# Relation between inverse trigonometric function

686

### Staff: Mentor

The second one is incorrect, and the other two are obvious.

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3. ### Jhenrique

686
And which is the correct form for the second?
Also, where can I find a full list (and correct)?

4. ### Jhenrique

686
Hey man, you'll let me in the doubt!?

5. ### craigi

583
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.

Last edited: Apr 4, 2014
6. ### Jhenrique

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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7. ### Jhenrique

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. ### craigi

583
Check this one.

Last edited: Apr 4, 2014

### Staff: Mentor

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. ### Jhenrique

686
1st I was trying undertand how create the relation between arc functions and arc functions hyp...

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?

Last edited: Apr 5, 2014
11. ### Curious3141

2,943
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.

Last edited: Apr 5, 2014
12. ### Jhenrique

686
So, how would be the complete list?

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