Let $x=\cos\theta$. Show that $\frac{d^{n-1}\sin^{2n-1}\theta}{dx^{n-1}}=\frac{(-1)^{n-1}}{n}\frac{(2n)!}{2^{n}n!}\sin n\theta$

Question

I need some help with the following exercise which I don't know how to start with. Any tip for a start would be already great. IT's from the AAR Special Functions book.

Let $x=\cos\theta$. Show that:

$\frac{d^{n-1}\sin^{2n-1}\theta}{dx^{n-1}}=\frac{(-1)^{n-1}}{n}\frac{(2n)!}{2^{n}n!}\sin n\theta$

Answer 1

Hint: Use the chain rule for single variable differentiation, and then try to use mathematical induction over n.