If you take the derivative of $\sin^2(x)$ and remember your double-angle formulas, you see that $$ \frac{\operatorname{d}}{\operatorname{d}\!x}\; \sin^2(x) = 2\sin(x)\cos(x) = \sin(2x)\,. $$ This looks surprisingly clean. You can say the rate of change of $\sin^2$ at a value is given by the $\sin$ of twice that value. Is this just a happy accident? Or is there some nice geometric/trigonometric intuition behind this that I'm not seeing?

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    $\begingroup$ $\sin^2{x} = \frac{1}{2}(1-\cos{2x})$? $\endgroup$ – Chappers Mar 29 '17 at 18:36
  • $\begingroup$ @Chappers Well yeah, if you use that identity and take derivative you get to $\sin(2x)$ immediately. But this doesn't help my geometric intuition at all. Ideally I would like to use this as a cool example to hammer home the idea that a derivative is a rate-of-change to my students. So it would be nice to have a way to see why the rate of change of $\sin^2(x)$ can be given by $\sin(2x)$. $\endgroup$ – Mike Pierce Mar 29 '17 at 18:37
  • $\begingroup$ @MikePierce It should help your intuition if you already have intuition about that identity and about the derivative of $-\cos$ being $\sin$. I think Chappers's comment is actually really close to an answer. You probably already have intuition about derivatives of $\sin$/$\cos$, and intuition for things like the double-angle formula can be found, say, at math.stackexchange.com/a/1342/26369 $\endgroup$ – Mark S. Mar 29 '17 at 18:41
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    $\begingroup$ Note that $\sin^2$, without having to pass to the power reduction formula, will have twice the frequency of the analogous $\sin$. Why? Basically because $\sin$ is odd, so upon squaring, the behavior on the half-periods $[-\pi,0]$ and $[0,\pi]$ become the same. Knowing the frequency already, you might suspect that actually $\sin^2$ is a sinusoid, and it turns out that it is. $\endgroup$ – Ian Mar 29 '17 at 18:53

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