Factorization $\cos\left(\tfrac{\pi z}{4}\right)-\sin\left(\tfrac{\pi z}{4}\right)$ 
Prove that 
  $$\cos\left(\frac{\pi z}{4}\right)-\sin\left(\frac{\pi z}{4}\right)=\prod_{n=1}^\infty\left(1+\frac{(-1)^nz}{2n-1}\right)$$

My try:
$$\cos\left(\frac{\pi z}{4}\right)-\sin\left(\frac{\pi z}{4}\right)=\cos\left(\frac{\pi z}{4}\right)-\cos\left(\frac{\pi}{2}-\frac{\pi z}{4}\right)=-\sqrt{2}\sin\left(\frac{\pi z-\pi}{4}\right) $$
Now, substitute the expression of the $\sin(\pi z)$ and get
$$\cos\left(\frac{\pi z}{4}\right)-\sin\left(\frac{\pi z}{4}\right)=-\sqrt{2}\pi\left(\frac{z-1}{4}\right)\prod_{n=1}^\infty\left(1-\frac{(z-1)^2}{16n^2}\right).$$
How should I proceed now?
 A: $\displaystyle \prod_{n=1}^\infty\left(1+\frac{(-1)^nz}{2n-1}\right) =\prod_{n=0}^\infty\left(1-\frac{z}{4n+1}\right)\prod_{n=1}^\infty\left(1+\frac{z}{4n-1}\right)=$
$\displaystyle =(1-z)\prod_{n=1}^\infty\left(\frac{4n(1-\frac{z-1}{4n})}{4n(1+\frac{1}{4n})}\right)\prod_{n=1}^\infty\left(\frac{4n(1+\frac{z-1}{4n})}{4n(1-\frac{1}{4n})}\right)= (1-z)\frac{\prod_{n=1}^\infty\left(1-(\frac{z-1}{4n})^2\right)}{\prod_{n=1}^\infty\left(1-(\frac{1}{4n})^2\right)}$
$\displaystyle =(1-z)\frac{\sin(\frac{z-1}{4}\pi)}{\frac{z-1}{4}\pi}\frac{\frac{1}{4}\pi}{\sin(\frac{1}{4}\pi)}=-\sqrt{2}\sin\left(\frac{\pi z-\pi}{4}\right)= -2\sin\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi z-\pi}{4}\right)$
$\displaystyle = \cos\left(\frac{\pi z-\pi}{4}+\frac{\pi}{4}\right) - \cos\left(\frac{\pi z-\pi}{4} -\frac{\pi}{4}\right) = \cos\left(\frac{\pi z}{4}\right) - \sin\left(\frac{\pi z}{4}\right)$
because of $\enspace\displaystyle \sin x\sin y=-\frac{1}{2}\left(\cos(x+y)-\cos(x-y)\right)$
Note:
The decisive step was $\enspace\displaystyle \frac{1}{\sin(\frac{1}{4}\pi)}=2\sin\left(\frac{\pi}{4}\right)\,$ .
