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    As I don't have the answers to the P2 paper my teacher has set us for hw, i never know if I'm doing the right method! So can someone check my method below and tell me if there's anywhere I've gone right and if it looks correct. Thanks

    Find the value of z to 2dp in range -pi<z<pi

    cosz + sinz = 1/3

    If square both sides:

    [cosz]^2 + [sinz]^2 + 2coszsinz = 1/9

    1 + sin2z = 1/9

    2z = -1.09; 2.05; -4.24; 5.19
    z = -0.55; 1.03; -2.12; 2.60

    The bit I'm not sure about is my final values. I'm find with working out the other values in the range if it's positive, but I get all in a muddle if it's negative, so not sure they're right!

    Thanks - No doubt i'll post some more questions later!
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    Your method looks fine to me and your answers are all within the range, so I think you've done it ok.
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    check your answers by putting them back in.
    You know that sin2z = -0.888888...
    so use your calculator to find out the values of,

    sin(-1.09)
    sin(2.05)
    sin(-4.24)
    sin(5.19)

    They're all approx. correct.

    the best way of finding out if you are right is to plug your answers back into the equation and seeing if you get 1/3 out.
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    (Original post by GeneralGrievous)
    the best way of finding out if you are right is to plug your answers back into the equation and seeing if you get 1/3 out.
    Thanks for everyone's help! And silly me for not thinking about checking my answers!
 
 
 
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