Let $y$ be the function defined by $$y(\theta)=2sin\frac{\theta}{2}\prod_{k=1}^{\infty}\frac{(1-e^{i\theta}q^k)(1-e^{-i\theta}q^k)}{(1-q^k)^{2}}$$ where $q = e^{2\pi i\tau}$
Show that $y$ has simple zeroes at each $\theta=2\pi m+2\pi n \tau$ for all integers $m$ and $n$.
Prove that
$y(\theta+2\pi \tau)=-e^{i \theta}q^{-1/2}y(\theta)$
$y(2\theta+4\pi \tau)=e^{4i \theta}q^{-2}y(2\theta)$
We have
\begin{align}\frac{y(\theta + 2\pi\tau)}{y(\theta)} &= \frac{\sin\left(\frac{\theta}{2} + \pi \tau\right)}{\sin \frac{\theta}{2}} \prod_{k = 1}^\infty \frac{(1 - e^{i\theta}q^{k+1})(1 - e^{-i\theta} q^{k-1})}{(1 - e^{i\theta}q^k)(1 - e^{-i\theta}q^k)}\\ &= \frac{\sin\left(\frac{\theta}{2} + 2\pi \tau\right)}{\sin \frac{\theta}{2}}\frac{1 - e^{-i\theta}}{1 - e^{i\theta}q}\\ &= \frac{\sin\left(\frac{\theta}{2} + 2\pi \tau\right)}{\sin \frac{\theta}{2}}\frac{e^{-i\theta/2}}{e^{i\theta/2}q^{1/2}} \frac{e^{i\theta/2} - e^{-i\theta/2}}{e^{-i\theta/2+\pi\tau} - e^{i\theta/2 + \pi \tau}}\\ &= \frac{\sin\left(\frac{\theta}{2} + 2\pi \tau\right)}{\sin \frac{\theta}{2}} e^{-i\theta}q^{-1/2}\left[-\frac{\sin \frac{\theta}{2}}{\sin\left(\frac{\theta}{2} + \pi \tau\right)}\right]\\ &= -e^{-i\theta}q^{-1/2}. \end{align}
Thus $y(\theta + 2\pi \tau) = -e^{-i\theta}q^{-1/2}y(\theta)$.
Using the above identity, we obtain
\begin{align}y(2\theta + 4\pi \tau) &= y((2\theta + 2\pi \tau) + 2\pi \tau)\\ &= -e^{-i(2\theta + 2\pi \tau)}q^{-1/2}y(2\theta + 2\pi \tau)\\ &= -e^{-i(2\theta + 2\pi \tau)}q^{-1/2}[-e^{-i(2\theta)}q^{-1/2}]y(2\theta)\\ &= e^{-4i\theta}q^{-1 - 1/2 - 1/2}y(2\theta)\\ &= e^{-4i\theta}q^{-2}y(2\theta). \end{align}