My final answer is then:
for M odd
\begin{equation}
P_x=\frac{1}{M}\sum_{n=0}^{M-1}{|Acos(\frac{2\pi n}{M}+\phi)|}^2=|A|^2cos^2(\phi)
\end{equation}
for M even
\begin{equation}
P_x=\frac{1}{M}\sum_{n=0}^{M-1}{|Acos(\frac{2\pi n}{M}+\phi)|}^2=|A|^2cos^2(\phi)+\frac{|A|^2}{2}...
Done similarly,
\begin{equation}
(3) \frac{|A|^2}{M}\sum_{n=0}^{M-1}|-2sin(\alpha)sin(\phi)cos(\alpha)cos(\phi)|
\end{equation}
\begin{equation}
\frac{|A|^2}{M}sin(\phi)cos(\phi)\sum_{n=0}^{M-1}|-2sin(\alpha)cos(\alpha)|
\end{equation}
using the same polar defintiions of sine and...
I'm not 100% sure that I have done this part correctly, my result is a two-case answer, where the period, M, is either and even or odd integer,
\begin{equation}
(2) \frac{|A|^2}{M}\sum_{n=0}^{M-1}|sin^2(\alpha)|
\end{equation}
The next term is somewhat simplified, since I already know what...
(I didn't mention earlier that M is the period of the sinusoid, and the average power equation sums the sampled terms during one period only, thus M=N)
Here is what I have so far:
\begin{equation}
x[n]=Acos(\frac{2\pi n}{M}+\phi)
\end{equation}
and
\begin{equation}...
I think I may have accidentally caused some confusion with the title of the thread. Did my tex/mathjax script load properly on your browser? I may want to clarify for what I am looking before going too much further..
(id est, did I fail to type some sort of script bookend to make my equations...