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Relation between inverse trigonometric function

by Jhenrique
Tags: function, inverse, relation, trigonometric
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Jhenrique
#1
Mar31-14, 10:22 PM
P: 686
Digging in the wiki, I found this relation between 'arc-functions' and 'arc-functions-hyperbolics"

[tex]\\ arcsinh(x)= i \arcsin(-ix) \\ arccosh(x)= i \arccos(+ix) \\ arctanh(x)= i \arctan(-ix)[/tex] https://it.wikipedia.org/wiki/Funzio...ento_complesso

Happens that I never see in anywhere a relation between those functions. This relationship is correct?
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D H
#2
Mar31-14, 11:19 PM
Mentor
P: 15,147
Quote Quote by Jhenrique View Post
This relationship is correct?
The second one is incorrect, and the other two are obvious.
Jhenrique
#3
Apr1-14, 01:38 AM
P: 686
Quote Quote by D H View Post
The second one is incorrect, and the other two are obvious.
And which is the correct form for the second?
Also, where can I find a full list (and correct)?

Jhenrique
#4
Apr3-14, 06:55 PM
P: 686
Relation between inverse trigonometric function

Hey man, you'll let me in the doubt!?
craigi
#5
Apr3-14, 07:17 PM
P: 419
Quote Quote by Jhenrique View Post
And which is the correct form for the second?
Also, where can I find a full list (and correct)?
cosh(ix) = cos(x)

therefore:

arcosh(x) = i arccos(x)

It's in the link you provided in the first post. You transcribed it incorrectly, that is all.

All you need to prove the others is:

sinh(ix) = i sin(x) and
tanh(ix) = i tan(x)

Give it a go, if can't work it out - ask again.
Jhenrique
#6
Apr3-14, 09:10 PM
P: 686
So, following your ideia, I got:

asin(x) = -i asinh(+i x)
acos(x) = -i acosh( x)
atan(x) = -i atanh(+i x)
acot(x) = -i acoth(-i x)
asec(x) = -i asech( x)
acsc(x) = -i acsch(-i x)

asinh(x) = -i asin(+i x)
acosh(x) = -i acos( x)
atanh(x) = -i atan(+i x)
acoth(x) = -i acot(-i x)
asech(x) = -i asec( x)
acsch(x) = -i acsc(-i x)

Correct?
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Jhenrique
#7
Apr4-14, 05:20 AM
P: 686
I started with

sin(z) = -i sinh(iz) (1)

and I applied the arcsin for get z

arcsin(sin(z)) = z

So I realized that z should appears in the right side of equation (1) and the way this happen is aplying -i arcsinh(ix) in the right side, so:

arcsin(sin(z)) = - i arcsinh(i -i sinh(iz)) = - i arcsinh(sinh(iz)) = -iiz = z
craigi
#8
Apr4-14, 05:38 AM
P: 419
Quote Quote by Jhenrique View Post
...
acosh(x) = -i acos( x)
Check this one.
D H
#9
Apr4-14, 05:58 AM
Mentor
P: 15,147
And that one is not correct with many definitions of inverse hyperbolic cosine and inverse cosine.

Jhenrique, you are ignoring the problems of branch cuts. You have not even defined your definitions of the analytic continuations of the inverse functions. There are many choices; infinitely many. What choices have you made?
Jhenrique
#10
Apr5-14, 05:33 AM
P: 686
Quote Quote by D H View Post
And that one is not correct with many definitions of inverse hyperbolic cosine and inverse cosine.

Jhenrique, you are ignoring the problems of branch cuts. You have not even defined your definitions of the analytic continuations of the inverse functions. There are many choices; infinitely many. What choices have you made?
1st I was trying undertand how create the relation between arc functions and arc functions hyp...

Quote Quote by craigi View Post
acosh(x) = -i acos( x)
Check this one.
x = cos(z) = cosh(iz)

acosh(cosh(iz)) = -i acos(cos(z))

iz = -iz .....

hummm
the formula worked for x = cosh(z) = cosh(iz)

So, which are the correct relations?
Curious3141
#11
Apr5-14, 06:38 AM
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P: 2,949
Quote Quote by Jhenrique View Post


x = cos(z) = cosh(iz)

acosh(cosh(iz)) = -i acos(cos(z))

iz = -iz .....
No, this is obviously wrong. If you end up with a mathematical absurdity like ##x = -x## for nonzero ##x##, you've made a mistake.

If you want to go from ##\cos z = \cosh iz## to a relationship between the inverse circular and hyperbolic functions, here's one way to proceed:

Put ##iz = \cosh^{-1} x##, where ##z = \frac{1}{i}\cosh^{-1} x = -i\cosh^{-1} x##.

Then the RHS becomes ##x##. The LHS is ##\cos(-i\cosh^{-1}x)##.

You now have ##\cos(-i\cosh^{-1}x) = x##. Take the inverse cosine on both sides and you end up with

##-i\cosh^{-1} x = \cos^{-1}(x)##

Multiply both sides by ##i## to get:

##\cosh^{-1} x = i\cos^{-1}(x)##

which is the exact relationship mentioned in the Italian Wiki page.
Jhenrique
#12
Apr10-14, 03:32 PM
P: 686
So, how would be the complete list?


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