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Relation between inverse trigonometric function 
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#1
Mar3114, 10:22 PM

P: 686

Digging in the wiki, I found this relation between 'arcfunctions' and 'arcfunctionshyperbolics"
[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? 


#2
Mar3114, 11:19 PM

Mentor
P: 15,055




#3
Apr114, 01:38 AM

P: 686

Also, where can I find a full list (and correct)? 


#4
Apr314, 06:55 PM

P: 686

Relation between inverse trigonometric function
Hey man, you'll let me in the doubt!?



#5
Apr314, 07:17 PM

P: 419

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. 


#6
Apr314, 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? 


#7
Apr414, 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)) = i·iz = z 


#8
Apr414, 05:38 AM

P: 419




#9
Apr414, 05:58 AM

Mentor
P: 15,055

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? 


#10
Apr514, 05:33 AM

P: 686

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? 


#11
Apr514, 06:38 AM

HW Helper
P: 2,940

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. 


#12
Apr1014, 03:32 PM

P: 686

So, how would be the complete list?



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