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    havent revised
    not gonna revise
    oh well
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    crap crap crap can't do any of it
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    (Original post by jlcf)
    First re-arrange equation to in terms of y:

    y = [ arcsin(x/4) - 6 ] / 2

    Then sub y into the dy/dx:

    dy/dx= 1 / 8cos(2y+6) =>

    = 1 / 8cos{2[arcsin(x/4)-6]/2 +6} times 2 and divide 2 => 1 / cancels out
    -6 and +6 => 0

    = 1 / 8cos[arcsin(x/4)]

    Now arcsin(x/4) is the same as sin feta = x/4 ... draw triangle with opposite as 4 and hypotenuse as x

    Find adjacent which is √4² - x² => √16 - x²

    Cos feta is √16 - x² / 4 (ajacent over hypothenuse)

    = 1 / 8cos[arcsin(x/4)] = 1 / 8 (√16 - x² / 4 )

    => 1 / 2 √16 - x²

    Sorry hard to explain! Hope it helps!
    Ah ok cheers..at least I understand it a bit better!

    Hoping for nothing like this tomorrow :O
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    help with these questions needed.

    https://eiewebvip.edexcel.org.uk/Rep...e_20070118.pdf


    2b and c thanks.
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    (Original post by houseo2)
    help with these questions needed.

    https://eiewebvip.edexcel.org.uk/Rep...e_20070118.pdf


    2b and c thanks.
    2b complete the square
    2c numerator>0, look at denominator
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    (Original post by houseo2)
    help with these questions needed.

    https://eiewebvip.edexcel.org.uk/Rep...e_20070118.pdf


    2b and c thanks.
    For 2b you can also differentiate and find the minimum point of the curve. It will have a y value more than zero, proving that f(x) is always >0
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    who else is sitting the exam tomorrow? i did it in january and i got 83 but im resitting to try and boost my chance of an a*
    i have done all the past papers and im feeling alot more confident about it than in january!
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    Feeling ready for this exam - should get over 90UMS. :yeah:
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    I am sitting it =] You did really well! I am sitting it for the first time and I don't have high expectations ahaha!
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    (Original post by Mr ABC)
    2b complete the square
    2c numerator>0, look at denominator
    i know howto complete square but am unsure for this question could show how to do it
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    (Original post by game well and truly over)
    The sine of 2y+6 is x over 4. Doesnt matter what you call the angle, in this case, 2y plus 6. Call it theta or whatever. Draw a triangle whose opposite side is length x, with hypotenuse 4. Then the length of the adjacent side is the sq root of (16 minus x squared). The cosine is therefore root(16-x squared) over 4. The dreivative is therefore the inverse. See now? Much easier than all those complex explanations involving arcsin etc.
    i understand up until the inverse bit you said.
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    trigonometry
    how should we go about these questions? in the exam i usually panic and cant think properly
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    (Original post by houseo2)
    i know howto complete square but am unsure for this question could show how to do it
    if i remember right, it was x^2 + x + 1

    complete the square, =(x+0.5)^2 +0.75

    any real number squared >=0, so clearly the above >= 0.75
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    I'm just wondering in our working at differnt stages, if we don't have an exact value how many sig fig should we work with to avoid accuracy errors?

    Thanks
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    After solving too many questions on trigonometry I found out it's very easy and most questions on this topic are the same.
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    (Original post by FZka)
    After solving too many questions on trigonometry I found out it's very easy and most questions on this topic are the same.
    same in what sense?
    As in the "prove" questions and "solve"?
    I dont see double angles appearing a lot in the papers, what do you think? Also, I havent learnt the sum and product rules, do you think its important to know them?
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    3 sf
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    Just want to get this exam out the way now! I must remember to revise inverse trig graphs and graphs of cot/sec/cosec :eek: !
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    how do you do this question

    cosec^2 2x - cot2x = 1

    solve for 0 - 180 degrees

    7 marks
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    (Original post by dunnoaboutme)
    how do you do this question

    cosec^2 2x - cot2x = 1

    solve for 0 - 180 degrees

    7 marks
    What paper? Who questions are easier for these.
 
 
 
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