Question #84027

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Question #84027

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Answer 1 · 2017-03-30T15:56:58

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

Explanation:

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

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

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

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

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Question #84027

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