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Prove a formula for the sum from k=1 to n of sin kx

by RoRi

Prove

    \[ \sum_{k=1}^n \sin kx = \frac{\sin \frac{nx}{2} \sin \frac{(n+1)x}{2}}{\sin \frac{x}{2} } \]

for x \neq 2 m \pi, \ \ m \in \mathbb{Z}.

Proof. First, we recall from this exercise

    \begin{align*} - 2 \sin \frac{x}{2} \sin (kx) &= \cos \left( \frac{x}{2} + kx \right) - \cos \left( \frac{x}{2} - kx \right) \\ &= \cos \left( (2k+1) \frac{x}{2} \right) - \cos \left( (1-2k) \frac{x}{2} \right) \\ &= \cos \left( (2k+1) \frac{x}{2} \right) - \cos \left( (2k-1) \frac{x}{2} \right). \end{align*}

Adding, with the right side telescoping, as in the previous exercise, we have

    \begin{align*} \sum_{k=1}^n \sin kx &= \frac{ \cos \left(\left(n + \frac{1}{2}\right)x\right) - \cos \frac{x}{2}}{-2 \sin \frac{x}{2}} \\ &= \frac{ \cos (nx) \cos \frac{x}{2} - \sin (nx) \sin \frac{x}{2} - \cos \frac{x}{2}}{-2 \sin \frac{x}{2} } \\ &= \frac{ \cos \frac{x}{2} (\cos (nx) -1) - \sin (nx) \sin \frac{x}{2}}{-2 \sin \frac{x}{2} } \\ &= \frac{-2 \cos \frac{x}{2} \sin^2 \frac{nx}{2} - 2 \sin \frac{nx}{2} \cos \frac{nx}{2} \sin \frac{x}{2}}{-2 \sin \frac{x}{2}} \\ &= \frac{\sin \frac{nx}{2} \left( \cos \frac{x}{2} \sin \frac{nx}{2} + \cos \frac{nx}{2} \sin \frac{x}{2} \right)}{\sin \frac{x}{2} } \\ &= \frac{\sin \frac{nx}{2} \sin \frac{(n+1)x}{2}}{\sin \frac{x}{2}}. \qquad \blacksquare \end{align*}

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