Relations similar to $\sin(\pi/18) + \sin(5\pi/18) = \sin(7\pi/18)$ This I checked numerically (and can be proven analytically easily). I guess there are many similar relations between the numbers 
$$\{ \sin\frac{m\pi}{q}, 1\leq m \leq q \} , $$
where $q$ is some integer. 
Can anyone give some reference? 
 A: More generally, let $T_n(x) + 1$ have a factor $Q(x)$ (over the rationals) where $T_n$ is the $n$'th Chebyshev polynomial of the first kind.  A root of $Q(x)$ will be $\cos(j \pi/n)$ for some integer $j$.  Using trigonometric identities, we can then express $Q(\cos(\pi/n)) \sin(\pi/n)$ as a rational linear combination of sines of multiples of $j \pi/n$.  In your case
$$ T_{18}(x)+1 = 2 x^2 (4 x^2-3)^2 (64 x^6 - 96 x^4 + 36 x^2 - 3)^2 $$
and $$(64 \cos^6(\theta) - 96 \cos^4(\theta) + 36 \cos^2(\theta) - 3) \sin(\theta) = \sin(7\theta) - \sin(5 \theta) - \sin(\theta)$$
leading to your identity with $\theta = \pi/18$.
Similarly, using $$T_{20}(x) + 1 = 2\, \left( 2\,{x}^{2}-1 \right) ^{2} \left( 256\,{x}^{8}-512\,{x}^{6}+
304\,{x}^{4}-48\,{x}^{2}+1 \right) ^{2}
$$
we get
$$ \sin \left( {\frac {9\,\pi}{20}} \right) -\sin \left( {\frac {7\,\pi}{
20}} \right) -\sin \left( \frac{\pi}4 \right) +\sin \left( {\frac {3\,\pi}{
20}} \right) +\sin \left( \frac{\pi}{20} \right) = 0
$$
A: I have studied this problem and found two kinds of solutions.
[Solution 1]
$ \cos\left(\frac{\pi k}{3 n}\right)=\cos \left(\frac{\pi (n-k)}{3 n}\right)+\cos \left(\frac{\pi (k+n)}{3 n}\right) $
Example:
$ \cos\left(\frac{\pi }{15}\right)=\cos\left(\frac{4 \pi }{15}\right)+\cos \left(\frac{6 \pi }{15}\right) $
$ \cos \left(\frac{2 \pi }{15}\right)=\cos \left(\frac{3 \pi }{15}\right)+\cos \left(\frac{7 \pi }{15}\right) $
$ \cos \left(\frac{3 \pi }{15}\right)=\cos \left(\frac{5 \pi }{15}\right)+\cos \left(\frac{6 \pi }{15}\right) $
$ \cos \left(\frac{ \pi }{21}\right)=\cos \left(\frac{6 \pi }{21}\right)+\cos \left(\frac{8 \pi }{21}\right) $
$ \cos \left(\frac{2 \pi }{21}\right)=\cos \left(\frac{5 \pi }{21}\right)+\cos \left(\frac{9 \pi }{21}\right) $
$ \cos \left(\frac{3 \pi }{21}\right)=\cos \left(\frac{4 \pi }{21}\right)+\cos \left(\frac{10 \pi }{21}\right) $
$ \cos \left(\frac{ \pi }{27}\right)=\cos \left(\frac{8 \pi }{27}\right)+\cos \left(\frac{10 \pi }{27}\right) $
$ \cos \left(\frac{2 \pi }{27}\right)=\cos \left(\frac{7 \pi }{27}\right)+\cos \left(\frac{11 \pi }{27}\right) $
$ \cos \left(\frac{3 \pi }{27}\right)=\cos \left(\frac{6 \pi }{27}\right)+\cos \left(\frac{12 \pi }{27}\right) $
$ \cos \left(\frac{4 \pi }{27}\right)=\cos \left(\frac{5 \pi }{27}\right)+\cos \left(\frac{13 \pi }{27}\right) $
[Solution 2]
$ \cos \left(\frac{\pi  (6 k-3)}{5 (6 k-3)}\right)=\cos \left(\frac{\pi  (10 k-5)}{3 (10 k-5)}\right)+\cos \left(\frac{(2 \pi ) (6 k-3)}{5 (6 k-3)}\right) $
Example:
$ \cos \left(\frac{ \pi }{5}\right)=\cos \left(\frac{  \pi }{3}\right)+\cos \left(\frac{2 \pi }{5}\right) $
$ \cos \left(\frac{3 \pi }{15}\right)=\cos \left(\frac{ 5 \pi }{15}\right)+\cos \left(\frac{6 \pi }{15}\right) $
