Sum of Cosines of Arithmetic Sequence of Angles/Formulation 1

Theorem

Let $\alpha \in \R$ be a real number such that $\alpha \ne 2 \pi k$ for $k \in \Z$.

Then:

\(\ds \sum_{k \mathop = 0}^n \map \cos {\theta + k \alpha}\)

\(=\)

\(\ds \cos \theta + \map \cos {\theta + \alpha} + \map \cos {\theta + 2 \alpha} + \map \cos {\theta + 3 \alpha} + \dotsb\)

\(\ds \)

\(=\)

\(\ds \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} } \map \cos {\theta + \frac {n \alpha} 2}\)

Proof

From Sum of Complex Exponentials of i times Arithmetic Sequence of Angles: Formulation 1:

$\ds \sum_{k \mathop = 0}^n e^{i \paren {\theta + k \alpha} } = \paren {\map \cos {\theta + \frac {n \alpha} 2} + i \map \sin {\theta + \frac {n \alpha} 2} } \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} }$

It is noted that, from Sine of Multiple of Pi, when $\alpha = 2 \pi k$ for $k \in \Z$, $\map \sin {\alpha / 2} = 0$ and the right hand side is not defined.

From Euler's Formula, this can be expressed as:

$\ds \sum_{k \mathop = 0}^n \paren {\map \cos {\theta + k \alpha} + i \map \sin {\theta + k \alpha} } = \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} } \paren {\map \cos {\theta + \frac {n \alpha} 2} + i \map \sin {\theta + \frac {n \alpha} 2} }$

Equating real parts:

$\ds \sum_{k \mathop = 0}^n \map \cos {\theta + k \alpha} = \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} } \map \cos {\theta + \frac {n \alpha} 2}$

$\blacksquare$

Sources

• 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $143 \ \text{(a)}$