Algebraically find the fundamental period of a $\cos^2(2\pi t)$?

Question

How do I find algebraically the fundamental period $T$ of $\cos^2(2\pi t) $? I understand that the condition for periodicity is $x(t) = x(t+T)$ and that $\ \omega=2\pi f ={2 \pi \over T} $ but I don't know how to get the period of the sinusoid without graphing it.

Answer 1

I ended up realizing that the easiest way to go about it would be to convert the $ cos^2(2\pi t) $ into a term without a square via the double angle formulas, so that it would turn into $ 1 \over 2 $ * $ cos(4\pi t) $ and some constant. Using the angular frequency formula above, the period then is 0.5s.

Answer 2

Hint:

from $$ \cos^2(x+T)=\cos^2 x $$

we have:

$$ \cos (x+T)= \cos x \quad \lor \quad \cos (x+T)=-\cos x $$

so $T=k\pi$. In your equation : $x=2\pi t$ and you want ( for $k=1$) $x+\pi=2\pi(t+\tau)$ where $\tau$ is the perion ( in variable $t$). Substitute and find $\tau$.