Proof of a trigonometric expression Let $f(x) = (\sin \frac{πx}{7})^{-1}$. Prove that $f(3) + f(2) = f(1)$.
This is another trig question, which I cannot get how to start with. Sum to product identities also did not work.

Let $7\theta=\pi$
$$\frac1{\sin3\theta}+\frac1{\sin2\theta}-\frac1{\sin\theta}$$
$$=\frac{2\sin2\theta\sin\theta+2\sin3\theta\sin\theta-2\sin2\theta\sin3\theta}{2\sin3\theta\sin2\theta\sin\theta}$$
Now use $\displaystyle2\sin A\sin B=\cos(A-B)-\cos(A+B)$
and as $\displaystyle\cos(\pi-x)=-\cos x, \cos(7-r)\theta=\cos(7\theta-r\theta)=\cos(\pi-r\theta)=-\cos r\theta$
