# How to prove that $\pi \frac{e^{it\frac{\pi}{2}}-e^{it\frac{3\pi}{2}}}{1 - e^{2\pi it}} = \frac{\pi}{2\cos\left(\frac{\pi t}{2}\right)}$

How to prove that : $$\pi \frac{e^{it\frac{\pi}{2}}-e^{it\frac{3\pi}{2}}}{1 - e^{2\pi it}} = \frac{\pi}{2\cos\left(\frac{\pi t}{2}\right)}$$

I start with $$e^{it\frac{\pi}{2}}-e^{it\frac{3\pi}{2}} = e^{it\frac{\pi}{2}}-e^{it\frac{-\pi}{2}} = 2i \sin\left( \frac{\pi}{2} \right)$$

So $$\pi \frac{e^{it\frac{\pi}{2}}-e^{it\frac{3\pi}{2}}}{1 - e^{2\pi it}} = 2\pi i \frac{\sin\left( \frac{\pi}{2} \right)}{1 - e^{2\pi it}}$$

I would want to make something appear with $$\frac{2i}{1 - e^{2\pi it}}$$

with a de Moivre formula but I don't see how to do it :/

Setting $\displaystyle e^{\dfrac{i\pi t}2}=u\implies u^2=e^{i\pi t};u^3=e^{\dfrac{3i\pi t}2};u^4=e^{2i\pi t}$
$$\frac{e^{\dfrac{it\pi}2}-e^{\dfrac{3it\pi}2}}{1 - e^{2\pi it}} =\frac{u-u^3}{1-u^4}=\frac{u(1-u^2)}{(1-u^2)(1+u^2)}=\frac u{1+u^2}$$ if $\displaystyle1-u^2\ne0\iff u^2\ne1\iff e^{i\pi t}\ne1=e^{2m\pi i}\iff t\ne2m$ where $m$ is any integer
$$\text{Again, }\frac u{1+u^2}=\frac1{u+u^{-1}}$$
$\displaystyle e^{\dfrac{i\pi t}2}=u\iff u^{-1}=e^{-\dfrac{i\pi t}2}\implies u+u^{-1}=2\cos{\dfrac{\pi t}2}$ (using Euler Formula)
• A - is missing in $u^{-1}$ but this is exactly what I needed :) Thanks ! Commented Apr 15, 2014 at 14:48