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AQA A2 Mathematics MPC4 Core 4 - 9th June 2015 [Discussion & unofficial markscheme] Watch

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    (Original post by datpr0)
    No, they're not equivalent. You would have to write cos(4x) as cos(2x + 2x) and go from there.
    How would you do something like cos(2x +2x)? And do you think that kind of thing might come up - I thimk we've only really done it with single 2x using double angle formulae..
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    (Original post by datpr0)
    Well that's because they're a pain in the a**! ...and everything else is pretty straight forward once you've done a few past papers.
    Yeah I know I hate them too, but I mean that it seems everyone commenting thinks that it's just a personal weakness when really they're just awful haha
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    Can anyone tell me where the ln x comes from question 8 c in june2014?
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    I don't understand what the markscheme has done for 7c.
    I calculated B and then worked out the distance between AB, however they state that AB^2 = 2AP^2
    Where does that come from? :/

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    (Original post by Jankie)
    With vectors, how do you use the AB = OA+OB? Like how does that work? Thanks!
    I could explain it, but it would take too long, so i wont explain it. Sorry for being useless.
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    (Original post by Haza100)
    Would also like to know this.

    Would something like Cos(4x) be equivalent to
    2Cos^2(2x) - 1
    1 - 2Sin^2(2x)
    cos^2(2x) - sin^2(2x)
    I'm pretty sure those are all correct.
    I just did a trig question with 4x in them and it worked.
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    (Original post by Becky J)
    How would you do something like cos(2x +2x)? And do you think that kind of thing might come up - I think we've only really done it with single 2x using double angle formulae..
    Well, remember the double angle formulae are derivatives of the formulae given in the formula booklet regarding sin(A+B), cos(A+B) and tan(A+B). That is why you rewrite cos4x as cos(2x + 2x), because it is now in the form cos(A+B) which you know from the formula booklet equals cosAcosB - sinAsinB (in this case cos2xcos2x - sin2xsin2x).

    Which simplifies to cos^2(2x) - sin^2(2x). That's interesting, lol !
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    Is the value of tanB negative because B is obtuse? How does that work?


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    (Original post by datpr0)
    No, they're not equivalent. You would have to write cos(4x) as cos(2x + 2x) and go from there.
    I'm probably missing something totally obvious (which is a little worrying haha), can you explain in a little more detail how you would identify what cos(4x) is equal to?

    Simply using the DA formulae I get the following:
    Cos(4x) == Cos(2x + 2x) == Cos(2x) * Cos(2x) - Sin(2x) * Sin(2x) == Cos^2(2x) - Sin^2(2x)
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    (Original post by emsca)
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    Is the value of tanB negative because B is obtuse? How does that work?


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    Think about cast circle and where acute angles, obtuse angles and reflex angles are. Here this might help some what. Sine you are given a positive sin value it must be in the second quadrant.

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    (Original post by Jimmy20002012)
    Think about cast circle and where acute angles, obtuse angles and reflex angles are. Here this might help some what. Sine you are given a positive sin value it must be in the second quadrant.

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    Thank you so much!


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    (Original post by Haza100)
    I'm probably missing something totally obvious (which is a little worrying haha), can you explain in a little more detail how you would identify what cos(4x) is equal to?

    Simply using the DA formulae I get the following:
    Cos(4x) == Cos(2x + 2x) == Cos(2x) * Cos(2x) - Sin(2x) * Sin(2x) == Cos^2(2x) - Sin^2(2x)
    Yes, I reached the same conclusion. I believe it's because you can split 4x into two equal parts. A better example would be cos3x/sin3x/tan3x, which they would be more likely to ask as it doesn't simplify so nicely.
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    (Original post by emsca)
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    Is the value of tanB negative because B is obtuse? How does that work?


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    Can someone explain this to me without the use of a CAST diagram?
    (My college never taught them)
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    With a question like this is it just convention to label the sides of the parallelogram clockwise starting from bottom left? I tried this question and I labelled my diagram in a different way from the markscheme and got a different answer. How are you supposed to know which corner is D?
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    (Original post by saad97)
    I don't understand what the markscheme has done for 7c.
    I calculated B and then worked out the distance between AB, however they state that AB^2 = 2AP^2
    Where does that come from? :/

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    Oh wow where does that come from?? I got |AP|=|BP|=6sqrt(14)
    What paper is this? which year?
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    (Original post by mamacita)
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    With a question like this is it just convention to label the sides of the parallelogram clockwise starting from bottom left? I tried this question and I labelled my diagram in a different way from the markscheme and got a different answer. How are you supposed to know which corner is D?
    Is this c4?? Aqa c4?? :O No way...
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    question - http://prntscr.com/7eopwv
    ms - http://prntscr.com/7eoq1x
    my workings - http://prntscr.com/7eoyez

    would I get the marks?
    from june 2014.
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    (Original post by bakedbeans247)
    Is this c4?? Aqa c4?? :O No way...
    Noooo it's from an OCR paper but the one's I was doing have the same content as AQA, I would imagine they'd give us a diagram?
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    (Original post by bakedbeans247)
    Oh wow where does that come from?? I got |AP|=|BP|=6sqrt(14)
    What paper is this? which year?
    It's June 2007 Q7.
    My answer was sqrt(1085) :/
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    Quick question! Does 2p^t = q^t equal

    2 ln p^t = ln q^t

    I'm just unsure what to do with the 2!

    Thanks
 
 
 
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