# 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.)

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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, $c = 1$.

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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 $\sin^2 d+\cos^2d=1$

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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.

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Well, this is clear when $c=1$, but what about otherwise? – Cameron Buie Jul 4 '12 at 17:51
@CameronBuie : If it depends on $\alpha$ and $\beta$ only through their sum, then the identity I wrote above holds, and that simplifies to $\cos(\alpha+\beta)$. Apply the usual identity and get $c=1$. So in other words, it works ONLY if $c=1$. – Michael Hardy Jul 4 '12 at 17:56
@MichaelHardy: I'm almost tempted to downvote, because I think the way you've written the answer that's not clear at all. – Ben Millwood Jul 4 '12 at 18:23
@BenMillwood : OK, I've expanded on it. – Michael Hardy Jul 4 '12 at 19:30