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

1 Answer

Jun 17, 2016

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

Explanation:

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#

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