# How do you evaluate cos(2cos^-1(1/7))?

Jun 17, 2016

$- \frac{47}{49}$, about $- 0.9592$. See below.

#### Explanation:

Let $\setminus \theta = \setminus \arccos \left(\frac{1}{7}\right)$. Thus $\setminus \cos \setminus \theta = \frac{1}{7}$. From the double-angle formula for cosine:

$\setminus \cos \left(2 \setminus \arccos \left(\frac{1}{7}\right)\right)$
$= \setminus \cos \left(2 \setminus \theta\right)$
$= 2 \setminus {\cos}^{2} \setminus \theta - 1$
$= 2 {\left(\frac{1}{7}\right)}^{2} - 1 = - \frac{47}{49}$