prove without using calculators
cos(П/7) - cos(2П/7) + cos(3П/7) = 1/2
where П = pi
2006-09-07 03:30:38 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
well its the answer;
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.
Good Luck , I hope it would be enough could help you.
2006-09-07 04:13:50 · answer #1 · answered by sweetie 5 · 3⤊ 0⤋
2
2006-09-07 10:31:58 · answer #2 · answered by S K 7 · 0⤊ 4⤋
u can have the sum of them all
then inverse the equasion into sin
and it will be proved
2006-09-07 10:41:20 · answer #3 · answered by gggnm 3 · 1⤊ 3⤋
i dont know how to make pi symbol but i know question.
2006-09-07 11:06:30 · answer #4 · answered by CHIMPU 2 · 0⤊ 3⤋
I really can...but it's a lengthy answer to write which I can't here...
2006-09-07 10:38:45 · answer #5 · answered by Ajit 2 · 0⤊ 4⤋