# Recomposing trig and hyp functions

1. Aug 5, 2008

### Marin

Hi!

I want to recompose the expression of the form:

arcsin(cosx) , arsinh(coshx), arcsin(coshx), arsinh(cosx)

but I can't figure out how this is supposed to work :( I expect square root expressions for the first two and I'm not sure one can recompose the last two without using complex numbers.

2. Aug 5, 2008

### uart

Firstly understand that problems of the form arcsin(cos(x)) are quite different to problems of the form cos(arcsin(x)).

For the first type simply use the "complement" relation that cos(x) = sin(Pi/2-x) to write,

$$\sin^{-1}(\cos(x)) = \sin^{-1}(\sin(\frac{\pi}{2}-x)) = \frac{\pi}{2}-x$$

{or 90-x if you prefer to use degrees}.

Since you "were expecting square root type expressions" I assume you where actually thinking about the opposite type of problem such as cos(arcsin(x)).

Here you need to use the relation that $\cos(x) = \sqrt{1-(\sin x)^2}$

Making this substitution you get,

$$\cos(\sin^{-1}(x)) = \sqrt{1-(\sin (\sin^{-1} x))^2}$$
$$= \sqrt(1-x^2)$$

Last edited: Aug 5, 2008
3. Aug 5, 2008

### Marin

Thanks, uart!

And what about the hyperbolic functions? There is no Pythagoras for hyperb. functions, is there?

And the mixed type in the field of the real numbers is quite unclear to me :(

4. Aug 5, 2008

### uart

For hyp-trig functions the equivalent of the pythagorean relation is

$$\cosh^2(x) - \sinh^2(x) = 1$$

Yeah for the mixed ones I thinked I'd have to go complex and use cosh(x) = cos(i x) and sinh(x) = -i sin(i x). Maybe someone else can offer a better solution there.

Last edited: Aug 5, 2008
5. Aug 5, 2008

### Marin

So it's:

$$\cosh(\arsinhx)=\sqrt{1+\sinh(\arsinhx)}=\sqrt{1+x^2}$$

() is suppose to have \arsinhx inside - what's wrong with the Latex typing it?

And what do we do when we have: arsinh(coshx) hmmmm?