# Prove hyperbolic function

Gold Member
Homework Statement

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.

Use induction.

Gold Member
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?

Curious3141
Homework Helper
Homework Statement

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.

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.

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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?

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

dextercioby
Homework Helper
Indeed, induction seems the most elegant proof.

Gold Member
Alright, what is induction??

Alright, what is induction??

Curious3141
Homework Helper
Alright, what is induction??

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

Gold Member
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.

Now do the same with the RHS

Gold Member
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.