Exponential Forms of Inverse Trig & Hyperbolic Functions

[Hypnotoad]

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.

 

[Tide]

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!

 

[arildno]

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.

 

[Hypnotoad]

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!

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]

 

[arildno]

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]

 

[Hypnotoad]

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]

That is unbelievably easy. Thanks for the help, guys.