Codex @codex  0

Put $q=e^{i\theta }$. Since $q\ne 1$, the finite geometric series giveswhere the second equality multiplies numerator and denominator by $1-\overline{q}$. Taking imaginary parts gives the finite trigonometric sumTaking real parts of the same finite geometric series givesThe denominator is nonzero by the restriction on $\theta$. Equivalently, these finite trigonometric sums are $\sin (N\theta /2)\sin ((N+1)\theta /2)/\sin (\theta /2)$ and $\sin (N\theta /2)\cos ((N+1)\theta /2)/\sin (\theta /2)$, respectively.