Cambridge Poly p.136 Q10(a) (1 Viewer)

[ssglain]

Poly: Cambridge p.136 Q10(a)

Show that cos(5@) = 16(cos@)^5 - 20(cos@)^3 + 5(cos@)
Hence solve P(x) = 16x^5 - 20x^3 + 5x - 1 = 0
& Deduce the exact values of cos(2pi/5) and cos(4pi/5)

I've done similar deductions with nice quartics where the quadratic formula would be sufficient to work out exact values of (cos#)^2, etc., but I'm a bit stumped by the 5th degree poly. Some hints would be greatly appreciated.

Edit:
I had another attempt and made myself a nice quartic - but it doesn't seem to be producing the right answers. Please check my steps:

The roots of P(x) = 0 are x = 1, cos(2pi/5), cos(4pi/5). Clearly some of these are multiple roots.
P'(x) = 80x^4 - 60^2 + 5 = 5(16x^4 - 12x^2 +1)
P'(1) =/= 0, P'[cos(2pi/5)] = 0, P'[cos(4pi/5)] = 0
So x = cos(2pi/5), cos(4pi/5) are roots of P'(x) = 0 and double roots of P(x) = 0.

Then y = x^2 = [cos(2pi/5)]^2 and y = x^2 = [cos(4pi/5)]^2 should both satisfy 16y^2 - 12y + 1 = 0

y = (1/8)(3 + √5) or (1/8)(3 - √5)
Therefore: cos(2pi/5) = √[(1/8)(3 - √5)], cos(4pi/5) = - √[(1/8)(3 + √5)]
(Since: 0 < [cos(2pi/5)]^2 < [cos(4pi/5)]^2 )

The answers should have been
cos(2pi/5) = (1/4).(-1 + √5), cos(4pi/5) = - (1/4).(1 + √5)

Edit 2:
The calculator seems to indicate that my ugly surds are, in fact, of the correct values. How did Mr & Mrs Arnold's get those more pleasant-looking ones though?

I tried playing with the sum & product of roots of 16y^2 - 12y + 1 = 0. It didn't work.

 

[Bank$]

sorry i didnt check your method its just i dnt like maths on computer screens lol anyway here is my solution:

EDIT: wow that came out massive lol

 

[ssglain]

That is a much neater method. It's nice to learn something new. Thanks a lot for you input!