1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Exponential forms

  1. Oct 19, 2004 #1
    I'm try to prove some trig identities and I need to know how to write the following functions in their exponential forms:

    [tex]sin^{-1}z[/tex]
    [tex]cos^{-1}z[/tex]
    [tex]sinh^{-1}z[/tex]
    [tex]cosh^{-1}z[/tex]

    I know how the sine and cosine functions are expressed with exponents, but I'm not sure how that would translate to the inverses of the functions.
     
  2. jcsd
  3. Oct 19, 2004 #2

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    If the trig functions can be written in exponential form then the inverse functions would be expressible in logarithmic form!

    E.g. Let

    [tex]\sin z = \frac {e^{i z} - e^{-iz}}{2i}[/tex]

    then solve for [itex]e^{iz}[/itex] in terms of [itex]\sin z[/itex]. Once you have your expression for [itex]e^{iz}[/itex] then just find the natural logarithm of both sides and you have your inverse function!
     
  4. Oct 19, 2004 #3

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Just a reminder:
    Since the complex logarithm is a multi-valued function (i.e, not a "usual" type of function), you must choose a branch of it to gain a "proper" inverse function.
     
  5. Oct 19, 2004 #4
    Maybe I could get a little more help. If I start with
    [tex]\sin z = \frac {e^{i z} - e^{-iz}}{2i}[/tex]

    then I take the 2i term to the left and then take the natural log of both sides I get the following:

    [tex]ln2i*sinz=ln(e^{2iz}-1)-iz[/tex]

    [tex]ln(2i*sinz)+iz=ln(e^{2iz}-1)[/tex]

    [tex]-2zsinz+1=e^{2iz}[/tex]

    [tex]e^{iz}=(-2zsinz+1)^{\frac{1}{2}}[/tex]

    But now I don't see how to use that. What I am trying to do is show that [tex]\sin^{-1}z=-iln({iz+(1-z^2)^{\frac{1}{2}}})[/tex]
     
  6. Oct 19, 2004 #5

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Set y=sin(z).
    Then:
    [tex]2iye^{iz}=e^{2iz}-1[/tex]
    [tex](e^{iz})^{2}-2iye^{iz}-1=0[/tex]
    Or:
    [tex]e^{iz}=\frac{2iy\pm\sqrt{-4y^{2}+4}}{2}[/tex]
    Or:
    [tex]z=sin^{-1}(y)=-iln(iy\pm\sqrt{1-y^{2}})[/tex]
     
  7. Oct 19, 2004 #6

    That is unbelievably easy. Thanks for the help, guys.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Exponential forms
  1. Exponential decay (Replies: 3)

  2. Complex Exponentials (Replies: 6)

Loading...