How to graph the trigonometric function when period is more than the range? I have to graph this trigonometric equation for a given range,

$y = -\sin⁡ \left(\dfrac{x}{3}\right) - 2, ~-\dfrac{\pi}{2} \le x \le \dfrac{\pi}{2}$

But the period is coming out to be $6 \pi$. So the question is how? 
I would appreciate any help.
 A: Simply sketch lightly two vertical lines, each intersecting the $x$ axis: one line $x = \pi/2$ that's perpendicular to the x-axis, the other, also a perpendicular line being $x = -\pi/2$. 
Then sketch the portion of the curves period that lies between those vertical lines. Feel free to sketch in the whole period, lightly, but then darken the portion contained on and between the two lines. The points where the graph intersect the two vertical lines should be darkened in, too, to illustrate that they are endpoints which are included in you closed interval.
From Wolfram Alpha: below is the full period + of the curve $\;y = \sin\left(\frac x3\right) - 2$:

And what follows is a graph of the two vertical lines at $x = \pi/2$, $x = -\pi/ 2$, with the portion of the sine curve $\;y = \sin\left(\frac x3\right) - 2$ that lies between (and intersects) those lines. Note that the portion of the curve that lies within the given range resembles a line: if you look at the vertical line $x = 0$ as a common reference point for both graphs, you can see which portion of the curve is the focus of the second graph: Indeed, the curve below which is sandwiched between our two vertical lines includes an inflection point of the given sine curve, and explains why it "looks like a straight line."

A: If $-\pi/2\leq x \leq \pi/2$ then $-\pi/6\leq x/3 \leq \pi/6$. Let $u=x/3$ so you have to plot $-\sin(u)-2$ for $-\pi/6\leq u \leq \pi/6$. You'll get a (non scaled) part of the $\sin$ graph.
