How can I simplify this complex number to get a real number? $$\large \frac {e^{i \frac{\pi a}{2}}[1-e^{i\pi a}]} {[1-e^{i2\pi a}]}$$
I am trying to arrive at $$\large \frac {1}{2\cos\left(\frac{\pi a}{2}\right)}$$
I've tried dividing top and bottom by one of the exponential terms, and also tried expanding out the Euler formulas for each exponential term to see if I get some cancellations and an obvious representation for cosine, but I have not been able to do it
Any hints or solutions are welcome.
Thanks!
 A: Note that $$\frac{1-e^{i\pi a}}{1-e^{i2\pi a}}=\frac{1}{1+e^{i\pi a}}$$ 
So $$e^{\frac{i\pi a}{2}}\frac{1-e^{i\pi a}}{1-e^{i2\pi a}}=\frac{e^{\frac{i\pi a}{2}}}{1+e^{i\pi a}}$$$$=\frac{1}{e^{\frac{-i\pi a}{2}}+e^{\frac{i\pi a}{2}}}$$$$=\frac{1}{2\cos\left(\frac{\pi a}{2}\right)}$$
A: This involves a little judicious factoring of the numerator and denominator.  so, here we go   ...
$$\begin{align}
\frac{e^{i\pi a/2}(1-e^{i\pi a})}{1-e^{i2\pi a}}&=\frac{e^{i\pi a/2}e^{i\pi a/2}(e^{-i\pi a/2}-e^{i\pi a/2})}{e^{i\pi a}(e^{-i\pi a}-e^{i\pi a})}\\\\
&=\frac{-2i \sin(\pi a/2)}{-2i \sin(\pi a)}\\\\
&=\frac{ \sin(\pi a/2)}{2 \sin(\pi a/2)\cos(\pi a/2)}\\\\
&=\frac{1}{2\cos(\pi a/2)}
\end{align}$$
A: $$e^{2ix}-1=e^{ix}(e^{ix}-e^{-ix})=e^{ix}2i\sin x$$
Set $x=\pi a, 2\pi a$ to get  
$$\large \frac {e^{i \frac{\pi a}{2}}[1-e^{i\pi a}]} {[1-e^{i2\pi a}]}=\dfrac{\sin\pi a}{\sin2\pi a}$$
Now, $\sin2\pi a=2\sin\pi a\cdot\cos\pi a$
A: Notice, $$\frac{e^{i\frac{\pi\alpha}{2}}(1-e^{i\pi\alpha})}{1-e^{i2\pi\alpha}}$$
$$=\frac{e^{\frac{i\pi\alpha}{2}}(1-e^{i\pi\alpha})}{1-(e^{i\pi\alpha})^2}$$
$$=\frac{e^{\frac{i\pi\alpha}{2}}(1-e^{i\pi\alpha})}{(1-e^{i\pi\alpha})(1+e^{i\pi\alpha})}$$
$$=\frac{e^{\frac{i\pi\alpha}{2}}}{(1+e^{i\pi\alpha})}$$
$$=\frac{1}{e^{\frac{-i\pi\alpha}{2}}(1+e^{i\pi\alpha})}$$
$$=\frac{1}{e^{\frac{-i\pi\alpha}{2}}+e^{\frac{i\pi\alpha}{2}}}$$
$$=\frac{1}{2\left(\frac{e^{\frac{i\pi\alpha}{2}}+e^{\frac{-i\pi\alpha}{2}}}{2}\right)}$$
$$=\frac{1}{2\cos\left(\frac{\pi\alpha}{2}\right)}$$
A: Recall the identity: $a^2-b^2=(a+b)(a-b)$
Now, $$e^{\frac{i\pi a}{2}}\left(\frac{1-e^{i\pi a}}{1-e^{i2\pi a}}\right)=e^{\frac{i\pi a}{2}}\left(\frac{1-e^{i\pi a}}{(1+e^{i\pi a})(1-e^{i\pi a})}\right)$$
$$=\frac{e^{\frac{i\pi a}{2}}}{1+e^{i\pi a}}$$
$$=\frac{1}{e^{\frac{-i\pi a}{2}}+e^{\frac{-i\pi a}{2}}\cdot e^{i\pi a}}$$
$$=\frac{1}{e^{\frac{-i\pi a}{2}}+e^{\frac{i\pi a}{2}}}$$
$$=\frac{1}{2\left(\frac{e^{\frac{-i\pi a}{2}}+e^{\frac{i\pi a}{2}}}{2}\right)}$$
$$=\frac{1}{2\cos\left(\frac{\pi a}{2}\right)}$$

proved

