# Prove hyperbolic function

1. Dec 3, 2011

### sharks

The problem statement, all variables and given/known data

Frankly, i've been sitting starring at that problem for long enough, and it just can't be solved through the direct use of the standard hyperbolic identities. I need a hint.

2. Dec 3, 2011

### micromass

Staff Emeritus
Use induction.

3. Dec 3, 2011

### sharks

That's a long shot... I just started this topic so i'm very new to it and i have no idea what induction means either. Can you elaborate a little?

4. Dec 3, 2011

### Curious3141

Really? Have you tried expressing cosh x and sinh x in terms of e^x and e^(-x)?

No need for induction. This is a simple direct proof. And induction only covers non-negative integral n, whereas you can more easily prove the general result for all complex n directly.

Last edited: Dec 3, 2011
5. Dec 3, 2011

### micromass

Staff Emeritus
See our proofs FAQ: https://www.physicsforums.com/showthread.php?t=523874 [Broken] for a discussion on induction.

To make it simpler, try and set n=2. Can you prove it for n=2?

Last edited by a moderator: May 5, 2017
6. Dec 3, 2011

### Dick

No, no induction needed. Just put in the definition of cosh and sinh in terms of exponentials.

7. Dec 3, 2011

### dextercioby

Indeed, induction seems the most elegant proof.

8. Dec 3, 2011

### sharks

Alright, what is induction??

9. Dec 3, 2011

### micromass

Staff Emeritus

10. Dec 3, 2011

### Curious3141

Have you tried the direct method? Start by expressing sinh and cosh as exponentials, then it's just algebra!

11. Dec 3, 2011

### sharks

First, i've expressed the L.H.S into two parts:
(coshx + sinhx)^n and (coshx - sinhx)^n

For the first one:
(coshx + sinhx)^n = e^(nx)

For the second one:
(coshx - sinhx)^n = e^(-nx)

I am aware that i somehow need to bring that power n down, but can't see how.

12. Dec 3, 2011

### micromass

Staff Emeritus
Now do the same with the RHS

13. Dec 3, 2011

### sharks

For the R.H.S. i split it into 2 parts:

(cosh nx + sinh nx) and (cosh nx - sinh nx)

(cosh nx + sinh nx) = [e^(nx) + e^(-nx)]/2 + [e^(nx) - e^(-nx)]/2 = [e^(nx)]/2

(cosh nx - sinh nx) = [e^(nx) + e^(-nx)]/2 - [e^(nx) - e^(-nx)]/2 = [e^(-nx)]/2

OK, i see it now. They are both the same. It did not occur to me that i had to tinker with the R.H.S also, as normally, i should have been able to get the R.H.S. using only the L.H.S.