Coincidental Trigonometric Identity for Two Particular Values 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 $b$?


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 $\pi$.
