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

Question

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?

Answer 1

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

Answer 2

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 $\cos\theta =\frac1{\sqrt{1+x^2}}$.