extracting the middle term of $ (z \cos \theta + w\sin \theta )^m(- z\sin \theta + w\cos \theta )^m $

Question

Is there a systematic way to extract the middle term of the following expression?

$$ (z \cos \theta + w\sin \theta )^m(- z\sin \theta + w\cos \theta )^m $$

This is homogeneous polynomial of degree $2m$, so I am looking for the $(zw)^m$ term.

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Example $m = 2$:

$$ (z \cos \theta + w\sin \theta )^2(- z\sin \theta + w\sin \theta )^2 = \dots + (zw)^2(\cos^4 \theta - 4 \cos^2 \theta \sin^2 \theta + \sin^4 \theta) + \dots $$

The middle terms can be simplified using the cosine double angle formulas.

\begin{eqnarray}\cos^4 \theta - 4 \cos^2 \theta \sin^2 \theta + \sin^4 \theta &=& (\cos^2 \theta - \sin^2 \theta)^2 - 2 \cos^2 \theta \sin^2 \theta \\ &=& \cos^2 2\theta - \tfrac{1}{2}\sin^2 2\theta \\ &=& \boxed{\tfrac{3}{2}\cos^2 2\theta - \tfrac{1}{2}} \end{eqnarray}

Is it true this middle term can always be expanded as a polynomial in $x=\cos 2\theta$?

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I learned of this trick while studying the Wigner D-matrices. It says the Legendre polynomials are certain matrix elements in the representation theory of $SU(2)$. Checking this fact was surprisingly hard.

Answer 1

We use the binomial theorem, to see that $$ (z\cos\theta+w\sin\theta)^m = \sum_{k=0}^m {m\choose k}z^{m-k}\cos^{m-k}\theta \,w^k\sin^k\theta $$ and $$ (-z\sin\theta+w\cos\theta)^m=\sum_{k=0}^m {m\choose k}(-z)^k\sin^k\theta\,w^{m-k}\cos^{m-k}\theta. $$ Written this way, I think it is clear that when we multiply these factors together, the coefficient in front of $z^mw^m$ becomes $$ \sum_{k=0}^m (-1)^k{m\choose k}^2 (\cos^{2}\theta)^{m-k}(\sin^2\theta)^k. $$ Next, since $\cos^2\theta=\frac{1}{2}(1+\cos2\theta)$ and $\sin^2\theta=\frac{1}{2}(1-\cos2\theta)$ this simplifies into $$ \sum_{k=0}^m (-1)^k{m\choose k}^2 \bigl[\tfrac{1}{2}(1+\cos2\theta)\bigr]^{m-k}\bigl[\tfrac{1}{2}(1-\cos2\theta)\bigr]^k, $$ a polynomial in $\cos2\theta$.

A check of the case $m=2$

In this case, the formula gives $$ \begin{align} &(-1)^0{2\choose 0}^2\bigl[\tfrac{1}{2}(1+\cos2\theta)\bigr]^2\\ &+(-1)^1{2\choose 1}^2\tfrac{1}{2}(1+\cos2\theta)\tfrac{1}{2}(1-\cos2\theta)\\ &+(-1)^2{2\choose 2}^2\bigl[\tfrac{1}{2}(1-\cos2\theta)\bigr]^2\\ &=\frac{3}{2}\cos^22\theta-\frac{1}{2}. \end{align} $$

Answer 2

Continuing mickep's answer, these are, indeed, the Legendre Polynomials $$ \sum_{k=0}^m (-1)^k{m\choose k}^2 \bigl[\tfrac{1}{2}(1+\cos2\theta)\bigr]^{m-k}\bigl[\tfrac{1}{2}(1-\cos2\theta)\bigr]^k = P_m(\cos(2\theta)) $$