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I noticed that $$\sin(a+b)\sin(a-b) = \cos a \cos b\qquad (1)$$ when $$(a,b)=\left(\frac{2\pi}{5},\frac{\pi}{3}\right)$$

Is there an underlying reason for this coincidence?

Concretely, I would like to answer these questions:

  • Does the solution set $(a,b)$ for $(1)$ permit a simple description?
  • If $a$ is commensurable with $\pi$, does the same hold for $b$?
pre-kidney
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    The left-hand side $=\cos^2 b- \cos^2 a$. Hence we have: $\cos b = x \cos a$, where $x^2-x-1=0$. – njguliyev Sep 05 '13 at 08:47
  • @njguliyev How are you getting $\cos^2{b} - \cos^2{a}$? I get $\sin^2{a}\cos^2{b} - \sin^2{b}\cos^2{a}$. – Eric Auld Sep 05 '13 at 09:46
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    @EricAuld, they are equal: $\sin^2 a = 1- \cos^2 a$. – njguliyev Sep 05 '13 at 09:51
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    @njguliyev, thanks for the insight. So the "nice" form of $(1)$ isn't actually that special, because it boils down to characterizing solutions of $x^2-xy-y^2$, a pretty arbitrary polynomial (Fibonacci numbers are irrelevant in this context). It seems you've answered the first bullet as "yes", and the second bullet is most certainly "no". – pre-kidney Sep 06 '13 at 04:37
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    @njguliyev Why dont you turn your comment into an answer? – Vishal Gupta Sep 08 '13 at 05:49

2 Answers2

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The left-hand side of $(1)$ equals $\cos^2 b−\cos^2 a$. Hence we have: $\cos b = x \cos a$, where $x^2−x−1=0$.

The answer to your second question looks like "no". At least Wolfram Alpha says nothing about the rationality of $\frac{1}{\pi}\arccos \frac{\sqrt{5}-1}{2}$.

njguliyev
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    Thanks for manipulating the equation into a form that allows us to parametrize its solutions (and thereby answering my first question). Building on this, I was able to answer my second question; see below. – pre-kidney Sep 09 '13 at 06:24
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The answer to my second question is "no". Suppose some $q\in\mathbb{Q}\cap [0,1)$ satisfies $$\cos q\pi = x, \qquad x=\frac{1-\sqrt{5}}{2}$$ Then the quantity $z=x+i\sqrt{-x}=e^{iq\pi}$ must be a root of unity.

However, we may explicitly compute $$z^n=(a_n x + b_n) + (c_n x + d_n)i\sqrt{-x}$$ (using $x^2=x+1$) whereupon a simple induction shows that $a_n, b_n, c_n, d_n\in\mathbb{Z}^+$ are all increasing sequences for $n\geq 1$, and in particular $a_n \geq 1$ for $n\geq 1$.

Since $x$ is irrational we have $\Re(z^n)=a_n x + b_n \not = 1$ for $n\geq 1$, so $z$ is not a root of unity, hence no such $q$ exists. It follows that taking $a=0$ in the original equation leads to a value for $b$ that is not commensurable with $\pi$.

pre-kidney
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