Sum of Cosines of Multiples of Angle

From ProofWiki
Jump to navigation Jump to search

Theorem

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

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


Proof

By the Simpson's Formula for Cosine by Sine:

$2 \cos \alpha \sin \beta = \sin \left({\alpha + \beta}\right) - \sin \left({\alpha - \beta}\right)$

Thus we establish the following sequence of identities:

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


Summing the above:

$\displaystyle 2 \sin \frac x 2 \left({\frac 1 2 + \sum_{k \mathop = 1}^n \cos \left({k x}\right)}\right) = \sin \frac {\left({2 n + 1}\right) x} 2$

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