# Sum of Sines of Multiples of Angle/Proof 2

## Theorem

 $\displaystyle \sum_{k \mathop = 1}^n \sin k x$ $=$ $\displaystyle \sin x + \sin 2 x + \sin 3 x + \cdots + \sin n x$ $\displaystyle$ $=$ $\displaystyle \frac {\sin \frac {\paren {n + 1} x} 2 \sin \frac {n x} 2} {\sin \frac x 2}$

## Proof

Let $x$ be a real number that is not a integer multiple of $2 \pi$.

Let $k$ be a non-negative integer.

We have, from Euler's Formula:

$\map \exp {i k x} = i \sin k x + \cos k x$

Summing from $k = 0$ to $k = n$, we have:

$\displaystyle \sum_{k \mathop = 0}^n \map \exp {i k x} = i \sum_{k \mathop = 0}^n \sin k x + \sum_{k \mathop = 0}^n \cos k x$

As $\sin k x$ and $\cos k x$ are both real for real $k, x$, we have:

 $\displaystyle \sum_{k \mathop = 0}^n \sin k x$ $=$ $\displaystyle \map \Im {\sum_{k \mathop = 0}^n \map \exp {i k x} }$ $\displaystyle$ $=$ $\displaystyle \map \Im {\paren {i \sin \frac {n x} 2 + \cos \frac {n x} 2} \frac {\map \sin {\frac {\paren {n + 1} x} 2} } {\sin \frac x 2} }$ Sum of $\map \exp {i k x}$ $\displaystyle$ $=$ $\displaystyle \frac {\sin \frac {\paren {n + 1} x} 2 \sin \frac {n x} 2} {\sin \frac x 2}$

$\blacksquare$