MHB Prove this hyperbolic identities

[arroww]

6b) tanh^2(x) + 1/cosh^2(x) = 1

Could someone help start me off? I know that you have to sub in (e^x + e^-x)/2 for cosh and (e^x - e ^-x)/(e^x + e ^-x) for tanh. Then I'd add these together, but I'm not sure how I'd solve/simplify them arithmetically after that.

Help would be appreciated! thanks.

 

[Deveno]

I would think it easier to use the fundamental hyperbolic trigonometric identity:

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

EDIT: If you've never seen this proved, here it is:

$\cosh^2 x - \sinh^2 x = \left(\dfrac{e^x + e^{-x}}{2}\right)^2 - \left(\dfrac{e^x - e^{-x}}{2}\right)^2$

$=\dfrac{e^{2x} + 2 + e^{-2x}}{4} - \dfrac{e^{2x} - 2 + e^{-2x}}{4} = \dfrac{2}{4} - \dfrac{-2}{4} = 1$

 

[arroww]

I would think it easier to use the fundamental hyperbolic trigonometric identity:

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

That was actually the previous question. So I'm not sure what you mean?

 

[Deveno]

Divide the fundamental identity by $\cosh^2 x$, what do you get?

 

[arroww]

Divide the fundamental identity by $\cosh^2 x$, what do you get?

$\frac{cosh^2 x - sin^2x}{cosh^2x}$ ?

 

[Deveno]

You only divided HALF of the identity (the stuff on one side of the equals sign). What about the other half?

 

[arroww]

You only divided HALF of the identity (the stuff on one side of the equals sign). What about the other half?

Oh, so it's: $\frac{cosh^2 x - sin^2x}{cosh^2x} = \frac {1}{cosh^2x}$

 

[Deveno]

Now split up the sum on the left into 2 fractions.

 

[arroww]

Now split up the sum on the left into 2 fractions.

$\frac{cosh^2 x}{cosh^2x} - \frac{sin^2x}{cosh^2x}$

so then you'd get, - $\frac{sin^2x}{cosh^2x}$ ?

 

[Deveno]

Can you think of another way to write:

$\dfrac{\sinh x}{\cosh x}$?

And, don't "skip steps". Write down the WHOLE IDENTITY. Don't confuse "parts" and "whole" (a word is not a sentence).

 

[arroww]

Can you think of another way to write:

$\dfrac{\sinh x}{\cosh x}$?

And, don't "skip steps". Write down the WHOLE IDENTITY. Don't confuse "parts" and "whole" (a word is not a sentence).

Hm I'm not sure. The only think I can think of is subbing $\frac{e^x + e^{-x}}{2}$ for coshx and $\frac{e^x - e^{-x}}{2}$ for sinhx.

 

[Deveno]

I don't think you're getting what I'm driving at.Objective: we want to prove

$\tanh^2 x + \dfrac{1}{\cosh^2 x} = 1$

Well, we KNOW (because we just proved it) that:

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

is true.

Now equality still holds if we divide two sides of an equation by the same quantity, so:

$\dfrac{\cosh^2 x - \sinh^2 x}{\cosh^2 x} = \dfrac{1}{\cosh^2 x}$

Splitting the left-hand side into two terms, we get:

$\dfrac{\cosh^2 x}{\cosh^2 x} - \dfrac{\sinh^2 x}{\cosh^2 x} = \dfrac{1}{\cosh^2 x}$

Now, what is:

$\dfrac{\cosh^2 x}{\cosh^2 x}$, and what is:

$\dfrac{\sinh^2 x}{\cosh^2 x}$? (hint: this occurs in what we're trying to prove).

 

[arroww]

I don't think you're getting what I'm driving at.Objective: we want to prove

$\tanh^2 x + \dfrac{1}{\cosh^2 x} = 1$

Well, we KNOW (because we just proved it) that:

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

is true.

Now equality still holds if we divide two sides of an equation by the same quantity, so:

$\dfrac{\cosh^2 x - \sinh^2 x}{\cosh^2 x} = \dfrac{1}{\cosh^2 x}$

Splitting the left-hand side into two terms, we get:

$\dfrac{\cosh^2 x}{\cosh^2 x} - \dfrac{\sinh^2 x}{\cosh^2 x} = \dfrac{1}{\cosh^2 x}$

Now, what is:

$\dfrac{\cosh^2 x}{\cosh^2 x}$, and what is:

$\dfrac{\sinh^2 x}{\cosh^2 x}$? (hint: this occurs in what we're trying to prove).

$\dfrac{\cosh^2 x}{\cosh^2 x}$ is the RS (1). So does that mean $\dfrac{\sinh^2 x}{\cosh^2 x}$ is the LS? though I'm not sure how it is.

 

[Deveno]

No, the left side has two terms. The question I'm asking when I say what is:

$\dfrac{\cosh^2 x}{\cosh^2 x}$

is a much simpler one: What is the fraction $\dfrac{a}{a}$ no matter WHAT (non-zero, of course) expression "$a$" is?

 

[arroww]

No, the left side has two terms. The question I'm asking when I say what is:

$\dfrac{\cosh^2 x}{\cosh^2 x}$

is a much simpler one: What is the fraction $\dfrac{a}{a}$ no matter WHAT (non-zero, of course) expression "$a$" is?

It's 1.

 

[Deveno]

Right. so that means that:

$\dfrac{\cosh^2 x}{\cosh^2 x} + \dfrac{\sinh^2 x}{\cosh^2 x} = \dfrac{1}{\cosh^2 x}$

is really the equation:

$1 + \dfrac{\sinh^2 x}{\cosh^2 x} = \dfrac{1}{\cosh^2 x}$.

So now, we look at:

$\dfrac{\sinh^2 x}{\cosh^2 x}$, which is really:

$\dfrac{(\sinh x)^2}{(\cosh x)^2} = \left(\dfrac{\sinh x}{\cosh x}\right)^2$.

Do you know of another name for:

$\dfrac{\sinh x}{\cosh x}$?

 

[arroww]

Right. so that means that:

$\dfrac{\cosh^2 x}{\cosh^2 x} + \dfrac{\sinh^2 x}{\cosh^2 x} = \dfrac{1}{\cosh^2 x}$

is really the equation:

$1 + \dfrac{\sinh^2 x}{\cosh^2 x} = \dfrac{1}{\cosh^2 x}$.

So now, we look at:

$\dfrac{\sinh^2 x}{\cosh^2 x}$, which is really:

$\dfrac{(\sinh x)^2}{(\cosh x)^2} = \left(\dfrac{\sinh x}{\cosh x}\right)^2$.

Do you know of another name for:

$\dfrac{\sinh x}{\cosh x}$?

Isn't it tanhx?

 

[Deveno]

Yes! So, what is our equation now equivalent to?

 

[arroww]

Yes! So, what is our equation now equivalent to?

Oh, so rearranging it gets:
1 - tanhx = 1/cosh^2x
1 = 1/cosh^2x + tanhx

 

[Deveno]

Not quite.

We have:

$\dfrac{\sinh^2 x}{\cosh^2 x} = \left(\dfrac{\sinh x}{\cosh x}\right)^2 = (\tanh x)^2 = \tanh^2 x$

(you have forgotten we have a "squared" in there).

 

[arroww]

Not quite.

We have:

$\dfrac{\sinh^2 x}{\cosh^2 x} = \left(\dfrac{\sinh x}{\cosh x}\right)^2 = (\tanh x)^2 = \tanh^2 x$

(you have forgotten we have a "squared" in there).

Oh, oops. ;p
Alright, so it's pretty much the same thing when rearranging:
1 - tanh^2x = 1/cosh^2x
1 = 1/cosh^2x + tanh^2x

 

[Deveno]

And therein, lies the tale of a proof, yes?