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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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