4u Polynomial Question (1 Viewer)

[AreYouAlright?]

Hey guys,

We as a class have worked through this question up to the point shown on the attached image of the working out. We were able to show the second part but we thought there must be a simpler or more logical way to do so, does anyone have a method to address the second part in this question?

Any help would be greatly appreciated.

 

[shafqat]

use the fact that sum of roots = 0

 

[AreYouAlright?]

Could you please demonstrate

I still can't see how you would show that, can you please write a proof for it?

 

[shafqat]

from i you know that the solutions of polynomial are cos(-pi/9), cos(7pi/9) and cos (13pi/9)
so from polynomial, some of roots = coeff of x^2/1 = 0
so cos(-pi/9) + cos(7pi/9) + cos (13pi/9) = 0
as cosx is even, cos(-pi/9) = cos(pi/9)
also, remember that cosA = -cos(pi - A) [can be proved expanding RHS using cos(a-b) formula]
applying this to the other 2 terms in above equation,

cos(pi/9) - cos(2pi/9) - cos(4pi/9) = 0
rearranging, cos(pi/9) = cos(2pi/9) + cos(4pi/9)