How to prove #cos^2(pi/8)+cos^2 ((3pi)/8)+cos^2((5pi)/8)+cos^2((7pi)/8)=2# ?

1 Answer
Abhishek K.
Jul 11, 2018

Please see below

Explanation:

Let #pi/8=x# then #8x=pi and 4x=pi/2#

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

#=cos^2x+cos^2(3x)+cos^2(5x)+cos^2(7x)#

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

#=cos^2x+cos^2(3x)+cos^2(pi/2+x)+cos^2(pi/2+3x)#

#=cos^2x+cos^2(3x)+sin^2(x)+sin^2(3x)=1+1=2=RHS#

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