Sum of Trigonometric Functions over Power

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Theorem

\(\text {(1)}: \quad\) \(\ds \sum_{n \mathop = 0}^\infty \frac {\cos n} {K^n}\) \(=\) \(\ds \frac {K \paren {K - \cos 1} } {K^2 - 2 K \cos 1 + 1}\)
\(\text {(2)}: \quad\) \(\ds \sum_{n \mathop = 0}^\infty \frac {\sin n} {K^n}\) \(=\) \(\ds \frac {K \sin 1} {K^2 - 2 K \cos 1 + 1}\)
\(\text {(3)}: \quad\) \(\ds \sum_{n \mathop = 0}^\infty \frac {\sin n} {K^n}\) \(=\) \(\ds \paren {\frac {\sin 1} {K - \cos 1} } \sum_{n \mathop = 0}^\infty \frac {\cos n} {K^n}\)


Proof

First, let:

\(\ds \mathbf A\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {\cos n} {K^n}\)
\(\ds \mathbf B\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {\sin n} {K^n}\)

Now, consider:

$\mathbf A + i \, \mathbf B$

where $i$ is the imaginary unit:

\(\ds \mathbf{A} + i \, \mathbf B\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {\cos n} {K^n} + \sum_{n \mathop = 0}^\infty \frac {i \sin n} {K^n}\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {\cos n + i \sin n} {K^n}\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {e^{i n} } {K^n}\) Euler's Formula
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \paren {\frac {e^i} K}^n\)
\(\ds \) \(=\) \(\ds \frac K {K - e^i}\) Sum of Infinite Geometric Sequence
The plan now is to equate real and imaginary parts of this to re-obtain $\mathbf A$ and $\mathbf B$:
\(\ds \) \(=\) \(\ds \frac K {K - \cos 1 - i \sin 1}\) Euler's Formula
\(\ds \) \(=\) \(\ds \frac K {K - \cos 1 - i \sin 1} \cdot \frac {K - \cos 1 + i \sin 1} {K - \cos 1 + i \sin 1}\) multiplying by complex conjugate of denominator
\(\ds \) \(=\) \(\ds \frac {K \paren {K - \cos 1 + i \sin 1} } {K^2 - 2 K \cos 1 + \cos^2 1 + \sin^2 1}\)
\(\ds \) \(=\) \(\ds \frac {K \paren {K - \cos 1 + i \sin 1} } {K^2 - 2 K \cos 1 + 1}\) Sum of Squares of Sine and Cosine
\(\ds \) \(=\) \(\ds \paren {\frac {K \paren {K - \cos 1} } {K^2 - 2 K \cos 1 + 1} } + i \paren {\frac {K \sin 1} {K^2 - 2 K \cos 1 + 1} }\)


So equating real and imaginary parts:

\(\ds \mathbf A = \sum_{n \mathop = 0}^\infty \frac {\cos n} {K^n}\) \(=\) \(\ds \frac {K \paren {K - \cos 1} } {K^2 - 2 K \cos 1 + 1}\)
\(\ds \mathbf B = \sum_{n \mathop = 0}^\infty \frac {\sin n} {K^n}\) \(=\) \(\ds \frac {K \sin 1} {K^2 - 2 K \cos 1 + 1}\)

This proves $(1)$ and $(2)$ respectively.


$(3)$ is proved by finding the quotient of the two results:

\(\ds \frac {\sum_{n \mathop = 0}^\infty \frac {\sin n} {K^n} } {\sum_{n \mathop = 0}^\infty \frac {\cos n} {K^n} }\) \(=\) \(\ds \frac {K \sin 1} {K^2 - 2 K \cos 1 + 1} \cdot \frac {K^2 - 2 K \cos 1 + 1} {K \paren {K - \cos 1} } = \frac {\sin 1} {K - \cos 1}\)
\(\ds \leadsto \ \ \) \(\ds \sum_{n \mathop = 0}^\infty \frac {\sin n} {K^n}\) \(=\) \(\ds \paren {\frac {\sin 1} {K - \cos 1} } \sum_{n \mathop = 0}^\infty \frac {\cos n} {K^n}\)

$\blacksquare$

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