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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.

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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 $7\not\mid r$

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consider regular 7-gon. applying Ptolemy theorem for $\square ABCD$,
we get $BD\cdot AC=AB\cdot CD+AD\cdot BC$
$BC=AC=2R\sin\dfrac{3\pi}{7}$ , $~BD=CD=2R\sin\dfrac{2\pi}{7}$ , $~AB=AD=2R\sin\dfrac{\pi}{7}$
we're done, where $R$ is the circumradius of regular 7-gon.

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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 $\displaystyle\cos(\pi-x)=-\cos x, \cos(7-r)\theta=\cos(7\theta-r\theta)=\cos(\pi-r\theta)=-\cos r\theta$

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