FP3 question on de moivre's Watch
I really dont get how to do the last part. I got the equation (which is right):
64s^7 - 112s^5 + 56s^3 - 7s - 1 = 0
Any help appreciated, thnx in advance!
Edit: Also another question: show that, if cos4x = coz3x, then x is (2/7)*r*pi where r is a natural number
Soz Im being really stupid today!
As for the second question:
cos(4x) = cos(3x)
=> cos(4x) - cos(3x) = 0
=> -2 sin((7/2)x)sin((1/2)x) = 0
sin((7/2)x) = 0 => (7/2)x = 2npi => x = (2/7) * (2n) * pi
sin((1/2)x) = 0 => (1/2)x = 2kpi => x = (2/7) * (14k) * pi
Both of which have the form required. (Note that actually r is an integer.)
Back to your first question. We essentially want to play with the 7th roots of -1. We can see that:
7x = -pi/2 + 2kpi
=> x = ((4k-1)pi)/14
And let's take k=-3,2,...,2,3. Now remember that the sum of the 7th roots of -1 is zero, and hence the sum of their imaginary parts is also zero. Therefore:
sin(-13pi/14) + sin(-9pi/14) + sin(-5pi/14) + sin(-pi/14) + sin(3pi/14) + sin(7pi/14) + sin(11pi/14) = 0
However because sin(-x)=-sin(x) and sin(pi-x)=sin(x), the above equation is reduced to:
-(sin(13pi/14) + sin(pi/14)) - (sin(9pi/14) + sin(5pi/14)) + (sin(3pi/14) + sin(11pi/14)) + sin(7pi/14) = 0
=> 2sin(pi/14) + 2sin(5pi/14) - 2sin(3pi/14) = sin(7pi/14)
But sin(7x) = -1, and we know that x=-pi/14 is a solution to that. So sin(7pi/14) = -sin(-7pi/14) = -(-1) = 1. Hence:
2sin(pi/14) + 2sin(5pi/14) - 2sin(3pi/14) = 1
Rearranging and dividing by 2 gives us the equality we want.
There might be an easier way, but I just can't see one right now.