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Explain why the graph of $y=\frac{4x}{x^2+1}$ and $y=2\sin(2\arctan x)$ are the same.

The first equation is of the form of Newton's Serpentine. When you graph the second equation it appears to overlap the first equation.

I'm not sure whether these two equations are identities or just very close approximations.

I tried to manipulate both equations to get the other but failed.

How does one explain why these two equations are identical?

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take $$x=\tan \alpha\\ -\frac{\pi}{2} <\alpha<\frac{\pi}{2}$$so

$$y=\frac{4x}{x^2+1}=\\ y=\frac{4\tan \alpha}{(\tan \alpha)^2+1}\\=\frac{4\tan \alpha}{\frac{1}{\cos^2 \alpha}}\\=4\tan \alpha .\cos ^2 \alpha\\=4\sin \alpha \cos \alpha\\=2\sin(2\alpha)$$

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Since $\sin(2A) = 2 \sin A \cos A$,

\begin{align}2\sin(2 \arctan x) &= 4 \sin (\arctan x) \cos(\arctan x) \\ &=4\left( \frac{x}{\sqrt{1+x^2}} \right)\left( \frac{1}{\sqrt{1+x^2}} \right) \\ &=\frac{4x}{1+x^2}\end{align}

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    $\begingroup$ Did you mean $\cos(\arctan x) = \dfrac{1}{\sqrt{\color{red}{1} + x^2}}$? $\endgroup$ – N. F. Taussig Jul 22 '17 at 10:17
  • $\begingroup$ yikes, thanks for catching the mistake. $\endgroup$ – Siong Thye Goh Jul 22 '17 at 14:37
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The expression on the left sort of begs for the substitution $x = \tan \theta$ so that you get

$$\frac{4x}{x^2 + 1} =\frac{4\tan \theta}{\sec^2 \theta} = 4\cos\theta\sin\theta = 2\sin2\theta = 2 \sin (2\arctan x)$$

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We use the identity $$\sin(2\theta) = \frac{2\tan\theta}{1 + \tan^2\theta}$$ Let $\arctan x = \theta$. Then $x = \tan\theta, -\frac{\pi}{2} < x < \frac{\pi}{2}$. Hence, $$2\sin(2\arctan x) = 2\sin(2\theta) = 2 \cdot \frac{2\tan\theta}{1 + \tan^2\theta} = 2 \cdot \frac{2x}{1 + x^2} = \frac{4x}{1 + x^2}$$

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