period vs time period of sine wave It's weird I'm still confused about this, but usually when we figure out the period of a sine wave from its graph, it's in radians. But the true period should be in time, like how fast we are revolving around the circle. For example if we see two hills and a trough in $4\times\pi$, we say the period is $4\pi$. But depending on the frequency, the actual period is different? 
Take another example, $\cos(2{\pi}ft)$ where f=1/sec. Then it will take one second to go around the circle once. But on the graph you see the period as $2\pi$?
 A: The $\sin(x)$ function is what it is: it is zero when $x$ is a multiple of $2\pi$, with peaks of $\pm 1$ at $x=(2k \pm \frac{1}{2})\pi$, and so on. 
The important thing to note is that $x$ is just an argument, and can represent anything you like. For instance, if we were spraying a sine wave on a wall, $x$ would represent the distance along the wall. However, presumably a period of $2\pi$ cm would render us with too small a sine wave, while $2\pi$ metres may be too large. We would like something in between, but we don't want to change our units of measurement (cm or m), because that's what we have on our measuring tape. So we can scale the argument $x$ by multiplying it by some factor $\alpha$: what we are doing effectively is stretching out or compressing the wave along the horizontal axis (if $\alpha<1$ and $\alpha>1$, respectively).
Now, along the same lines, for signal processing, we would like the argument $x$ to represent the time, $t$, with some period $T$. But we want $t$ and $T$ to be in seconds, and  we certainly don't want to change the definition of what "a second" means! Therefore, we need to scale the argument $x$ such that at $t=T$, it evaluates to $2\pi$, and so on. So we let: $x=\frac{2\pi}{T}t$ in the function. Take some time to convince yourself why this works, it shouldn't be very hard. 
Now for convenience we often let $f=\frac{1}{T}$ so that we can write $\sin(2\pi f t)$. In signal processing, it is even more common to also get rid of the $2\pi$ by defining $\omega = \frac{2\pi}{T} = 2\pi f$ such that now we can simply write $\sin(\omega t)$. Convince yourself that $f$ corresponds to "periods per second" while $\omega$ corresponds to "radians per second" (hint: how many radians in a circle?).
A note about scaling: we are not changing the basic shape of the function as long as we scale its argument and/or its output linearly. So in general, from $\sin(x)$ we can get $A\sin(\alpha x + \beta)+B$. $\alpha$ and $A$ stretch it out or compress it along the horizontal and vertical axes, respectively, while $\beta$ and $B$ shift it along these axes. I would suggest that you play around with this idea on some plotting software until you get a firm, visual grasp of it.  Note also that as soon as we modify something non-linearly, what we have left is no longer a sine wave (try plotting $\sin(x^2)$ and $\sin^2(x)$).
A: It makes sense to think of $\sin(x)$ of having a period of $2\cdot\pi$ because assuming $x$ is radians then if you now the value of $\sin(x)$ for that interval then because the function just repeats outside this range we know $\sin(x + n \cdot2 \cdot \pi) = \sin (x)$ where $n$ is any integer.
When we are concerned about a sine function in time we may write $\sin(\omega\cdot t)$ where $\omega$ is the frequency expressed in radians per second.
Alternatively we may write $\sin(2\cdot\pi\cdot f\cdot t)$ where $f$ is the frequency expressed in Hz.
The $2\pi$ comes from the fact that there are $2\pi$ radians in a circle and the period is just $\frac{1}{f}$.
A: I suppose this question comes from confusion on scaling and the use of units.
The function $\sin(x)$ expects a real number $x$ in radians. It has a period of $2\pi\,{\rm rad}$, i.e. $$\sin(x+2\pi\,{\rm rad}) =\sin(x)$$ for all values of $x$ in radians.
If you want to produce a function that takes an input in seconds, you can't just evaluate $\sin(1\,{\rm sec})$, since the sine function doesn't even know or care what seconds are or what time is. What you can do, is introduce a factor $\omega$ with the unit ${\rm rad}/{\rm sec}$ and look at $$\sin(\omega t).$$
For example if $\omega=2\pi\,{\rm rad}/{\rm sec}$, you can evaluate at $t=1\,{\rm sec}$ and get
$$
\sin(\omega \cdot 1\,{\rm sec}) = \sin(2\pi\,{\rm rad}/{\rm sec} \cdot 1\,{\rm sec}) = \sin(2\pi\,{\rm rad})=0.
$$
See how the factor $\omega$ converts a time in seconds to a value in radians? This factor $\omega$ is called the angular frequency, it tells you how many radians the argument of $\sin$ is increasing per unit of time.
Another way to achieve the same thing would be $$\sin(2\pi\,{\rm rad}\cdot f\, t),$$ where $f$ has unit $1/{\rm sec}$ now. If we choose $f=1\,1/{\rm sec}$ and evaluate again at $t=1\,{\rm sec}$, we get
$$
\sin(2\pi\,{\rm rad}\cdot f\, t) = 
\sin(2\pi\,{\rm rad}\cdot 1\,1/{\rm sec} \cdot 1\,{\rm sec}) = \sin(2\pi\,{\rm rad})=0.
$$
This $f$ is called the frequency, it tells us how many periods the argument of the sine function goes through per unit of time.
A little confusion comes in, because in mathematics we don't use units at all, so for the sine function, we just take unitless real numbers as input and interpret it as radians, thus the unit ${\rm rad}$ becomes meaningless and can be omitted. So we usually write the periodic property of the sine function as
$$\sin(x+2\pi) = \sin(x)$$
without involing the unit ${\rm rad}$.
I used the unit ${\rm rad}$ in my explanation purely to stress the point that this question is a question about units of values.
