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?

It probably uses: $1- \frac{1}{1-\theta}= \frac{\theta}{1-\theta}$ so it's shifting by $\pi$ — Sort of Damocles, Jun 06 '19 at 13:00

score 11 · Accepted Answer · answered Jun 06 '19 at 12:58

11

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 $\sin(\alpha + \beta)$.

answered Jun 06 '19 at 12:58

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A strange trigonometric identity

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