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Inverse trigonometric functions

  1. Jun 23, 2015 #1
    I am familiar with the importance of the following inverse circular/hyperbolic functions:
    ##\sin^{-1}##, ##\cos^{-1}##, ##\tan^{-1}##, ##\sinh^{-1}##, ##\cosh^{-1}##, ##\tanh^{-1}##.
    However, I don't really get the point of ##\csc^{-1}##, ##\coth^{-1}##, and so on.
    Given any equation of the form ##f(x) = a##, where ##f## is the reciprocal of any circular/hyperbolic function, we can always write the equation as ##\frac{1}{f(x)} = \frac{1}{a}## before solving it with the aid of the six aforementioned inverse functions.
    For instance, ##\csc{x} = 2##, ##0 < x < 2\pi##.
    ##\csc{x} = 2##
    ##\sin{x} = \frac{1}{2}##
    ##x = \frac{\pi}{6}, \frac{5\pi}{6}##
    Do we use them to avoid cluttered notation in calculus?
    For example, for ##x \geq 1##:
    $$\int \frac{1}{1 - x^2} dx = \coth^{-1}{x} + C = \frac{1}{2} \ln{\frac{x + 1}{x - 1}} + C$$
     
    Last edited by a moderator: Jun 23, 2015
  2. jcsd
  3. Jun 23, 2015 #2

    Mark44

    Staff: Mentor

    If the cosecent function is defined, it might be useful in some circumstances to have an inverse. Same for the hyperbolic cotangent.
     
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