How do you evaluate?

#cos^2((pi)/13)+cos^2((2pi)/13)+cos^2((3pi)/13)+cos^2((4pi)/13)+cos^2((5pi)/13)=?#

1 Answer
George C.
Jan 16, 2018

#cos^2 (pi/13) + cos^2 ((2pi)/13) + cos^2 ((3pi)/13) + cos^2 ((4pi)/13) + cos^2((5pi)/13) + cos^2((6pi)/13)=11/4#

Explanation:

The #13#th roots of #1# are:

#cos((2npi)/13)+isin((2npi)/13)#

for #n = -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6#

Noting that:

#{ (cos(-theta) = cos theta), (sin(-theta) = -sin theta) :}#

we can take the roots in pairs to find:

#x^13-1 = (x-1)(x^2-2cos((2pi)/13)x+1)(x^2-2cos((4pi)/13)x+1)(x^2-2cos((6pi)/13)x+1)(x^2-2cos((8pi)/13)x+1)(x^2-2cos((10pi)/13)x+1)(x^2-2cos((12pi)/13)x+1)#

#color(white)(x^13-1) = (x-1)(x^12+x^11+x^10+...+x+1)#

Equating the coefficients of #x^11#, we find:

#-2(cos((2pi)/13)+cos((4pi)/13)+cos((6pi)/13)+cos((8pi)/13)+cos((10pi)/13)+cos((12pi)/13)) = 1#

So:

#cos((2pi)/13)+cos((4pi)/13)+cos((6pi)/13)+cos((8pi)/13)+cos((10pi)/13)+cos((12pi)/13)=-1/2#

Now:

#cos 2 theta = 2cos^2 theta-1#

Hence:

#cos^2 theta = 1/2(cos 2theta+1)#

So we find:

#cos^2 (pi/13) + cos^2 ((2pi)/13) + cos^2 ((3pi)/13) + cos^2 ((4pi)/13) + cos^2((5pi)/13) + cos^2((6pi)/13)#

#= 1/2(cos((2pi)/13)+cos((4pi)/13)+cos((6pi)/13)+cos((8pi)/13)+cos((10pi)/13)+cos((12pi)/13)) + 3#

#=-1/4+3#

#=11/4#

Related questions
Impact of this question
1136 views around the world
You can reuse this answer
Creative Commons License