Relation between inverse trigonometric function

In summary, there is a relationship between inverse circular functions and inverse hyperbolic functions, where the inverse hyperbolic function is equal to the inverse circular function multiplied by i. The full list of this relationship is:- arcsinh(x)= i * arcsin(-ix)- arccosh(x)= i * arccos(+ix)- arctanh(x)= i * arctan(-ix)- arccoth(x)= i * arccot(-ix)- arcsech(x)= i * arcsec(+ix)- arccsch(x)= i * arccsc(-ix)
  • #1
Jhenrique
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  • #2
Jhenrique said:
This relationship is correct?
The second one is incorrect, and the other two are obvious.
 
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  • #3
D H said:
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)?
 
  • #4
Hey man, you'll let me in the doubt!?
 
  • #5
Jhenrique said:
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.
 
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  • #6
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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  • #7
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
Jhenrique said:
...
acosh(x) = -i acos( x)

Check this one.
 
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  • #9
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
D H said:
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...

craigi said:
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?
 
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  • #11
Jhenrique said:
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.
 
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  • #12
So, how would be the complete list?
 

1. What are inverse trigonometric functions?

Inverse trigonometric functions are mathematical functions that are used to find the angle of a right triangle based on the ratios of its sides. They are the inverse, or opposite, of the regular trigonometric functions such as sine, cosine, and tangent.

2. What is the relation between inverse trigonometric functions and regular trigonometric functions?

The inverse trigonometric functions are used to "undo" the regular trigonometric functions. They can be thought of as a way to find the missing angle in a right triangle when given the ratios of its sides, whereas the regular trigonometric functions find the ratios of the sides when given the angle.

3. How do you find the values of inverse trigonometric functions?

The values of inverse trigonometric functions can be found by using a calculator or by using mathematical tables. These functions also have specific properties and identities that can be used to simplify their values.

4. What are the domain and range of inverse trigonometric functions?

The domain of inverse trigonometric functions is the set of all possible input values, or angles, for which the function can be defined. The range is the set of all possible output values, or angles, that the function can produce.

5. How are inverse trigonometric functions used in real life?

Inverse trigonometric functions have many real-life applications, such as in engineering, physics, and astronomy. They are used to solve problems involving angles and distances, as well as in the design and construction of buildings, bridges, and other structures.

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