All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Trigonometric Fun (Posted on 2004-12-06)
Show that cos(π/7) - cos(2π/7) + cos(3π/7) = 1/2.

 See The Solution Submitted by SilverKnight Rating: 3.6667 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Solution | Comment 1 of 7
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

 Search: Search body:
Forums (0)