# legendre polynomial problem

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?

## 1 Answer

I may have figured it out actually, very simple.

If

$$x=cos\theta$$

then

$$dx/d\theta=-sin\theta$$

then just inverse