TITLE: A difficult trigonometry problem
QUESTION [6 upvotes]: How to prove that $$\left(\sin{\frac{9\pi}{70}}+\sin{\frac{29\pi}{70}}-\sin{\frac{31\pi}{70}}\right)\left(\sin{\frac{\pi}{70}}-\sin{\frac{11\pi}{70}}-\sin{\frac{19\pi}{70}}\right)=\frac{\sqrt{5}-4}{4}?$$
I don't have any idea.

REPLY [4 votes]: Here is an answer, though I'll be amazed if no-one can provide something more elegant.  Let $\alpha=e^{i\pi/70}$.  Then $\alpha^{70}=-1$ and we have
$$LHS=\Bigl(\frac{\alpha^9-\alpha^{-9}}{2i}+etc\Bigr)\Bigl(etc\Bigr)\ ,$$
so
$$-4(LHS)=(\alpha^9+\alpha^{61}+etc)(etc)\ .$$
Multiplying everything out on the RHS gives the sum of all $\alpha^k$, where $k$ belongs to the following sets:
$$\eqalign{
  &\{2,22,42,62,82,102,122\}\cr
  &\{18,38,58,78,98,118,138\}\cr
  &\{42,98\}\quad\hbox{twice}\cr
  &\{10,30,50,90,110,130\}\quad\hbox{three times.}\cr}$$
Now the numbers $\alpha^k$ for the first set are seven points equally spaced around the unit circle, so the sum is zero.  Same goes for the second set.  Same for the fourth, except that $70$ is missing - no it wasn't a typo!  So the sum of $\alpha^k$ for the fourth set is $-\alpha^{70}$, that is, $1$.  So we have
$$-4(LHS)=2(\alpha^{42}+\alpha^{98})+3=4\cos\frac{3\pi}5+3\ ,$$
and since it is known that $4\cos(3\pi/5)=1-\sqrt5\,$, we are done.