Periods of Sine, Cosine and Tangent

Question

I would like to know how to calculate $(-1)^n$ and $(-1)^n(-1)^n$ being periodic and get an explanation for the functions sine, cosine, and tangent periods. I know that for a signal to be periodic there has to be such $T$, that satisfies: $$f(n)=f(n+T)$$ It may explain the reason why sine and cosine periods are $2\pi$, once $(−1)^n$ period is $2$, and why the tangent period is $\pi$, once $(−1)^n⋅(−1)^n$ has period $1$.

Answer 1

I know that for a signal to be periodic there has to be such T, that satisfies:

By that definition there may be multiple periods.

$\sin(n + 4,567,282\pi) = \sin n$ so $\sin n$ is periodic.

And $4,567,282\pi$ is a period of $\sin$ but in might not be the smallest.

Likewise if $f(n+T) = f(n)$ then $f(n+97T)=f(n)$ so $f$ is periodic but $97$ isn't the smallest period and, for all we know, $T$ might not be either.

Usually when saying a function is "periodic" we want to know what the smallest period is (which we usually say is "the" period).

If $T = 2k\pi$ then $\sin (n+T) = \sin n$ but the smallest such $T$ (greater than $0$) so that $\sin (n+T)=\sin n$ is $T= 2\pi$.

It is true that for any $n$ that $\sin (n + 2\pi) = \sin n$.

I won't prove this, but for any $T< 2\pi$ there will always be some $n$ where $\sin (n+T) \ne \sin n$. For example if $T = \pi$ then $\sin(n+\pi) =-\sin n$ and those aren't usually equal.

So why is $\tan$ different. Well, $\tan$ is a different function that $\sin$ so there is no reason it should be the same.

But notice: $\tan (n+\pi) = \frac {\sin (n + \pi)}{\cos (n+\pi)} = \frac {-\sin n}{-\cos n} = \frac {\sin n}{\cos n}$ so $\pi$ is a period, even though it isn't a period for $\sin$ or $\cos$.