Placing transcendental equation into simpler form

Question

I have come across a simple transcendental equation related to Stokes flow in a corner.

I am trying to place the transcendental equation $$n\tan n\alpha = (n-2)\tan(n-2)\alpha$$ into the form [with $x = 2(n-1)\alpha$] $$\frac{\sin x}{x} = -\frac{\sin 2\alpha}{2\alpha}.$$

I have tried double angle formulas but can't seem to reduce it to this form . Any ideas? Thank you!

Answer 1

The first equation can be written as $$ n = \frac{2 \tan((n-2)\alpha)}{\tan((n-2)\alpha) - \tan(n\alpha)} $$ The second, with $x = 2(n-1)\alpha$, is $$ n = \frac{\sin(2\alpha) - \sin(2(n-1)\alpha)}{\sin(2\alpha)} $$ So your task is to show the two right sides are equal. Expand everything in terms of $\sin(\alpha)$, $\cos(\alpha)$, $\sin(n\alpha)$ and $\cos(n\alpha)$ and use $\sin^2 + \cos^2 = 1$ and it should work.