A strange trigonometric identity

Question

In this paper, equation (4.5), the authors state the trigonometic identity

$$ \sin\left( \frac{n\pi }{1-\theta} \right) = (-1)^{n} \sin\left( \frac{n\pi \theta}{1-\theta}\right) $$ Nothing like it is on Wikipedia's list of trigonometric identities. How can we prove this?

Answer 1

Hint:

Just write $$\sin\left( \frac{n\pi }{1-\theta} \right) =\sin\left( \frac{n\pi(1-\theta + \theta) }{1-\theta} \right) = \sin\left(n\pi + \frac{n\pi\theta}{1-\theta} \right) $$

Now, apply the addition formula for $\sin(\alpha + \beta)$.