# Can the graph of $\tan^{-1}{\left(\frac{\sin x}{x}\right)}$ be expressed as $Ce^{-kx}\cos(\omega x + \phi)$?

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): 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 $$C,k,\omega, \text{and } \phi$$?

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

## 2 Answers

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.

• Is there an explanation as to why the damping/envelope is represented by that expression? – Ujkan Sulejmani Apr 5 '17 at 16:49
• @UjkanSulejmani: look again, this is obvious. – Yves Daoust Apr 5 '17 at 17:40

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).