We know that period of $\cos(ax+b)$ assuming $a\neq0$ is $T=\frac{2\pi}{a}$ my question is why is it that $\cos^{n}(ax+b)$assuming $a\neq0$ has a period of $\displaystyle\frac{\pi}{a}$ if $n$ is odd and $\displaystyle\frac{2\pi}{a}$ if $n$ is even?
Update:
Well it seems we can express the powers of sine in terms of a binomial expansion: $$ \sin^{2m}(x)~=~\frac{(-1)^m}{2^n}\left[2^*\sum_{k=0}^m~\binom{n}{k}(-1)^k\cos((n-2k)x)\right] $$ $$ \sin^{2m+1}(x)~=~\frac{(-1)^m}{2^{n-1}}\left[\sum_{k=0}^m~\binom{n}{k}(-1)^k\sin((n-2k)x)\right] $$