hi again!

how would you solve these functions:

2cosh2x - sinh2x = 2

and prove

thanks

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- Jan 21st 2007, 03:08 AMdadonhyperbolic functions2
hi again!

how would you solve these functions:

2cosh2x - sinh2x = 2

and prove

thanks - Jan 21st 2007, 04:05 AMCaptainBlack
You need some brackets here I think you mean:

One way is if you know the "Hyperbolic Pythagoras's" theorem:

when it follows at once, but that would be to simple, so we must suppose

that you are expected to expand the hyperbolic funtions in terms of

exponentials. So:

,

and:

,

Then:

,

and:

,

So:

,

and the required result follows on multiplying both sides by

RonL - Jan 21st 2007, 04:08 AMdadon
sorry the question was express cosh2x and sinhx in exponential form and hence solve:

2cosh2x - sinh2x = 2 - Jan 21st 2007, 04:20 AMtopsquark
I am wondering if the first one isn't an identity, but to solve the equation for x?

Now

So the equation becomes:

<-- Multiply both sides by

Now, for ease of notation, let . Then :

So the equation is now:

or <-- Put the x back: :

or

or

or

So the solution set for x is

-Dan - Jan 21st 2007, 06:14 AMSoroban
Hello, dadon!

By the way, natural log is , not .

Quote:

Prove:

Let

From the definition of , we have: .

On the left side, multiply top and bottom by

Then: . . . .

. . . .

. .

. . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . .

Therefore: .