Finite sums of the sine harmonics

For the nonperiodic case, assume $x\mathrm{\notin }2\pi \mathbb{Z}$ and define the auxiliary complex sums recursively by $S_0=0$ and $S_{j+1}=S_j+e^{i(j+1)x}$. Multiplication by $1-e^{ix}$ and the exponential addition law [L7] telescope directly to $(1-e^{ix})S_N=e^{ix}-e^{i(N+1)x}$. The half-angle identity $1-e^{ix}=-2i{e}^{ix\mathrm{/}2}\sin (x\mathrm{/}2)$ and [L3] show that the multiplier is nonzero.

For the periodic case, assume $x\in 2\pi \mathbb{Z}$. Then every $nx$ is a multiple of $2\pi$, so $\sin (nx)=0$ by [L3] and the finite sine sum is zero.

For the nonperiodic case, divide the identity in step 1.1 by its nonzero multiplier and use [L1] and [L2] to obtain $S_N=e^{i(N+1)x\mathrm{/}2}\frac{\sin (Nx\mathrm{/}2)}{\sin (x\mathrm{/}2)}\mathrm{.}$

For the nonperiodic case, take imaginary parts in step 2.1 and apply the product-to-sum consequence of [L2] to get ${\sum }_{n=1}^N\sin (nx)=\frac{\cos (x\mathrm{/}2)-\cos ((N+1\mathrm{/}2)x)}{2\sin (x\mathrm{/}2)}\mathrm{.}$

For the nonperiodic case, [L6] bounds the numerator in step 3.1 by $2$, so the absolute value of the sum is at most $1\mathrm{/}\mathrm{\mid }\sin (x\mathrm{/}2)\mathrm{\mid }$.

The nonperiodic branch gives the displayed formula and bound by steps 3.1 and 4.1, while the periodic branch gives the separate zero value by step 1.2; the two cases exhaust all real $x$.