I've been working on this for a while. The answer in the book is $\frac{2x}{x^2 + 1}$ Here's my workings:

$\sin(2\tan^{-1} x)$

Let $\alpha = \tan^{-1}x \Rightarrow \tan \alpha = x$

$\sin(2\alpha) = 2\sin\alpha\cos\alpha = 2\tan\alpha\cos^2\alpha = 2x\cos^2\alpha$

I'm not sure how to proceed to turn that $cos^2\alpha$ into $\frac{1}{x^2 + 1}$

  • $\begingroup$ $\sin(\Arctan x)=\frac{x}{\sqrt{x^2+1}}$ and $\cos(\Arctan x)=\frac{1}{\sqrt{x^2+1}}$... $\endgroup$ – long tom Apr 24 '13 at 21:20

Hint: $\cos^2(\alpha) = \dfrac{\cos^2(\alpha)}{\cos^2(\alpha) + \sin^2(\alpha)} $

  • $\begingroup$ Your hint combined with Mike's helped me to see the solution. Maybe my algebra is lacking, but I had to invert that fraction to allow me to manipulate it, then invert again. $\endgroup$ – PeteUK Apr 24 '13 at 21:38

Making a right-triangle picture can be helpful too, either for getting to the result quickly or checking up on the analytical result. With $\tan \alpha = x = \frac{x}{1}$, we would have $x$ for the leg opposite $\alpha$ and $1$ for the adjacent leg; the hypotenuse is then $\sqrt{x^2 + 1}$ . We wish to evaluate $\sin 2\alpha$ , which is thus

$$2 \sin \alpha \cos \alpha = 2 \cdot \frac{x}{\sqrt{x^2 + 1}} \cdot \frac{1}{\sqrt{x^2 + 1}} = \frac{2x}{x^2 + 1} .$$

  • $\begingroup$ Your post says $\alpha = \tan x = \tan(\frac{x}{1})$, but should that be $\tan \alpha = x = \frac{x}{1}$? $\endgroup$ – PeteUK Apr 24 '13 at 21:55
  • $\begingroup$ D'oh! Thanks! That is what comes of typing too fast... $\endgroup$ – colormegone Apr 24 '13 at 21:58
  • $\begingroup$ Thanks for the clarification and a great post! $\endgroup$ – PeteUK Apr 24 '13 at 21:58

Hint: What's the reciprocal of $\cos^2 \alpha$?



let $y=arctan(x)$

so $x=tan(y)$

$sin(2y)=2sin(y)cos(y)=2sin(arctan(x))\cdot cos(arctan(x))$

$sin(arctan(x)) = \frac{x}{\sqrt{1+x^2}}$ $cos(arctan(x))= \frac{1}{\sqrt{1+x^2}}$

$2\cdot\frac{x}{\sqrt{1+x^2}} \cdot \frac{1}{\sqrt{1+x^2}}=\frac{2x}{{1+x^2}}$


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