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
Explanation:
The
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
-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