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    http://mei.org.uk/files/papers/c410ju_weio.pdf

    q3 - i dont understand how to do it
    i got to half sin 2 theta but no idea from there
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    (Original post by liverpool2044)
    http://mei.org.uk/files/papers/c410ju_weio.pdf

    q3 - i dont understand how to do it
    i got to half sin 2 theta but no idea from there
    All you have to do here is just show that, given the parameterisation, show that x^2+4y^2=1. So you know that x=\cos 2\theta and y=\sin \theta \cos \theta= \frac{1}{2} \sin 2\theta. Sub those into x^2+4y^2 and you should get 1 easily from \cos^2 2\theta +\sin^2 2\theta=1.
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    (Original post by IrrationalRoot)
    All you have to do here is just show that, given the parameterisation, show that x^2+4y^2=1. So you know that x=\cos 2\theta and y=\sin \theta \cos \theta= \frac{1}{2} \sin 2\theta. Sub those into x^2+4y^2 and you should get 1 easily from \cos^2 2\theta +\sin^2 2\theta=1.
    when i do this i get 2sin^2 2theta?
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    (Original post by liverpool2044)
    when i do this i get 2sin^2 2theta?
    Which part do you get 2\sin^2 2\theta for? Is that what you got for y? If that's the case, then remember that \sin 2\theta=2\sin \theta \cos \theta \Rightarrow \sin \theta \cos \theta=\frac{1}{2} \sin 2\theta.
    So x^2+4y^2=\cos^2 2\theta+4(\frac{1}{2} \sin 2\theta)^2 and you should be able to finish it from there.
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    (Original post by IrrationalRoot)
    Which part do you get 2\sin^2 2\theta for? Is that what you got for y? If that's the case, then remember that \sin 2\theta=2\sin \theta \cos \theta \Rightarrow \sin \theta \cos \theta=\frac{1}{2} \sin 2\theta.
    So x^2+4y^2=\cos^2 2\theta+(\frac{1}{2} \sin 2\theta)^2 and you should be able to finish it from there.
    4(0.5sin2theta)^2? im not getting the value
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    (Original post by liverpool2044)
    4(0.5sin2theta)^2? im not getting the value
    y = sin(theta)cos(theta)
    therefore, 2y = 2sin(theta)cos(theta)
    therefore, 2y = sin(2theta)
    you have x = cos(2theta) and y = sin(2theta)
    cos(2theta)^2 + sin(2theta)^2 = 1
    I imagine you can do the rest (just subbing in y and x appropriately)
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    (Original post by liverpool2044)
    4(0.5sin2theta)^2? im not getting the value
    x^2+4y^2=\cos^2 2\theta+4(\frac{1}{2} \sin 2\theta)^2=\cos^2 2\theta +4(\frac{1}{4} \sin^2 2\theta)^2=\cos^2 2\theta+\sin^2 2\theta=1, as required.

    The last part follows from the identity  \cos^2 x+\sin^2 x \equiv 1.
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    ahh makes sense now thanks guys
 
 
 
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