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After graphing $\sin x$, I thought of trying something interesting. I wanted to plot the angle $\theta$ that a point $(x,\sin x )$ makes with the origin on the $y$-axis, against $x$ on the $x$-axis.

$$\tan\theta = \frac{\sin x}{x}\Rightarrow \theta=\tan^{-1}{\left(\frac{\sin x}{x}\right)}$$

Graphing $y = 20\times\theta$ (multiply by 20 for graphical purposes): Diagram 1

Part of it reminded me of the graph for the damped oscillator (specifically, the $x>0$ part). That made me wonder if it was possible to find constants $C,k,\omega,$ and $\phi$ such that $$\theta = Ce^{-kx}\cos(\omega x + \phi)$$

However, after toying with Grapher for a while, $y = \theta$ didn't seem to decrease exponentially.

That led me to this question: is there any analytical way to find real constants $C,k,\omega,$ and $\phi$ such that $\theta = Ce^{-kx}\cos(\omega x + \phi)$?

Furthermore, are there any complex constants $C,k,\omega, \text{and } \phi$?

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  • $\begingroup$ No for real constants: the derivatives are quite different. $\endgroup$ – Bernard Apr 5 '17 at 8:54
  • $\begingroup$ Certainly not, there is a serious mismatch for negative $x$. $\endgroup$ – Yves Daoust Apr 5 '17 at 14:47
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The damping that you see is

$$\frac{\arctan\dfrac{\sin x}x}{\sin x}.$$

It is very close to the hyperbola $\dfrac1x$, as you remain in the linear part of the arc tangent.

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  • $\begingroup$ Is there an explanation as to why the damping/envelope is represented by that expression? $\endgroup$ – Ujkan Sulejmani Apr 5 '17 at 16:49
  • $\begingroup$ @UjkanSulejmani: look again, this is obvious. $\endgroup$ – Yves Daoust Apr 5 '17 at 17:40
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The envelope of $\frac{\sin x}{x}$ is $\pm \frac{1}{x}$, and for large $x$, $\arctan \frac{1}{x} \approx \frac{1}{x}$. So the decay of the function is inverse to $x$, not exponential, and no constant $k$ exists (real or complex; an imaginary part to $k$ will just introduce a sinusoidal oscillation in the envelope, in addition to the exponential decay or growth).

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