# Periods of Sine, Cosine and Tangent

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 $$1$$.

• Why are you having questions. $\sin(n + 2\pi) = \sin n$ so $2\pi$ is a period of $\sin$ (but it might not be the smallest one) and $(-1)^{n+2} = (-1)^n$ so $2$ is a period of $f(n)= (-1)^n$. .... so what's your question? – fleablood Apr 24 '20 at 3:36

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 $$\cos$$.