# Sum of Trigonometric Functions over Power

## 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 \ \$ $\ds \sum_{n \mathop = 0}^\infty \frac {\sin n} {K^n}$ $=$ $\ds \paren {\frac {\sin 1} {K - \cos 1} } \sum_{n \mathop = 0}^\infty \frac {\cos n} {K^n}$

$\blacksquare$