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

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Answer 1 · 2016-06-17T23:21:29

$-47/49$ , about $-0.9592$ . See below.

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

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

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