Show that cos(π/7) - cos(2π/7) + cos(3π/7) = 1/2.

Consider the 7 complex roots of the equation z⁷ + 1 = 0.
These have real parts cos(pi), cos(9*pi/7), cos(11*pi/7), cos(13*pi/7), cos(pi/7), cos(3*pi/7), cos(5*pi/7).
The sum of the roots is equal to the coefficient of z⁶, i.e., equal to zero. Hence the sum of the real parts is zero.
That is, cos(pi/7) + cos(3*pi/7) + cos(5*pi/7) + cos(9*pi/7) + cos(11*pi/7) + cos(13*pi/7) = -cos(pi) = 1.
But cos(x) = cos(-x) = cos(2*pi-x).
Hence 2(cos(pi/7) + cos(3*pi/7) + cos(5*pi/7)) = 1.
Then, cos(pi-x) = -cos(x), and so cos(5*pi/7) = -cos(2*pi/7).
Therefore cos(pi/7) - cos(2*pi/7) + cos(3*pi/7) = 1/2.

Posted by Nick Hobson on 2004-12-06 21:10:32