# How to Determine Cosine, and Sine Given $\tan^{-1}x$

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

• 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. – Jean-Claude Arbaut Nov 26 '18 at 21:56
• @Jean-ClaudeArbaut I left it purposefully to let the OP figure out the details. – Math Lover Nov 26 '18 at 21:57
• That's a good point :) +1 – Jean-Claude Arbaut Nov 26 '18 at 21:58
• @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 $\arctan x\in\mathopen]-\pi/2,\pi/2\mathclose[$. – egreg Nov 26 '18 at 21:59
• @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. – 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 $$\cos\theta =\frac1{\sqrt{1+x^2}}$$.