The problem was, "for $0 < x < 1$, express and simplify in therms of $x$: $\sin[2\tan^{-1}(x)]$".

My professor figured out that, $\sin[2\tan^{-1}(x)] = \frac{2x}{x^2+1}$, using the Double Angle Sine identity.

The Double Sine identity is, $\sin2\theta = 2\sin\theta\cos\theta$.

How did he know what values to use for Cosine and Sine in this equation?


Hint: $$\sin^2(y)+\cos^2(y) = 1 \implies \tan^{2}(y)+1=\frac{1}{\cos^2(y)}.$$ Consequently, if $$\tan^{-1}(x) = y \implies x=\tan(y) \implies \cos(y) = \frac{1}{\sqrt{x^2+1}}, \sin(y)=?$$

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  • $\begingroup$ There is a missing step: as is, $\cos y$ is only known to be $\pm\frac{1}{\sqrt{x^2+1}}$. However, $\mathrm{arctan}\,x\in]-\pi/2,\pi/2[$, so the cosine is nonnegative. $\endgroup$ – Jean-Claude Arbaut Nov 26 '18 at 21:56
  • $\begingroup$ @Jean-ClaudeArbaut I left it purposefully to let the OP figure out the details. $\endgroup$ – Math Lover Nov 26 '18 at 21:57
  • $\begingroup$ That's a good point :) +1 $\endgroup$ – Jean-Claude Arbaut Nov 26 '18 at 21:58
  • $\begingroup$ @Jean-ClaudeArbaut If you want to use the “perverse notation” (as D. E. Knuth calls it), use \arctan x\in\mathopen]-\pi/2,\pi/2\mathclose[, that renders $\arctan x\in\mathopen]-\pi/2,\pi/2\mathclose[$. $\endgroup$ – egreg Nov 26 '18 at 21:59
  • $\begingroup$ @egreg Thanks, good to know. Why perverse? That's the notation I learned and used since high school and through all my university curriculum. I find it clear, and more readable than alternative notations, especially in handwriting. $\endgroup$ – Jean-Claude Arbaut Nov 26 '18 at 22:04

Consider a right triangle whose other two angles are $\theta =\tan^{-1}x$ and $\frac{\pi}2-\theta$.

If the legs have lengths $1$ and $x$ (say the side opposite $\theta$ has length $x$) then the hypotenuse has length $\sqrt{1+x^2}$.

It follows that $\sin\theta=\frac x{\sqrt{1+x^2}}$ and $\cos\theta =\frac1{\sqrt{1+x^2}}$.

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