trigonometry - 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.

Answer

Let $7\theta=\pi, 4\theta=\pi-3\theta\implies \sin4\theta=\sin(\pi-3\theta)=\sin3\theta$

$$\frac1{\sin3\theta}+\frac1{\sin2\theta}$$

$$=\frac1{\sin4\theta}+\frac1{\sin2\theta}$$

$$=\frac{\sin4\theta+\sin2\theta}{\sin4\theta\sin2\theta}$$

$$=\frac{2\sin3\theta\cos\theta}{\sin4\theta\sin2\theta}\text{ Using } \sin2C+\sin2D=2\sin(C+D)\cos(C-D)$$

$$=\frac{2\cos\theta}{2\sin\theta\cos\theta}$$

$$=\frac1{\sin\theta}$$

All cancellations are legal as $\sin r\theta\ne0$ for $7\not\mid r$