Graphing $\sin(|x|)$? I'm confused on how the graph is in both quadrant II and III. If $|x|$ is evaluated first wouldn't all the answers be positive, so that when the range of $|x|$ is plugged into $\sin$ wouldn't the range of $\sin$ be all positive too?
 A: Just draw the graph of $y=\sin x$ for $x \ge 0$, and then reflect it about the $y$-axis. This procedure works for any function of the form $y = g(|x|)$.
A: Here's a graph generated by WolframAlpha:

A: 
If |x| is evaluated first wouldn't all the answers be positive

True

so that when the range of |x| is plugged into sin wouldn't the range of sin be all positive too?

What about $x = -\frac{3\pi}{2}$? We have $|x|  = \left | - \frac{3\pi}{2} \right | =  \frac{3\pi}{2} $, but $\sin(\left | - \frac{3\pi}{2} \right | ) = -1$
The actual plot of both $\sin(x)$ and $\sin(|x|)$ is shown below.
The process described by TonyK works because applying the transformation $y = f(-x)$ to $f(x)$ reflects the curve about the vertical axis. Now consider what the process does to $|x|$ for $x < 0$

A: No. $\sin(|x|)$ will be positive around zero because of the vertex there. But for $\pi < x<2\pi$, we still have $\sin|x|=\sin x<0$.
A: Recall that:
$$|x|= \begin{cases}
x & \text{if } x \ge0 \\
-x & \text{if } x < 0\\
\end{cases} $$
Hence, it follows that:
$$\sin(|x|)= \begin{cases}
\sin(x) & \text{if } x \ge0 \\
\sin(-x) & \text{if } x < 0\\
\end{cases} $$
In other words, the graph remains the same when $x \ge 0$; for the case when $x<0$, you must reflect the graph of the function in the $y$-axis.
