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    How do you get the TI-83 to display the graph of

    y = arcsec(ex) ?

    I ended up getting the graph of y=ln(1/cosx)

    [in place of x=ln(1/cosy)]

    and transposing it... There must be a correct way to display it!

    Aitch
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    1 / cos^-1(exp(x)) ??? did that work?
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    (Original post by Aitch)
    How do you get the TI-83 to display the graph of

    y = arcsec(ex) ?

    I ended up getting the graph of y=ln(1/cosx)

    [in place of x=ln(1/cosy)]

    and transposing it... There must be a correct way to display it!

    Aitch
    secy=e^x
    cosy=e^-x
    y=arcos(e^-x).
    Will it graph it in the simpler form?
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    (Original post by DaveManUK)
    1 / cos^-1(exp(x)) ??? did that work?
    No, it didn't! I thought that would be correct. That's exactly what I input. I have it in front of me. The graph it displays looks like half of y = -1/x

    Question relates to Jun 03 Edexcel P5 Q6b.


    Aitch
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    (Original post by Aitch)
    No, it didn't! I thought that would be correct. That's exactly what I input. I have it in front of me. The graph it displays looks like half of y = -1/x

    Question relates to Jun 03 Edexcel P5 Q6b.


    Aitch
    its the cos^-1 button not cos ^ (-1).
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    (Original post by Gaz031)
    secy=e^x
    cosy=e^-x
    y=arcos(e^-x).
    Will it graph it in the simpler form?
    Yes! It works!

    Thanks

    Aitch
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    Are you allowed a TI-83 for P5? Doesn't that do symbolic algebra and stuff?

    I'm happy with my little Casio fx-570MS anyway: does vectors, complex numbers, matrices and so on...
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    When x is near to 0 the gradient is ∞ so the curve is parallel to the y axis.
    As x increases, e^x increases. The gradient decreases. e^-x goes to 0 so as x tends to infinty y tends to arcos0=pi/2. Also at x=0, y=0.
    I'd guess the curve would be a 'hill' that is almost parallel to the y axis around the origin but levels off to have an asymptote y=pi/2.
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    (Original post by Gaz031)
    When x is near to 0 the gradient is ∞ so the curve is parallel to the y axis.
    As x increases, e^x increases. The gradient decreases. e^-x goes to 0 so as x tends to infinty y tends to arcos0=pi/2. Also at x=0, y=0.
    I'd guess the curve would be a 'hill' that is almost parallel to the y axis around the origin but levels off to have an asymptote y=pi/2.
    Yes. Have you done this paper? (Jun 03) It was the only bit I got stuck on!

    Aitch
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    (Original post by drw25)
    Are you allowed a TI-83 for P5? Doesn't that do symbolic algebra and stuff?

    I'm happy with my little Casio fx-570MS anyway: does vectors, complex numbers, matrices and so on...
    No, it doesn't do symbolic algebra. I think you need the TI-89 or TI-92 for that. (If you want to be arrested by the Edexcel police, that is...)

    Aitch
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    (Original post by Aitch)
    Yes. Have you done this paper? (Jun 03) It was the only bit I got stuck on!

    Aitch
    I will have done it at sometime. I did the P5 review exercise yesterday and am currently redoing the P6 one, as there is no way i'm going to spend both Monday and Tuesday revising for my general studies exams on Wednesday :rolleyes: I wish my general studies papers were multiple choice so I didn't have to write an essay.
    At least most of my exams are next week so there isn't too long to wait.
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    Which FP? module does Pure 6, correspond to? I thought it must be FP2 / FP3

    Which module involves stuff on de moivre's theorem? I thought it might be P6, but I can't find it in the MEI OCR P6 text book. :rolleyes:
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    (Original post by Vijay1)
    Which FP? module does Pure 6, correspond to? I thought it must be FP2 / FP3

    Which module involves stuff on de moivre's theorem? I thought it might be P6, but I can't find it in the MEI OCR P6 text book. :rolleyes:
    With edexcel FP3=P6 and contains de moivre's theorem.
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    (Original post by Gaz031)
    With edexcel FP3=P6 and contains de moivre's theorem.
    Thanks, That must mean its not on the Mei syllabus :rolleyes:
 
 
 
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