I am trying to express this equation in terms of theta:
$$(1-x^2)y''(x)-2xy'(x)+n(n+1)y(x)=0$$
where
$$x=cos\theta$$
I know I can begin with Legendre's equation:
$$(1-x^2)\frac{d^2P_n}{dx^2}-2x\frac{dP_n}{dx}+n(n+1)P_n(x)=0$$
For some reason,
$$\frac{dP_n}{dx}=\frac{d\Theta_n}{d\theta}\frac{d\theta}{dx}=-\frac{1}{sin\theta}\frac{d\Theta_n}{d\theta}$$
I do not understand where the
$$\frac{1}{sin\theta} $$
comes from.
Can someone explain this to me?
I may have figured it out actually, very simple.
If
$$x=cos\theta$$
then
$$dx/d\theta=-sin\theta$$
then just inverse