How to prove ?

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

2 Answers
Hriman
Jul 5, 2018

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

Narad T.
Jul 5, 2018

Please see the proof below

Explanation:

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

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