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    What is the difference between arccosHx and arcosHx? [FP3]
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    (Original post by Mihael_Keehl)
    What is the difference between arccosx and arcosx? [FP3]
    None, that I'm aware of. There is also acosx (at least on my calculator). All represent the inverse cosine function.
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    Absolutely none!
    They are also the same as \cos^{-1}(x).
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    (Original post by ghostwalker)
    None, that I'm aware of. There is also acosx (at least on my calculator). All represent the inverse cosine function.
    (Original post by Mpagtches)
    Absolutely none!
    They are also the same as \cos^{-1}(x).
    http://qualifications.pearson.com/co...cal-Tables.pdf

    In the FP3 section, they are listed as having different derivates smh...
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    (Original post by Mihael_Keehl)
    http://qualifications.pearson.com/co...cal-Tables.pdf

    In the FP3 section, they are listed as having different derivates smh...
    Can't see it. Are you misreading arcosh x ?
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    (Original post by ghostwalker)
    Can't see it. Are you misreading arcosh x ?
    sorry yes, I have reading problems
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    (Original post by Mihael_Keehl)
    sorry yes, I have reading problems
    No problem.
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    (Original post by ghostwalker)
    No problem.
    Sorry but what is the answer
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    xxx

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    (Original post by Mihael_Keehl)
    Sorry but what is the answer
    arccos is the inverse of the cosine (a trigometric function)

    arcosh is the inverse of the hyperbolic cosine, cosh
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    (Original post by ghostwalker)
    arccos is the inverse of the cosine (a trigometric function)

    arcosh is the inverse of the hyperbolic cosine, cosh
    I meant ARCOSH AND ARCCOSH sorry if I am being unclear. But thanks for the info.
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    (Original post by Mihael_Keehl)
    I meant ARCOSH AND ARCCOSH sorry if I am being unclear. But thanks for the info.
    There is no arccosh in that document.

    Can you reference, get an image of, the two lines you're having problems with.
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    (Original post by ghostwalker)
    There is no arccosh in that document.

    Can you reference, get an image of, the two lines you're having problems with.
    So it is safe to say that arCCosh x does not exist?

    And finally why are the hyperbolic idenities omititng a c, why AR instead of ARC.

    Thank you for your perneiall help.
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    (Original post by Mihael_Keehl)
    So it is safe to say that arCCosh x does not exist?
    In a similar fashion to the cosine, you could have:

    arcosh, arccosh, or acosh, all representing the same thing, the inverse hyperbolic cosine.

    And finally why are the hyperbolic idenities omititng a c, why AR instead of ARC.
    No idea, but I refer you to this article (notation section) on wiki which expounds on the matter.
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    (Original post by ghostwalker)
    In a similar fashion to the cosine, you could have:

    arcosh, arccosh, or acosh, all representing the same thing, the inverse hyperbolic cosine.



    No idea, but I refer you to this article (notation section) on wiki which expounds on the matter.
    Kind Regards, thanks
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    This confuses me too. With the inverse trig functions you put 'arc' in front of them, even for cos which gives you a double c.

    arccos(x)

    However with the inverse hyperbolic functions you put 'ar' in front instead of 'arc'. Since you already have the h which tells you that it is hyperbolic I don't understand why they can't use 'arc' for hyperbolic functions.

    In my brain arcosh(x) should be arccosh(x).
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    (Original post by Louisb19)
    This confuses me too. With the inverse trig functions you put 'arc' in front of them, even for cos which gives you a double c.

    arccos(x)

    However with the inverse hyperbolic functions you put 'ar' in front instead of 'arc'. Since you already have the h which tells you that it is hyperbolic I don't understand why they can't use 'arc' for hyperbolic functions.

    In my brain arcosh(x) should be arccosh(x).
    From memory I believe it is "arsinh" because "ar" stands for an area.

    We call the inverse of sin x arcsin x because it returns the length of an arc of a unit circle corresponding to that angle.
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    Personally, I always use just "a".
 
 
 
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