I thought I had my head around rotation

$\sin\left(\frac{x}{2}\right)$ has time period $4\pi$,

$\sin(x)$ has time period $2\pi$,

$\sin(2x)$ has time period $\pi$

Yet they all repeat after $2\pi$; so all have a time period that covers $2\pi$ radians on a unit circle but have different time periods above?


3 Answers 3


If you introduce an intermediate function $w=bx$ then things may become clearer. We have

$y = \sin(w) = \sin(bx)$

In the $w$ domain the function $y(w)$ repeats with a period of $2\pi$. This period in the $w$ domain is the same regardless of the value of $b$.

But in the $x$ domain the function $y(x)$ repeats with a period of $\frac{2\pi}{b}$. So the period in the $x$ domain does depend on $b$.


Recall the general form of a sine function.

$$y = a\sin[b(x-h)]+k$$

Of course, the only part you’re asking about is $b$, so the rest can be ignored.

$$y = \sin b(x)$$

$b$ shows frequency, or the number of cycles for $0 \leq x \leq 2\pi$.

The period, or wavelength, is given by $\frac{2\pi}{b}$.

$\sin\big(\frac{x}{2}\big)$ will make half a cycle from $0$ to $2\pi$, meaning it has a period of $4\pi$.

$\sin(x)$ will make one full cycle from $0$ to $2\pi$, meaning it has a period of $2\pi$.

$\sin(2x)$ will make two cycles from $0$ to $2\pi$, meaning it has a period of $\pi$.

$\sin\big(\frac{x}{2}\big)$, $\sin(x)$, and $\sin(2x)$ ALL pass through $2\pi$, but the number of cycles in this range of domain will differ. (You seem to have understood the second part, but you’re confused over the first, which is completely irrelevant.)

Maybe checking the graphs could help clarify.


  • $\begingroup$ I get this. not sure it answers my question. I am looking at text that says a period represents 2*pi radians. When you wrap a signal around a unit circle they all complete a revolution in 2 *pi radians. But all have different time periods!?!? $\endgroup$ Nov 2, 2018 at 13:32
  • $\begingroup$ A period is the length of a full cycle - wavelength, you could say. Not sure what “a period represents $2\pi$ radians” means. $\endgroup$
    – KM101
    Nov 2, 2018 at 13:33
  • $\begingroup$ No, they do not necessarily complete a full cycle in $2\pi$ radians. That depends completely on the value of $b$. $\endgroup$
    – KM101
    Nov 2, 2018 at 13:37
  • $\begingroup$ For $\sin(x)$, one full cycle occurs every $\pi$ radians, for $\sin(\frac{x}{2})$, one full cycle occurs every $4\pi$, and for $\sin(2x)$, one full cycle occurs every $\pi$ radians. The process repeats infinitely in both directions. I think you’re confused about the $2\pi n$ concept, where $\sin(x)$ repeats every $2\pi$ radians. Not all sine functions repeat every $2\pi$ radians. That depends on the period. Maybe I’m just missing your question. $\endgroup$
    – KM101
    Nov 2, 2018 at 13:42
  • $\begingroup$ It means that $b$ cycles will cover $2\pi$ radians, which is true. $P = \frac{2\pi}{b} \implies 2\pi = P\cdot b$ $\endgroup$
    – KM101
    Nov 2, 2018 at 13:47

If $x=\pi,$ then $\sin\left(\frac x2\right) = 1.$

Now let’s see if this repeats after $2\pi.$ We can do this by increasing $x$ by $2\pi$ and seeing what the result is. Taking the old value $\pi$ and increasing by $2\pi$ we get $x=3\pi.$

If $x=3\pi,$ then $\sin\left(\frac x2\right) = -1.$ So $\sin\left(\frac x2\right)$ does not repeat after $2\pi.$

On the other hand, if $x=\frac\pi4$ then $\sin(2x)=1.$ But if $x=\frac{5\pi}4$ then $\sin(2x)=1$ again. You can confirm this for any other initial value of $x.$ So $\sin(2x)$ repeats after $\pi.$

That’s the usual interpretation: we regard $\sin\left(\frac x2\right),$ $\sin(x),$ and $\sin(2x)$ as three different functions of $x,$ which have three different minimum periods. You seem to be treating them all as something else, which we might say is $\sin(\theta),$ a single function of $\theta$, where $\theta$ might be $\frac x2,$ $x,$ or $2x.$ That can be a useful thing to do when you want to evaluate the functions, but we don’t consider this one function the same as the three functions of $x.$

  • $\begingroup$ @KM101 @ David K complextoreal.com/tutorials/… Pg 87 of chapter 3 in the link (Free book). "To be considered periodic, in the continuous domain a period represents 2*pi radians." $\endgroup$ Nov 2, 2018 at 13:42
  • $\begingroup$ That link leads to a 73-page document which I am not going to read. I note that a common assumption in Fourier analysis is that we already know the function has a period of $2\pi.$ So we would not try to perform such an analysis directly on a function like $\sin(x/2).$ We would instead transform the function into a different function that does have period $2\pi.$ That is another reason you might do a change of variables like $\theta=x/2.$ $\endgroup$
    – David K
    Nov 2, 2018 at 13:51

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