# How do you simplify cos(2 tan ^-1 x)?

##### 1 Answer
Jun 16, 2018

Use double angle formula to remove coefficient inside the cos, then rearrange standard trig definitions to make the trig function match the inverse trig function inside the bracket

#### Explanation:

Recall the double angle formula:
$\cos 2 \theta = 1 - 2 {\sin}^{2} \theta$

Then $\cos \left(2 \arctan x\right) = 1 - 2 {\sin}^{2} \arctan x$. NB I've written "arctan" here rather than "${\tan}^{- 1}$" because the combination of exponents meaning powers and function inverses is potentially confusing.

So we now have a trig function of an inverse trig function. If we can express our $\sin$ in terms of $\tan$, this will cancel right out.

By definition, $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\sin \theta}{\sqrt{1 - {\sin}^{2} \theta}}$, so
${\tan}^{2} \theta \left(1 - {\sin}^{2} \theta\right) = {\sin}^{2} \theta$
${\tan}^{2} \theta = {\sin}^{2} \theta \left(1 + {\tan}^{2} \theta\right)$
${\sin}^{2} \theta = {\tan}^{2} \frac{\theta}{1 + {\tan}^{2} \theta}$

By definition, $\tan \arctan x = x$, so $1 - 2 {\sin}^{2} \arctan x$ becomes $1 - \frac{2 {x}^{2}}{1 + {x}^{2}}$. Putting this over a common denominator makes $\frac{1 - {x}^{2}}{1 + {x}^{2}}$.

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