# How do you simplify the expression sin(2 arctan x) ?

May 24, 2018

$\sin \left(2 \arctan x\right) = \frac{2 x}{1 + {x}^{2}}$

#### Explanation:

Let's assume by $\arctan x$ we mean the multivalued inverse, all the angles whose tangent is $x$.

The double angle formula for sine is

$\sin \left(2 \theta\right) = 2 \sin \theta \cos \theta$

$\sin \left(2 \arctan x\right) = 2 \sin \left(\arctan x\right) \cos \left(\arctan x\right)$

We can think of $\arctan \left(\frac{x}{1}\right)$ as a right triangle whose opposite is $x$ and adjacent is $1$, so the hypotenuse is $\sqrt{1 + {x}^{2}} .$

The sign of sine and of cosine are each ambiguous when we know the tangent. But the sign of the product sine and cosine and the sign of the tangent (which is the quotient of sine and cosine) are the same.

$\sin \arctan x = \frac{\textrm{o p p}}{\textrm{h y p}} = \pm \frac{x}{\sqrt{1 + {x}^{2}}}$

$\cos \arctan x = \frac{\textrm{o p p}}{\textrm{h y p}} = \pm \frac{1}{\sqrt{1 + {x}^{2}}}$

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

Despite the ambiguity of the sign of the factors, the sign of the sine of the double angle is not ambiguous.