= Solution
Set $q=e^{i\theta}/2$. The required sine sum is the <imaginary part> of a finite <geometric series>:
$$
\sum_{n=1}^Nq^n=\frac{q(1-q^N)}{1-q}
=\frac{(q-q^{N+1})(1-\overline q)}{|1-q|^2}.
$$
The denominator is $(5-4\cos\theta)/4$. The numerator's <imaginary part> is
$$
\frac12\sin\theta-\frac{\sin((N+1)\theta)}{2^{N+1}}
+\frac{\sin(N\theta)}{2^{N+2}}.
$$
Dividing and multiplying numerator and denominator by $2^N$ proves
$$
\boxed{\sum_{n=1}^N\frac{\sin(n\theta)}{2^n}
=\frac{2^{N+1}\sin\theta+\sin(N\theta)-2\sin((N+1)\theta)}{2^N(5-4\cos\theta)}.}
$$
For real $\theta$, $5-4\cos\theta\ge1$, so there is no exceptional angle requiring a limiting interpretation. At $N=1$ the expression reduces to $\sin\theta/2$.
Back to article page
