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    (Original post by amberf)
    Also help on jan 07 q8

    express arcsinx in terms of y ??!!!
    how do you get arcsinx = ∏/2 - y ??
    cosy=x

    cosy = sin(\frac{\pi}{2} - y) (sin is a translation of the cos graph by 90 degrees)

    sin(\frac{\pi}{2} - y)= x

    \frac{\pi}{2} - y= arcsinx
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    (Original post by samir12)
    There is another way of doing the question:


      x = 4sin(2y + 6)

     sin(2y + 6) = \dfrac{x}{4}

     sin^2(2y + 6) = \dfrac{x^2}{16}

     cos^2(2y+ 6) = 1 - \dfrac{x^2}{16}

     cos^2(2y+ 6) = \dfrac{16}{16} - \dfrac{x^2}{16}

     cos(2y+6) = \dfrac{\sqrt{16 - x^2}}{4}

     \dfrac{dy}{dx} = \dfrac{1}{8cos(2y+6)}

     \dfrac{dy}{dx} = \dfrac{1}{2\sqrt{16 - x^2}}

    yh i got up to the latter of the last 2 steps but i couldnt figure out how they got
     \dfrac{dy}{dx} = \dfrac{1}{8cos(2y+6)}
    to
     \dfrac{dy}{dx} = \dfrac{1}{2\sqrt{16 - x^2}}[/QUOTE]

    then i realised they used pythagorus
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    Gosh I'm really scared. :emo:
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    they give the differentiation formulae for sec, cosec, cot right?
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    (Original post by MadMaths)
    yh i got up to the latter of the last 2 steps but i couldnt figure out how they got
     \dfrac{dy}{dx} = \dfrac{1}{8cos(2y+6)}
    to
     \dfrac{dy}{dx} = \dfrac{1}{2\sqrt{16 - x^2}}
    I don't know why someone would actually do that method in the actual exam lol.
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    (Original post by Kameo)
    I don't know why someone would actually do that method in the actual exam lol.
    exactly....i thought it was abit odd
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    can I just ask something

    what does 'root 3' + (root 3)/2 =

    how do I find it in surd or exact form
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    btw guys, do you think the exam tomorrow will be alot more difficult since the last one (june 09) was relatively easy
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    you have to get a common denominator, so times root 3 by 2. then you would have (2root3 + root 3)/2, simply, so 3root3/2
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    (Original post by W.H.T)
    btw guys, do you think the exam tomorrow will be alot more difficult since the last one (june 09) was relatively easy
    might be the case...
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    Ooor just put Root3 + (root3)/2 into your caculator?
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    (Original post by Kameo)
    you have to get a common denominator, so times root 3 by 2. then you would have (2root3 + root 3)/2, simply, so 3root3/2
    thank you
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    I'm really confused what dy/dx of
    sin^2 x is (sin squared x)
    cos^2 x

    can anyone help?
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    (Original post by Mattjames)
    I'm really confused what dy/dx of
    sin^2 x is (sin squared x)
    cos^2 x

    can anyone help?
     y = sin^2x = (sinx)^2

    \frac{dy}{dx}= 2(sinx)^1(cosx)

    I don't see where you got cos^2 from.
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    (Original post by samir12)
     y = sin^2x = (sinx)^2

    \frac{dy}{dx}= 2(sinx)^1(cosx)

    I don't see where you got cos^2 from.

    i saw a question saying 3sin^2 x and just presumed there would be cos^2 somewhere.
    so how do you work out the sin^2 x?
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    oh right, the second line didnt come up at first
    and yer, i just presumed there would be dy/dx (cos^2 x) sort of thing.
    • PS Helper
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    PS Helper
    hy....only 2 hours for the exam to start!!!!
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    (Original post by kosy91)
    hy....only 2 hours for the exam to start!!!!
    Where are you taking your exam ?
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    (Original post by kosy91)
    hy....only 2 hours for the exam to start!!!!
    WTF. Please tell me it's an afternoon exam
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    ya its afternoon
 
 
 
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