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    Using the following diagram and various trig identities, I have worked out that Area = 2r^2 [1 - sin(x)]



    But I don't know how to find sin(x) in terms of r or how to do it any other way. Any ideas?
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    i think you'll find it's actually cos(x)...
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    Length of side of square = 2r sin(x/2)

    Area = length squared
    =4 r^2 sin^2(x/2)

    sinx = 1 - 2 sin^2(x/2)
    =>2sin^2(x/2) = 1 - sin(x)
    =>A = 2r^2 [1 - sin(x)]
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    o_0 one moment

    yeah, r=5z/8

    r2 = z2/4 +(z-r)2

    this is from one of the little triangles, across horizontally from the centre to the edge of the square (z/2) , up, then down the radius
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    (Original post by e-unit)
    sinx = 1 - 2 sin^2(x/2)
    no, cosx = 1 - 2 sin^2(x/2)
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    (Original post by chewwy)
    no, cosx = 1 - 2 sin^2(x/2)
    SUGAR. I'm having a mid-life mathematical crisis.
 
 
 
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