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    Write the equation sin7x=-1 as an equation in s=sinx, and state its roots. Hence show that sin(pi/14) - sin(3pi/14) +sin (5pi/14) = 1/2

    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!
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    The first question you've written down is clearly wrong. Just plug that into a calculator and you'll see why.

    As for the second question:
    cos(4x) = cos(3x)
    => cos(4x) - cos(3x) = 0
    => -2 sin((7/2)x)sin((1/2)x) = 0

    So either:
    sin((7/2)x) = 0 => (7/2)x = 2npi => x = (2/7) * (2n) * pi
    or
    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.)
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    Thnx a million m8, soz my rep isnt worth anything . I made a mistake in writing out the first question, but have now edited it.
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    Don't worry about it.

    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.
 
 
 
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Updated: April 13, 2006
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