Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.)

share|improve this question
add comment

3 Answers

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, $c = 1$.

share|improve this answer
add comment

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

share|improve this answer
add comment

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.