How to sketch the graph of $\sin(x^2)$? Clearly, we see that $y = \sin(x^2)$ is not periodic as $0$, $\sqrt{\pi}$, $\sqrt{2\pi}$, etc. are not equally spaced and for each of these values, $y=0$.
Also $x^2$ changes much faster than $x$ does (w.r.t. slope)
But I don't understand how to sketch it.
 A: It still has a wavy shape, but that wave becomes more and more "smooshed" together as $x$ tends to $\pm\infty$. For $x\ge0$, let $f(x)=\sin x$ and $g(x)=\sin(x^2)$. Notice that for $a\ge0$, the function $f$ can be described as the set of points $(a,\sin a)$, and $g$ as the set of points $(\sqrt{a},\sin a)$. So, in theory, you could plot the graph of $g$ by taking the graph of $f$, and then shifting all the $x$-coordinates to their square roots while keeping the $y-$coordinates the same. This explains why $0,\sqrt{\pi},\sqrt{2\pi},\dots$ are all roots of $g$.
To investigate the "smooshing" effects further, suppose that the points $\mathrm{P}(a,\sin a)$ and $\mathrm{Q}(b,\sin b)$ both lie on the graph of $f$. Then, the corresponding points on the graph of $g$ are $\mathrm{P'}(\sqrt{a},\sin a)$ and $\mathrm{Q'}(\sqrt{b},\sin b)$. The horizontal distance between $P$ and $Q$ is $|b-a|$, whereas the horizontal distance between $\mathrm{P'}$ and $\mathrm{Q'}$ is $|\sqrt{b}-\sqrt{a}|$. Hence, the horizontal distance has been shrunk by a factor of
$$
\frac{|b-a|}{\left|\sqrt{b}-\sqrt{a}\right|}=\sqrt{a}+\sqrt{b} \, .
$$
Hence, increasing either $a$ or $b$ magnifies the compression effect that can be readily seen in the graph below. This also provides a visual explanation for why the graph appears to get steeper and steeper as $x\to\infty$. Consideration of the derivative $g'(x)=2x\cos(x^2)$ confirms this.



(It should be easy to fill in the details for what happens for $x<0$ because $g$ is even.)
