How to prove ?

Question

#cos^2(pi/8)+cos^2((3pi)/8)+cos^2((5pi)/8)+cos^2((7pi)/8)=2#

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Answer 1 · 2018-07-05T16:46:31

#cos^2(pi/8)+cos^2((3pi)/8)+cos^2((5pi)/8)+cos^2((7pi)/8)=2#

Using cofunction identities:
#cosx=sin(pi/2-x)#

#cos^2(pi/8)+sin^2(pi/2-(3pi)/8)+sin^2(pi/2-(5pi)/8)+cos^2((7pi)/8)=2#

#cos^2(pi/8)+sin^2(pi/8)+sin^2((-pi)/8)+cos^2((7pi)/8)=2#

#sin^2(-pi/8)# and #sin^2((7pi)/8)# will yield the same values, so we can just substitute:

#cos^2(pi/8)+sin^2(pi/8)+sin^2((7pi)/8)+cos^2((7pi)/8)=2#

Recall #sin^2x+cos^2x=1#

#1+1=2#

#2=2#

Answer 2 · 2018-07-05T16:55:01

Please see the proof below

Apply the half angle formula

#cos^2(a/2)=(1+cosa)/2#

Therefore,

#cos^2(pi/8)=1/2(1+cos(pi/4))=1/2(1+sqrt2/2)#

#cos^2(3/8pi)=1/2(1+cos(3/4pi))=1/2(1-sqrt2/2)#

#cos^2(5/8pi)=1/2(1+cos(5/4pi))=1/2(1-sqrt2/2)#

#cos^2(7/8pi)=1/2(1+cos(7/4pi))=1/2(1+sqrt2/2)#

So,

#cos^2(pi/8)+cos^2(3/8pi)+cos^2(5/8pi)+cos^2(7/8pi)#

#=1/2(1+sqrt2/2)+1/2(1-sqrt2/2)+1/2(1-sqrt2/2)+1/2(1+sqrt2/2)#

#=1/2+1/2+1/2+1/2#

#=2#

#QED#