Even Binomial Expansion For a given integer $m \ge 1$, consider the sum:
$$ \frac {1}{2}[(x+y)^{2m-1}-(x-y)^{2m-1}] \tag1$$
This sum represents the terms of  $(x+y)^{2m} $ in which  the powers of $x $ are even and the powers of  $y $ are odd. For example, for $m=1$, (1) becomes $y $, for $m=2$ it becomes $3x^{2}y+y^{3} $ and so on. Thus, in practice, for every  $m $ we chose, (1) will always depend on $x^{2} $ to some power.
My question is: is there a way of writting (1) in such a way where the dependence is explicitly on $x^{2} $ rather than $x $? 
 A: For $m>1$, the expression is equal to 
$$(2m-1)y\prod_{k=1}^{m-1}\left ( x^2+y^2\cot^2 \left ( \frac{k \pi}{2m-1} \right )\right ).$$
To prove this, note that after dividing by $y^{2m-1}$, we can assume $y=1$. Thus, we need to show that
$$(x+1)^{2m-1}-(x-1)^{2m-1}=2(2m-1) \prod_{k=1}^{m-1}(x^2+\cot^2 (k\pi/(2m-1))).$$
Since both sides are polynomials of degree $2m-2$ and have equal leading coefficients, it is sufficient to show both sides have the same roots. Let $x$ be a root of the polynomial on the left side. Therefore $(x+1)/(x-1)=\eta \neq 1$ is a root of unity with $\eta^{2m-1}=1$. We can choose a nonzero integer $-m< k<m$ such that $\eta=e^{2k \pi i/(2m-1)}=\cos(2k \pi/(2m-1))+i \sin(2k \pi/(2m-1))$. 
It follows that
$$\frac{x+1}{x-1}=\eta \Rightarrow x= \frac{\eta+1}{\eta-1}=\frac{\bar \eta - \eta}{2-\eta -\bar \eta}=\frac{-2 i \sin(2k \pi/(2m-1))}{2-2\cos(2k \pi /(2m-1))}=-i \cot(k \pi / (2m-1)).$$
The claim then follows form the fact that for $0<k<m$, we have $(x+i \cot(k\pi/(2m-1))(x+i \cot (-k\pi/(2m-1))=(x^2+\cot^2(k\pi/(2m-1))$.
