I'm just getting learning about how Legendre polynomials come about when considering product solutions in spherical coordinates with azimuthal symmetry. I'm trying a problem on my own, and I'm a bit confused as to how I should rearrange a particular term. My question is basically, how to write $\sin^2\theta$ as a Legendre polynomial in $\cos\theta$: $P_l(\cos\theta)$. I'm trying to solve for $$ A_l=\frac{\sigma_0}{2\epsilon_0 R^{l-1}}\int_0^\pi \sin^2\theta P_l(\cos\theta)\sin\theta \,d\theta.$$ I know that Legendre polynomials are orthogonal, so $$\int_0^\pi P_{q}(\cos\theta) P_l(\cos\theta)\sin\theta \,d\theta=0\quad \text{ if }q\neq l.$$ Now I tried rewriting $\sin^2\theta$ as $1-\cos^2\theta$. This is close to the Legendre polynomial $P_2=\frac{3\cos^2\theta-1}{2}$, but I can't get it exactly into this form because of the $3$ term in front of $\cos^2\theta$. So how can I rewrite $\sin^2\theta$ as a Legendre polynomial $P_l(\cos\theta)$?

  • $\begingroup$ You can try writing $$\sin^2(x)=\sin^2(x) \frac{\frac{4 \sin^2(x)}{3 \cos(2 x)+1}}{\frac{4 \sin^2(x)}{3 \cos(2 x)+1}}=\left (\frac{3\cos^2(x)-1}{2}\right )\left(\frac{4 \sin^2(x)}{3 \cos(2 x)+1}\right )$$ but I dunno if it helps. $\endgroup$ – UserX Oct 1 '14 at 9:22
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    $\begingroup$ You need a sum of Legendre polynomials, I hope this helps: $1-\cos^2\theta = -\frac{2}{3} \cdot \left(\frac{3\cos^2\theta-1}{2}\right) + \frac{2}{3}\cdot 1 = -\frac{2}{3}P_2(\cos\theta) + \frac{2}{3}P_0(\cos\theta)$ $\endgroup$ – andre Oct 1 '14 at 9:42
  • $\begingroup$ @andre Thanks :) $\endgroup$ – user153582 Oct 1 '14 at 9:46

So here is my comment as an answer.

In order to be able to use your orthogonality relation for $P_k(\cos\theta)$, you may write your term as a linear combination of the $P_k(\cos\theta)$.


\begin{equation} \sin^2\theta = 1-\cos^2\theta = -\frac{2}{3} \cdot \left(\frac{3\cos^2\theta-1}{2}\right) + \frac{2}{3}\cdot 1 = -\frac{2}{3}P_2(\cos\theta) + \frac{2}{3}P_0(\cos\theta). \end{equation}


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