Sum of Trigonometric Functions over Power

Theorems
A.

B.

C.

Proof
First, let:

$ \displaystyle \mathbf{A} = \sum_{n=0}^\infty \frac{\cos (n)}{K^n}$

and

$\displaystyle \mathbf{B} = \sum_{n=0}^\infty \frac{\sin (n)}{K^n}$

Now, consider $\mathbf{A} + i\cdot\mathbf{B}$, where i is the imaginary unit:

Now, real and imaginary parts must be equated to get $\mathbf{A}$ and $\mathbf{B}$, respectively; this can be done simply by reapplying Euler's formula, then multiplying by the conjugate to simplify right side.

And remembering that $\sin^2(x) + \cos^2(x) = 1$, the right side finally reduces to

Thus, it is clear that:

and:

Proving A and B, respectively. C is proved by finding the quotient of the two results: