Edexcel FP2 friday 19th June (OLD spec)

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  1. Smith987's Avatar
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    Re: Edexcel FP2 friday 19th June (OLD spec)
    (Original post by xReenax)
    Is anyone doing the new spec...it feels like im the only one in the world doing it
    Don't think so :P http://www.thestudentroom.co.uk/showthread.php?t=914512
  2. Zygroth's Avatar
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    Re: Edexcel FP2 friday 19th June (OLD spec)
    (Original post by romski)
    or 1/(3+5cos*2) ,<-- in this case you will use tanx=t

    1/(3+5cos) <,-- in this case you will use tanx/2=t
    :confused: Why? What's the difference?
  3. Smith987's Avatar
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    Re: Edexcel FP2 friday 19th June (OLD spec)
    Its to do with getting rid of the square roots in the equation which makes the substitution so easy.
  4. SimonM's Avatar
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    Re: Edexcel FP2 friday 19th June (OLD spec)
    (Original post by Smith987)
    we dont have to use the substitution t=tanx do we?!

    If they do ask us, will they tell us "use the substitution..." or do they assume we know it? I'm concerned now :'(
    I think you can be fairly confident that they will tell you
  5. Zygroth's Avatar
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    Re: Edexcel FP2 friday 19th June (OLD spec)
    Hmm am I right in saying that if you use tan x = t, then:
    \cos^2 x = \dfrac{1}{1+t^2}
    and
    \sin^2 x = \dfrac{t^2}{1+t^2}
  6. SimonM's Avatar
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    Re: Edexcel FP2 friday 19th June (OLD spec)
    (Original post by Zygroth)
    Hmm am I right in saying that if you use tan x = t, then:
    \cos^2 x = \dfrac{1}{1+t^2}
    and
    \sin^2 x = \dfrac{t^2}{1+t^2}
    Yes.

    1+t^2 = 1+\tan^2 x = \sec^2 x

    So:

    \displaystyle \frac{1}{1+t^2} = \frac{1}{\sec^2 x} = \cos^2 x

    \displaystyle \frac{t^2}{1+t^2} = 1 - \frac{1}{t^2+1} = 1 - \cos^2 x = \sin^2 x
  7. Smith987's Avatar
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    Re: Edexcel FP2 friday 19th June (OLD spec)
    think so.. and if you use tan(x/2), you end up getting cosx= blah blah, with roots involved, but then sinx = 2cos(x/2)sin(x/2) which gets rid of the roots problem, if that makes sense?
  8. Smith987's Avatar
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    Re: Edexcel FP2 friday 19th June (OLD spec)
    (Original post by SimonM)
    Yes.

    1+t^2 = 1+\tan^2 x = \sec^2 x

    So:

    \displaystyle \frac{1}{1+t^2} = \frac{1}{\sec^2 x} = \cos^2 x

    \displaystyle \frac{t^2}{1+t^2} = 1 - \frac{1}{t^2+1} = 1 - \cos^2 x = \sin^2 x
    or just in terms of a triangle? the hypoteneuse would be sqrt(1+x^2) and x is the angle, easy to see.
  9. SimonM's Avatar
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    Re: Edexcel FP2 friday 19th June (OLD spec)
    (Original post by Smith987)
    think so.. and if you use tan(x/2), you end up getting cosx= blah blah, with roots involved, but then sinx = 2cos(x/2)sin(x/2) which gets rid of the roots problem, if that makes sense?
    You mean

    \displaystyle \cos x = \frac{1 - \tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}?
  10. Smith987's Avatar
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    Re: Edexcel FP2 friday 19th June (OLD spec)
    Not exactly, I was trying to help out Zygroth, but I'm probs not making sense haha
  11. Zygroth's Avatar
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    Re: Edexcel FP2 friday 19th June (OLD spec)
    Lol someone explained to me the tan x/2 = t one already. I had to look it up again though, cause it was ~2 weeks ago...Just goes to show that starting revision early does absolutely nothing :p:
  12. Smith987's Avatar
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    Re: Edexcel FP2 friday 19th June (OLD spec)
    Yup its all about the last minute cramming!!
  13. SimonM's Avatar
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    Re: Edexcel FP2 friday 19th June (OLD spec)
    (Original post by Zygroth)
    Lol someone explained to me the tan x/2 = t one already. I had to look it up again though, cause it was ~2 weeks ago...Just goes to show that starting revision early does absolutely nothing :p:
    Let \tan \frac{x}{2} = t, then

    \displaystyle \sin x = \frac{2t}{1+t^2}, \cos x = \frac{1-t^2}{1+t^2}, \tan x = \frac{2t}{1-t^2}, \frac{dx}{dt} = \frac{2}{1+t^2}

    And use normal substitution.
  14. Smith987's Avatar
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    Re: Edexcel FP2 friday 19th June (OLD spec)
    Does anyone have the June 08 mark scheme? its not on freeexampapers
  15. hamster123's Avatar
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    Re: Edexcel FP2 friday 19th June (OLD spec)
  16. Smith987's Avatar
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    Re: Edexcel FP2 friday 19th June (OLD spec)
    Thanks very much
  17. alexsheppard11's Avatar
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    Re: Edexcel FP2 friday 19th June (OLD spec)
    (Original post by Smith987)
    Does anyone have the June 08 mark scheme? its not on freeexampapers
    it is:p:

    http://www.freeexampapers.com/FreeEx...0wNiBNUy5wZGY=
  18. Smith987's Avatar
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    Re: Edexcel FP2 friday 19th June (OLD spec)
    Ooops :P totally missed that one hehe
  19. napanda's Avatar
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    Re: Edexcel FP2 friday 19th June (OLD spec)
    oh my, not looking forward to this exam at all!! but it's my last

    xx
  20. Zygroth's Avatar
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    Re: Edexcel FP2 friday 19th June (OLD spec)
    Ugh, I'm really regretting my early revision. I have 2 papers left (for tomorrow). Dunno what to do now I can't do the Qs in the book lol :p: My algebra manipulation isn't good enough for most of conic sections Anyone wanna teach me?
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