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    (Original post by masryboy94)
    thank you both !!! now i get it

    (Original post by gaffer dean)
    Are you sure? It's important to Remember these identities:
    sin^2x=\frac{-1}{2}(cos2x-1), cos^2x= \frac{1}{2}(cos2x+1), cos4x=2cos^2x-1
    Yeah definitely remember those identities, or at least remember how to derive them from  sin^2x + cos^2x = 1 and  cos2x = cos^2x - sin^2x
    I'd recommend writing them down at the side of any working out so that you are sure of the trig identities that you use, and to possibly remind yourself that you can use them if you get stuck in a question
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    (Original post by gaffer dean)
    It's important to Remember these identities:
    sin^2x=\frac{-1}{2}(cos2x-1), cos^2x= \frac{1}{2}(cos2x+1), cos4x=2cos^2x-1
    how about when it is 2cos^2 2x-1 because i've seen it in a couple of questions and i didn't know which identity matched with that?
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    (Original post by gaffer dean)
    It's important to Remember these identities:
    sin^2x=\frac{-1}{2}(cos2x-1), cos^2x= \frac{1}{2}(cos2x+1), cos4x=2cos^2x-1
    btw isn't cos2x=2cos^2x-1 because you wrote cos4x=2cos^2x-1 ??
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    (Original post by masryboy94)
    how about when it is 2cos^2 2x-1 because i've seen it in a couple of questions and i didn't know which identity matched with that?
    (Original post by masryboy94)
    btw isn't cos2x=2cos^2x-1 because you wrote cos4x=2cos^2x-1 ??
    I've edited my post now, Sorry I was supposed to say: 2cos^2 2x-1=cos4x
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    (Original post by gaffer dean)
    I've edited my post now, Sorry I was supposed to say: 2cos^2 2x-1=cos4x
    ooo okay, yep that answers my question. THANK YOU !
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    (Original post by gaffer dean)
    I've edited my post now, Sorry I was supposed to say: 2cos^2 2x-1=cos4x
    is that the same for when 1-2sin^2 2x=cos4x ?
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    (Original post by masryboy94)
    is that the same for when 1-2sin^2 2x=cos4x ?
    yes because 2cos^2 2x-1=cos4x can be written as 2(1-sin^22x)-1 (Since cos^22x=1-sin^22x) = 2-2sin^22x-1=1-2sin^22x
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    (Original post by gaffer dean)
    yes because 2cos^2 2x-1=cos4x can be written as 2(1-sin^22x)-1 (Since cos^22x=1-sin^22x) = 2-2sin^22x-1=1-2sin^22x
    ahh brilliant thank you very much !
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    6x + 9y -2pie = 0 this is mark scheme answer

    i got

    -6x - 9y + 2pie =0

    I mean the only difference is that they transferred everything to the other side, to make positive but I did not? will I still get the mark
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    (Original post by masryboy94)
    ahh brilliant thank you very much !
    No Problem.
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    (Original post by gaffer dean)
    yeah you would still get the marks.
    but when I sub x=5 and y=9 for example
    I get postitive value and the other one negatvie value
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    (Original post by otrivine)
    but when I sub x=5 and y=9 for example
    I get postitive value and the other one negatvie value
    wait what's the question?
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    (Original post by gaffer dean)
    wait what's the question?
    the question was on jan 2010 question 3)c)
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    (Original post by otrivine)
    6x + 9y -2pie = 0 this is mark scheme answer

    i got

    -6x - 9y + 2pie =0

    I mean the only difference is that they transferred everything to the other side, to make positive but I did not? will I still get the mark
    Yes, it's the same thing.
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    (Original post by justinawe)
    Yes, it's the same thing.
    so why when I sub x=5 and y=9 on to either equation I get one being 104.something and the other being -104.something
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    (Original post by otrivine)
    so why when I sub x=5 and y=9 on to either equation I get one being 104.something and the other being -104.something
    What?

    They're both supposed to equal 0... they can't equal 104 or -104. Why did you sub in x=5 and y=9?

    6x + 9y -2\pi = -(-6x - 9y + 2\pi), this is why you get the positive for one and negative for the other.

    However, the equation of the line is 6x + 9y -2\pi = 0 . This means it can only equal 0, it can't equal 104 or -104 or whatever.

    see here,

    6x + 9y -2\pi = 0

    -(-6x - 9y + 2\pi) = 0

    -6x - 9y + 2\pi = -0

    but "negative zero" is still zero, so:

    -6x - 9y + 2\pi = 0
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    (Original post by justinawe)
    What?

    They're both supposed to equal 0... they can't equal 104 or -104. Why did you sub in x=5 and y=9?

    6x + 9y -2\pi = -(-6x - 9y + 2\pi), this is why you get the positive for one and negative for the other.

    However, the equation of the line is 6x + 9y -2\pi = 0 . This means it can only equal 0, it can't equal 104 or -104 or whatever.

    see here,

    6x + 9y -2\pi = 0


    -(-6x - 9y + 2\pi) = 0

    -6x - 9y + 2\pi = -0


    but "negative zero" is still zero, so:

    -6x - 9y + 2\pi = 0
    So would I get full marks
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    (Original post by otrivine)
    So would I get full marks
    yes
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    For C3 and C4, Here's a good way of remembering differential and integral of sinx, cosx, -sinx and -cosx (these will not be given in the formula booklet):
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    anyone know where I can get some differential questions in context questions? done the solomon ones, but would like some more -hate them aha!
 
 
 
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