When does this equation $\cos(\alpha + \beta) = \cos(\alpha) + \cos(\beta)$ hold? I come across this problem in an advanced maths textbook for grade 11 in my country. And it's marked a star, which means that it's a difficult exercise, and so, no solution for this problem is given.
I can solve problems asking for which conditions do $\sin(\alpha + \beta) = \sin(\alpha) + \sin(\beta)$, and $\tan(\alpha + \beta) = \tan(\alpha) + \tan(\beta)$ hold. They are pretty easy, and straight-forward. But for this problem ($\cos(\alpha + \beta) = \cos(\alpha) + \cos(\beta)$), I have tried using all kinds of formulas, from Sum of Angles, to Sum to Product, and Double Angles, but without any luck.
So, I think there should be some glitch here that I haven't been able to spot it out.
So I hope you guys can give me some hints, or just a little push as a start.
Any help would be greatly appreciated,
Thank you very much,
And have a good day, :D
 A: Write $x=\cos\alpha$, $y=\cos\beta$, and now write $\cos(\alpha+\beta)$ in terms of just $x$ and $y$.  Rearrange the terms of the equation $\cos(\alpha+\beta)=\cos\alpha+\cos\beta$ and square both sides.  You should now get an equation you can use to solve for $y$ in terms of $x$.  It doesn't look like the solution is going to be very pretty though!
A: The following contour plots hint us why the third equation defies many attempts.

Another possible explanation is that, if we substitute $x = \tan(\alpha/2)$ and $y = \tan(\beta/2)$ then the formulas
$$ \sin \alpha = \frac{2x}{1+x^2}, \quad \cos \alpha = \frac{1-x^2}{1+x^2}, \quad \tan \alpha = \frac{2x}{1-x^2} $$
show that
\begin{align*}
\sin(\alpha+\beta) = \sin\alpha+\sin\beta
&\quad \Longleftrightarrow \quad xy(x+y) = 0, \\
\tan(\alpha+\beta) = \tan\alpha+\tan\beta
&\quad \Longleftrightarrow \quad xy(x+y)(1-xy) = 0, \\
\cos(\alpha+\beta) = \cos\alpha+\cos\beta
&\quad \Longleftrightarrow \quad 3x^2y^2 - x^2 - 4xy - y^2 - 1 = 0.
\end{align*}
This may be another reason why our equation seems impossible to solve in simple terms. Finally, one interesting observation is that
$$ \cos(\alpha+\beta) + 1 = \cos\alpha + \cos\beta \quad \Longleftrightarrow \quad xy(1-xy) = 0 $$
can be easily solved.
A: There may be a better way using the aforementioned transforms with $x$ and $y$, but just looking at the trig as is you can find the "nice" solutions as $\alpha = 2 \pi n_1+pi, \hspace{.1cm} \beta = \frac{1}{3} (6 \pi n_2 \pm \pi)$ and $\alpha = \frac{1}{3} (6 \pi n_2 \pm \pi), \hspace{.1cm} \beta = 2 \pi n_1+\pi$ for some integers $n_1$ and $n_2$. 
I found those just from checking numbers and without much sophistication. Examining the result with a CAS I found there is a rather general and very nasty result involving tangent for a range of $\alpha$'s.   
A: (Answer to be updated)
I am taking $a=\cos\alpha$ and $b=\cos\beta$, so applying the $cosx$ and $sinx$ relation, I get
$$\sin\alpha= \sqrt{1-\cos^{2}\alpha}$$  or  $$\sin\alpha= \sqrt{1-a^{2}}$$
and similarly
$$\sin\beta= \sqrt{1-b^{2}}$$
as
$$\cos\left(x+y\right)=\cos x\cos y-\sin x\sin y$$
$$\cos\left(\alpha\right)=\cos x\cos y-\sin x.\sin y$$
