Really simple question about sine wave equation with $\pi$

I'm reading some EE material, and my trigonometry is really rusty.

There is an equation that looks like: $y(t) = \sin(2\pi \times 150 \times t)$

Why is there a $2\pi$ factor in the argument?

When I learned trig (many, many years ago), I learned equations that look like: $y(t) = \sin(150 t)$

where the $150$ was assumed to be in radians already.

Are these the same equation, and if so, when would I use one form but not the other?

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Well, the sine functions has period $2\pi$. It you want your sine to exhibit some other period, you multiply the argument by an appropriate factor. In this case, the period of $\sin(300\pi t)$ can be derived like so: $\sin(300\pi t+2\pi)=\sin(300\pi(t+1/150))$, so your sine has period $\dfrac1{150}$. Very wiggly! – J. M. Sep 25 '11 at 2:10

It is just a matter of units of measure. In $u(t)=\sin(\omega t)$, $\omega$ is an angular frequency, that is radians per unit time. In $v(t)=\sin(2\pi \nu t)$, $\nu$ is a frequency, that is number of cycles per unit time.
@stack: $\sin(150t)$ pulsates with an angular frequency of $150$ rad/sec or with a frequency of $150/(2\pi)$ Hertz. Like J.M. says, you only need to choose one of the two. – Giuseppe Negro Sep 25 '11 at 9:09
@stack: Yes, if there is $\pi$ then in $\sin(2\pi \nu t)$ the term $\nu$ is frequency in Hertz. Otherwiswe you have $\sin(\omega t)$ and $\omega$ is rad/sec. I stress that they are merely two different units of measure for the same thing, though. Just like measuring a length in meters or in feet and inches. The factor of conversion is $\omega=2\pi \nu$. – Giuseppe Negro Sep 26 '11 at 0:55