice. If you use the coefficient
$\sqrt{a^2+b^2}=2\cos\left(\frac x2\right)$ instead, it'll simplify things a little. In particular you should be able to prove that $\theta=\frac x2+2k\pi$, $k\in\mathbb Z$.
Last remark, but the cases $a^2+b^2=0$ and $\lvert C\rvert>1$ must be handled as exceptions if you want a complete proof of every possible solutions.
A: I'm not sure parametrizing the solution is a good idea, if you take a look at wolframalphas sketch of all solutions you'll see that x(t) and y(t) wouldn't even be continous. Maybe proving that all pairs of (x,y) is on a certain form would be a simpler solution.
