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Cosxsinx P4 watch

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    Hey guys sorry but i'm stuck in a bit of a :eek: with this q on my P4 paper...

    Ok i've got a curve y = 3sin2x + cos2x and prior to doing a volume of revolution bit i need to square y and write it in the form Asin4x + Bcos4x + c

    With me so far?

    Now this is how far i got

    (3sin2x + cos2x)^2 = 9sin^2(4x) + cos^2(4x) + 6sin2xcos2x

    By using 2sinxcosx = sin 2A i can get that to:

    9sin^2(4x) + cos^2(4x) + 3sin4x

    But where do i go from here? Please help!!
    Love Boo xx
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    why do you need to get it in that form?
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    also when you square, the first line should read

    (3sin2x + cos2x)^2 = 9sin^2(2x) + cos^2(2x) + 6sin2xcos2x

    which will lead to

    3sin4x - 4cos4x + 5

    or something like that
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    Because the exam question tells me to...
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    (3sin2x + cos2x)² = 9sin²(2x) + cos²(2x) + 6sin(2x)cos(2x)

    now use,

    sin²(2x) = ½(1 - cos4x)
    cos²(2x) = ½(1 + cos4x)
    2sin2xcos2x = sin4x
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    (3sin2x + cos2x)^2 = 9(sin^2 2x) + (cos^2 x) + 6sin2xcos2x

    = 9[(0.5)(1-cos4x)] + (0.5)(1+cos4x) + 3[2sin2xcos2x] <---using half angle identities

    = 9/2 - (9/2)cos4x + (1/2) + (1/2)cos4x + 3sin4x

    = 3sin4x - (9/2)cos4x + 5
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    (Original post by Bekaboo)
    Hey guys sorry but i'm stuck in a bit of a :eek: with this q on my P4 paper...

    Ok i've got a curve y = 3sin2x + cos2x and prior to doing a volume of revolution bit i need to square y and write it in the form Asin4x + Bcos4x + c

    y=9sin^2(2x)+cos^2(2x)+6cos2xsin 2x
    sin 2x=2cosxsinx so 3sin 4x=6cos2xsin2x
    sin^2 (2x)+cos^2 (2x)=1
    so 9sin^2(2x)+cos^2(2x)=8 sin^2 2x+1
    but cos 4x=cos^2 2x-sin ^2 2x=(1-2sin^2 (2x))
    so
    4(1-cos 4x)=8sin^2 (2x)
 
 
 
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