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    A curve C has parametric equations:
    x=cos A and y=cos 3A where A lies between 0 and π inclusively

    Expand cos(2A+A), and hence show that an equation satisfied by all points on C is y=4x³-3x

    Really quite stuck on this one.....any help?
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    (Original post by corkskrew)
    A curve C has parametric equations:
    x=cos A and y=cos 3A where A lies between 0 and π inclusively

    Expand cos(2A+A), and hence show that an equation satisfied by all points on C is y=4x³-3x

    Really quite stuck on this one.....any help?
    cos(2A+A)=cos2AcosA-sin2AsinA
    =cosA[2(cosA)^2-1]-sinA[2sinAcosA]
    =2(cosA)^3-cosA-2(sinA)^2cosA
    =2(cosA)^3-cosA-2[1-(cosA)^2].cosA
    =2(cosA)^3-cosA-2cosA+2(cosA)^3
    =4(cosA)^3-3cosA

    But y=cos(2A+A)=4(cosA)^3-3cosA
    But x=cosA
    so y=4x³-3x
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    (Original post by corkskrew)
    A curve C has parametric equations:
    x=cos A and y=cos 3A where A lies between 0 and π inclusively

    Expand cos(2A+A), and hence show that an equation satisfied by all points on C is y=4x³-3x

    Really quite stuck on this one.....any help?
    It's worth learning the 2 identities

    Sin3A = 3sinA - 4sin³A
    Cos3A = 4cos³A- 3cosA

    Here you have to do the expansion, but if you are allowed to avoid it...

    Put y=cos 3A and x=cosA in the second of these to get y=4x³-3x

    Aitch
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    Thanks for your help.
    Might add, I'm not the mathematician - go a friend round for revision and he came a cropper on a P3 paper
    Thanks again.

    A
 
 
 
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