Sum of Sines of Multiples of Angle

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Theorem

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

where $x$ is not an integer multiple of $2 \pi$.


Proof 1

By Simpson's Formula for Sine by Sine:

$2 \sin \alpha \sin \beta = \map \cos {\alpha - \beta} - \map \cos {\alpha + \beta}$

Thus we establish the following sequence of identities:

\(\displaystyle 2 \sin x \sin \frac x 2\) \(=\) \(\displaystyle \cos \frac x 2 - \cos \frac {3 x} 2\)
\(\displaystyle 2 \sin 2 x \sin \frac x 2\) \(=\) \(\displaystyle \cos \frac {3 x} 2 - \cos \frac {5 x} 2\)
\(\displaystyle \) \(\cdots\) \(\displaystyle \)
\(\displaystyle 2 \sin n x \sin \frac x 2\) \(=\) \(\displaystyle \cos \frac {\paren {2 n - 1} x} 2 - \cos \frac {\paren {2 n + 1} x} 2\)


Summing the above:

\(\displaystyle 2 \sin \frac x 2 \paren {\sum_{k \mathop = 1}^n \sin k x}\) \(=\) \(\displaystyle \cos \frac x 2 - \cos \frac {\paren {2 n + 1} x} 2\) Sums on right hand side form Telescoping Series
\(\displaystyle \) \(=\) \(\displaystyle -2 \map \sin {\dfrac {\frac x 2 + \frac {\paren {2 n + 1} x} 2} 2} \map \sin {\dfrac {\frac x 2 - \frac {\paren {2 n + 1} x} 2} 2}\) Prosthaphaeresis Formula for Cosine minus Cosine
\(\displaystyle \) \(=\) \(\displaystyle -2 \sin \dfrac {\paren {n + 1} x} 2 \sin \dfrac {-n x} 2\)
\(\displaystyle \) \(=\) \(\displaystyle 2 \sin \dfrac {\paren {n + 1} x} 2 \sin \dfrac {n x} 2\) Sine Function is Odd


The result follows by dividing both sides by $2 \sin \dfrac x 2$.


It is noted that when $x$ is a multiple of $2 \pi$ then:

$\sin \dfrac x 2 = 0$

leaving the right hand side undefined.

$\blacksquare$


Proof 2

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 $\map \exp {i k x}$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\sin \frac {\paren {n + 1} x} 2 \sin \frac {n x} 2} {\sin \frac x 2}\)

$\blacksquare$


Also see


Sources