How do you evaluate $cos (2 {cos}^{- 1} (\frac{1}{7}))$ $cos(2cos^-1(1/7))$ ?

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How do you evaluate $cos (2 {cos}^{- 1} (\frac{1}{7}))$ $cos(2cos^-1(1/7))$ ?

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Answer 1 · 2016-06-17T23:21:29

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

Explanation:

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

$cos (2 arccos (\frac{1}{7}))$ $\cos(2\arccos(1/7))$
$= cos (2 θ)$ $=\cos(2\theta)$
$= 2 {cos}^{2} θ - 1$ $=2\cos^2\theta-1$
$= 2 {(\frac{1}{7})}^{2} - 1 = - \frac{47}{49}$ $=2(1/7)^2-1=-47/49$

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