How can the R.H.S. of a hyperbolic function be manipulated to match the L.H.S.?

[DryRun]

Homework Statement
http://s1.ipicture.ru/uploads/20111203/B1Ax1OcU.jpg

Frankly, I've been sitting staring 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.

 

[micromass]

Use induction.

 

[DryRun]

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 Statement
http://s1.ipicture.ru/uploads/20111203/B1Ax1OcU.jpg

Frankly, I've been sitting staring 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.

 

[micromass]

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

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

 

[Dick]

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

 

[dextercioby]

Indeed, induction seems the most elegant proof.

 

[DryRun]

Alright, what is induction??

 

[micromass]

Alright, what is induction??

Read the link I posted above. Or check your textbook.

 

[Curious3141]

Alright, what is induction??

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

 

[DryRun]

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.

 

[micromass]

Now do the same with the RHS

 

[DryRun]

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.