Evaluating $\sum_{n=1}^{N} 2^{-n}\sin(n\theta)$

Question

Use de Moivre's theorem to find $$\sum_{n=1}^{N} 2^{-n}\sin(n\theta)$$

How to find sum of imaginary parts of geometric progression, or use exponential form of complex numbers to find the summation in terms of N, sine, and cosine??

Answer 1

This sum is the imaginary part of $\;\displaystyle\smash{ \sum_{n=0}^N\biggl(\frac{\mathrm e^{i\theta}}2\biggr)^{\mkern-5mu n}} $. Now use the formula for the sum of a geometric series: \begin{align} \sum_{n=0}^N\biggl(\frac{\mathrm e^{i\theta}}2\biggr)^{\mkern-5mu n}&=\frac{1-\biggl(\cfrac{\mathrm e^{i\theta}}2\biggr)^{\mkern-5mu N+1}}{1-\cfrac{\mathrm e^{i\theta}}2}=\frac{2^{N+1}-\mathrm e^{i(N+1)\theta}}{2^N(2-\mathrm e^{i\theta})}\\ &=\frac{\bigl(2^{N+1}-\mathrm e^{i(N+1)\theta}\bigr)\bigl(2-\mathrm e^{-i\theta}\bigr)}{2^N(2-\mathrm e^{i\theta})(2-\mathrm e^{-i\theta})}=\frac{2^{N+2}-2\mkern1.5mu\mathrm e^{i(N+1)\theta+}-2^{N+1}\mathrm e^{-i\theta}+\mathrm e^{iN\theta}}{2^N(4-4\cos\theta+1)}\\[1ex] &=\frac{2^{N+2}-2\mkern1.5mu\mathrm e^{i(N+1)\theta+}-2^{N+1}\mathrm e^{-i\theta}+\mathrm e^{iN\theta}}{2^N(5-4\cos\theta)}\\ \text{so the imaginary part is}\\[1ex] &\phantom{={}}\frac{2^{N+1}\sin\theta+\sin N\theta-2\sin(N+1)\theta}{2^N(5-4\cos\theta)}. \end{align}