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
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 13th 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