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    • Thread Starter

    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!

    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
    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.)
    • Thread Starter

    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.

    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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