# Question 84027

Mar 30, 2017

$32 {x}^{2} - 1$

#### Explanation:

Let $\theta = {\cos}^{\text{-} 1} \left(4 x\right)$
cos(2cos^("-"1)(4x)=cos(2theta)#

Use the double-angle formula $\cos \left(2 \alpha\right) = 2 {\cos}^{2} \alpha - 1$
$\cos \left(2 \theta\right) = 2 {\cos}^{2} \theta - 1$

$\theta$ is some angle the cosine of which is $4 x$, so ${\cos}^{2} \theta = {\left(4 x\right)}^{2}$
$2 {\cos}^{2} \theta - 1 = 2 {\left(4 x\right)}^{2} - 1$

$2 {\left(4 x\right)}^{2} - 1 = 2 \left(16 {x}^{2}\right) - 1$
$2 \left(16 {x}^{2}\right) - 1 = 32 {x}^{2} - 1$