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Question #84027

1 Answer

Mar 30, 2017

$32x^2-1$

Explanation:

Let $theta=cos^("-"1)(4x)$
$cos(2cos^("-"1)(4x)=cos(2theta)$

Use the double-angle formula $cos(2alpha)=2cos^2alpha-1$
$cos(2theta)=2cos^2theta-1$

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

$2(4x)^2-1=2(16x^2)-1$
$2(16x^2)-1=32x^2-1$

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