Cute trigonometric triviality For which values of the coefficient $c$ does the quantity
$$
\cos\alpha\cos\beta- c\sin\alpha\sin\beta
$$
depend on $\alpha$ and $\beta$ only through their sum?
(I'll post a quick answer below.  This will be the first time I've posted a question with intent to immediately post an answer.)
 A: Well, let's define the new quantities $s=\frac{\alpha+\beta}{2}$ and $d=\frac{\alpha-\beta}{2}$. With those quantities the expression reads $$E:=\cos(s+d)\cos(s-d)-c\sin(s+d)\sin(s-d).$$ The question now is: For which values of $c$ is this expression independent from $d$?
Let's apply the standard addition theorems to get
$$E =(\sin s\cos d+\cos s\sin d)(\sin s\cos d-\cos s\sin d)
-c(\cos s\cos d-\sin s\sin d)(\cos s\cos d+\sin s\sin d).$$
Using $(a+b)(a-b) = a^2-b^2$, we therefore get $$E=\sin^2 s\,\cos^2 d-\cos^2s\, \sin^2d - c(\cos^2 s\,\cos^2 d-\sin^2 s\,\sin^2 d).$$
Now we collect the functions of $s$ to get $$E = \sin^2 s(\cos^2 d+c\,\sin^2 d) - \cos^2s(\sin^2 d+c\,\cos^2 d).$$
Now it is easy to see that the only possibility that $E$ is independent from $d$ (and therefore in the original form depends only on the sum) is $c=1$, where $\sin^2 d+\cos^2d=1$
A: Let $f(\alpha,\beta) = \cos\alpha\cos\beta- c\sin\alpha\sin\beta$.  Since this depends on $\alpha$ and $\beta$ only through their sum we have
$$f(\alpha,\beta)=f(\alpha+\beta,0).$$
Then
$$
f(\alpha+\beta,0) = \cos(\alpha+\beta)\cos 0 - c\sin(\alpha+\beta)\sin 0 = \cos(\alpha+\beta).
$$
So by the usual identity, $c=1$.
Later edit: Another way would be to write
$$
\begin{align}
\cos\alpha\cos\beta - c\sin\alpha\sin\beta & = \Big(\cos\alpha\cos\beta - \sin\alpha\sin\beta\Big) - (1-c)\sin\alpha\sin\beta \\[8pt]
& = \cos(\alpha+\beta) - (1-c)\sin\alpha\sin\beta
\end{align}
$$
and then observe that the last term doesn't depend on $\alpha$ and $\beta$ only through their sum.  From one point of view, this seems like the obvious way to do it---far more so than what I did above, and yet what I did above seems simpler.
A: We want $f(\alpha, \beta) = g(\alpha + \beta)$. This means that $f(\alpha,-\alpha) = g(0) = f(0,0)$, $\forall \alpha$.
$f(\alpha,-\alpha) = f(0,0)$ for all $\alpha$. Hence, we get that $$\cos^2(\alpha) + c \, \sin^2(\alpha) = 1, \, \, \forall \alpha$$
$$c \, \sin^2(\alpha) = \sin^2(\alpha), \, \, \forall \alpha$$
Hence, $c = 1$.
