How to prove #8cos^4x=3+4cos2x+cos4x#? Hint #cos^4x=(1/2(1+cos2x))^2)#

2 Answers
Narad T.
May 17, 2017

See proof below

Explanation:

We need

#(a+b)^2=a^2+2ab+b^2#

#cos2x=2cos^2x-1#

#cos4x=2cos^2(2x)-1#

#cos^2(2x)=1/2(1+cos4x)#

#LHS=8cos^4x=8((1/2(1+cos2x))^2)#

#=2(1+2cos2x+cos^2(2x))#

#=2+4cos2x+2cos^2(2x)#

#=2+4cos2x+2(1/2(1+cos4x))#

#=2+4cos2x+1+cos4x#

#=3+4cos2x+cos4x#

#=RHS#

#QED#

Andy Y.
May 17, 2017

Replace the term #cos^4(x)# with the hint. Factor the result on the left. Also, on the right, replace the term #cos(4x)# with #4cos(2x)-1#.

Explanation:

You have been given

#8cos^4(x)=3+4cos(2x)+cos(4x)#

You have also been given the hint #cos^4(x)=(1/2[1+cos(2x)])^2#

Replacing the term #cos^4(x)# with the hint gives
#8(1/2[1+cos(2x)])^2=3+4cos(2x)+cos(4x)#
#8(1/2[1+cos(2x)])(1/2[1+cos(2x)])=3+4cos(2x)+cos(4x)#
#8(1/2+1/2cos(2x))(1/2+1/2cos(2x))=3+4cos(2x)+cos(4x)#
#8(1/4+1/2cos(2x)+1/4cos^2(2x))=3+4cos(2x)+cos(4x)#

Next, distribute the 8 through
#2+4cos(2x)+4cos^2(2x)=3+4cos(2x)+cos(4x)#

Next, subtract #4cos(2x)# from both sides
#2+4cos^2(2x)=3+cos(4x)#

The double angle formula for cosines is #cos(2a)=2cos^2(a)-1#. So on the right hand side, #cos(4x)=cos(2*2x)=4cos^2(2x)-1#. This gives us:
#2+4cos^2(2x)=3+4cos^2(2x)-1#
#2+4cos^2(2x)=2+4cos^2(2x)#

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