Using De Moivre's Theorem for summation 
I am able to get the required denominator but I keep ending with 4 terms in numerator. Can somebody help me out in this question?( I am using geometric sum identity) 
 A: $$\Im\sum_{n=1}^{10} \left(\frac{1}{2}e^{i\pi/10}\right)^n=\Im\left(\frac{1}{2}e^{i\pi/10}\frac{1-(\frac{1}{2}e^{i\pi/10})^{10}}{1-\frac{1}{2}e^{i\pi/10}}\right)$$
$$=\Im\left(\frac{1}{2}e^{i\pi/10}\frac{1-\frac{1}{1024}e^{i\pi}}{1-\frac{1}{2}e^{i\pi/10}}\right)$$
$$=\Im\left(\frac{1}{2}e^{i\pi/10}\frac{1+\frac{1}{1024}}{1-\frac{1}{2}e^{i\pi/10}}\right)$$
$$=\frac{1025}{1024}\Im\left(\frac{\frac{1}{2}e^{i\pi/10}}{1-\frac{1}{2}e^{i\pi/10}}\right)$$
$$=\frac{1025}{1024}\Im\left(\frac{\frac{1}{2}e^{i\pi/10}}{1-\frac{1}{2}e^{i\pi/10}}\frac{1-\frac{1}{2}e^{-i\pi/10}}{1-\frac{1}{2}e^{-i\pi/10}}\right)$$
$$=\frac{1025}{1024}\Im\left(\frac{\frac{1}{2}e^{i\pi/10}-\frac{1}{4}}{\frac{5}{4}-\frac{1}{2}(2\cos\frac{\pi}{10})}\right)$$
$$=\frac{1025}{1024}\frac{\frac{1}{2}\sin\frac{\pi}{10}}{\frac{5}{4}-\frac{1}{2}(2\cos\frac{\pi}{10})}$$
$$=\frac{1025\sin\frac{\pi}{10}}{2560-2048\cos\frac{\pi}{10}}$$
A: You can write your sum as imaginary part of $S=\sum_{n=1}^{10} \left(e^{i\pi/10}/2\right)^n$. This gives :
$$S=\frac{e^{i\pi/10}}{2}\frac{1-\frac{e^{i\pi}}{2^{10}}}{1-\frac{e^{i\pi/10}}{2}}=\frac{1025}{1024}\frac{e^{i\pi/10}}{2-e^{i\pi/10}} = \frac{1025}{1024}\frac{2e^{i\pi/10}-1}{5-4\cos(\pi/10)}$$
Now 
$${\rm Im}(S) = \frac{1025\sin(\pi/10)}{512(5-4\cos(\pi/10))}$$
