So one of my past exam papers has the question:
Show:
$$\sum_\limits{n=0}^\infty \frac{\sin(n\theta)}{2^n} = \frac{\sin(\theta)}{5-4\cos(\theta)}$$
My working:
$e^{in\theta} = \cos(n\theta) + i\sin(n\theta)$
So $\sin(n\theta)$ is the imaginary part of $e^{in\theta}$
Thus we get: $$\Im\left(\sum_{n=0}^\infty \left(\frac{e^{i\theta}}{2}\right)^n\right)$$
Using the summation formula for infinite geometric series gives:
$$\frac{1}{1-\frac{e^{i\theta}}{2}}$$
And after multiplying by the complex conjugate I get the imaginary part to be
$$\frac{2\sin(\theta)}{5-4\cos(\theta)}$$
Any help would be great, thanks!
Best Answer
Testing... if $\theta=\frac{\pi}{2}$, the original series is $\frac1{2^1}-\frac1{2^3}+\frac1{2^5}-\frac1{2^7}+\cdots =\left(\frac12-\frac18\right)\left(1+\frac1{2^4}+\frac1{2^8}+\cdots\right)$ which becomes $\frac38\cdot\frac{16}{15}=\frac25$
The official answer claims $\frac{1}{5-4\cdot 0}=\frac15$. Yours claims $\frac{2}{5-4\cdot 0}=\frac25$. Your calculation is right and the official answer is wrong.