# Question #a2e24

Apr 30, 2016

Use the definition of ${\tan}^{-} 1$ and the double angle formula for cosine.

#### Explanation:

$\theta = {\tan}^{-} 1 x$ if and only if $- \frac{\pi}{2} < \theta < \frac{\pi}{2}$ and $\tan \theta = x$

$\cos \left(2 \theta\right) = 2 {\cos}^{2} \theta - 1$

There are several ways to find ${\cos}^{2} \theta$ for $\tan \theta = x$.
Here are two:

Method 1: Sketch a triangle
You can sketch a right triangle with one angle $\theta$. Label the side opposite $\theta$ as length $x$ and the side adjacent has length $1$, so the hypotenuse has length $\sqrt{1 + {x}^{2}}$

We can see that $\cos \theta = \frac{1}{\sqrt{1 + {x}^{2}}}$

Method 2: Use a trigonometric identity

Recall that ${\tan}^{2} \theta + 1 = {\sec}^{2} \theta = \frac{1}{\cos} ^ 2 \theta$.

So ${x}^{2} + 1 = \frac{1}{\cos} ^ 2 \theta$.

Using either method we continue

${\cos}^{2} \theta = \frac{1}{1 + {x}^{2}}$, so

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

$= \frac{1 - {x}^{2}}{1 + {x}^{2}}$.

And we have

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