Sum of Cosines of Multiples of Angle

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Theorem

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

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


Proof

By the Simpson's Formula for Cosine by Sine:

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

Thus we establish the following sequence of identities:

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


Summing the above:

$\ds 2 \sin \frac x 2 \paren {\frac 1 2 + \sum_{k \mathop = 1}^n \map \cos {k x} } = \sin \frac {\paren {2 n + 1} x} 2$

as the sums on the right hand side form a telescoping series.

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.

$\blacksquare$


Also see


Sources

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