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I Inverse hyperbolic function expressed as inverse trigonometry function

  1. Jun 7, 2016 #1
    Consider ##y=\cos{-x}=\cos x=\cosh ix##.

    Thus, ##\pm x=\cos^{-1}y## and ##ix=\cosh^{-1}y##.

    So ##\cosh^{-1}y=\pm i\cos^{-1}y##.

    Renaming the variable ##y##, we have ##\cosh^{-1}x=\pm i\cos^{-1}x##.

    Next, we evaluate the derivative of ##\cosh^{-1}x## by converting it to ##\cos^{-1}x## using ##\cosh^{-1}x=i\cos^{-1}x##.

    ##\frac{d}{dx}\cosh^{-1}x=\frac{d}{dx}i\cos^{-1}x##
    ##=i\frac{-1}{\sqrt{1-x^2}}## --- (*)
    ##=\frac{-i}{i\sqrt{x^2-1}}## where I've used the rule ##\sqrt{-c}=i\sqrt{c}## for ##c>0##, since ##x>1##.
    ##=\frac{-1}{\sqrt{x^2-1}}##, which differs from the correct answer by a negative sign.

    This means we would get the correct answer had we used ##\cosh^{-1}x=-i\cos^{-1}x##.

    My question is how do we know when to add the negative sign.

    I realized that step (*) is wrong because the formula for the derivative of ##\cos^{-1}## is true only for ##-1<x<1##. However, since we get the right answer in the end, it seems like there is a reason why we could still use the formula for values of ##x## outside ##-1<x<1##, and an explanation when we should use ##\cosh^{-1}x=-i\cos^{-1}x## instead of ##\cosh^{-1}x=i\cos^{-1}x##.
     
  2. jcsd
  3. Jun 8, 2016 #2
    Hi Happiness:

    I am not sure I understand what you are asking, but it seems to me that the first equation omitted
    cosh ix = cosh -ix.​
    That is, both cos and cosh are symmetrical functions. Therefore when you take the inverse, you get two possible correct answers:
    + and -. This the same as taking the square root of a positive value and getting both a positive and negative result.

    Hope this helps.

    Regards,
    Buzz
     
  4. Jun 8, 2016 #3
    Hi Buzz

    Thanks for replying.

    I am asking why is ##\cosh^{-1}x=-i\cos^{-1}x##? And why isn't ##\cosh^{-1}x=+i\cos^{-1}x##? Why is the positive result rejected?
     
  5. Jun 8, 2016 #4
    Hi Happiness:

    Can you post a link to where you found that the positive result is rejected?

    Regards,
    Buzz
     
  6. Jun 9, 2016 #5
  7. Jun 9, 2016 #6
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