# hyperbolic functions2

• Jan 21st 2007, 02:08 AM
hyperbolic functions2
hi again!

how would you solve these functions:

2cosh2x - sinh2x = 2

and prove

$tanh^{-1}(\frac{x^2-a^2}{x^2+a^2}) = In(\frac{x}{a})$

thanks
• Jan 21st 2007, 03:05 AM
CaptainBlack
Quote:

hi again!

how would you solve these functions:

2cosh2x - sinh2x = 2

You need some brackets here I think you mean:

$2[\cosh^2(x)-\sinh^2(x)]=2$

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

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

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:

$\sinh(x)=\frac{e^x-e^{-x}}{2}$,

and:

$\cosh(x)=\frac{e^x+e^{-x}}{2}$,

Then:

$\sinh^2(x)=\frac{e^{2x}-2+e^{-2x}}{4}$,

and:

$\cosh^2(x)=\frac{e^{2x}+2+e^{-2x}}{4}$,

So:

$\cosh^2(x)-\sinh^2(x)=\frac{e^{2x}+2+e^{-2x}}{4} - \frac{e^{2x}-2+e^{-2x}}{4}=1$,

and the required result follows on multiplying both sides by $2$

RonL
• Jan 21st 2007, 03:08 AM
sorry the question was express cosh2x and sinhx in exponential form and hence solve:

2cosh2x - sinh2x = 2
• Jan 21st 2007, 03:20 AM
topsquark
Quote:

2cosh2x - sinh2x = 2

I am wondering if the first one isn't an identity, but to solve the equation for x?

$2cosh(2x) - sinh(2x) = 2$

Now
$cosh(2x) = \frac{1}{2}(e^{2x} + e^{-2x})$
$sinh(2x) = \frac{1}{2}(e^{2x} - e^{-2x})$

So the equation becomes:
$(e^{2x} + e^{-2x}) - \frac{1}{2}(e^{2x} - e^{-2x}) = 2$

$\frac{1}{2}e^{2x} + \frac{3}{2}e^{-2x} = 2$

$e^{2x} + 3e^{-2x} = 4$ <-- Multiply both sides by $e^{2x}$

$e^{4x} + 3 = 4e^{2x}$

$e^{4x} - 4e^{2x} + 3 = 0$

Now, for ease of notation, let $y = e^{2x}$. Then $e^{4x} = y^2$:

So the equation is now:
$y^2 - 4y + 3 = 0$

$(y - 3)(y - 1) = 0$

$y = 3$ or $y = 1$ <-- Put the x back: $y = e^{2x}$:

$e^{2x} = 3$ or $e^{2x} = 1$

$2x = ln(3)$ or $2x = ln(1) = 0$

$x = \frac{1}{2}ln(3)$ or $x = 0$

So the solution set for x is $0, \, \frac{1}{2}ln(3)$

-Dan
• Jan 21st 2007, 05:14 AM
Soroban

By the way, natural log is $\ln\;(LN)$, not $In$.

Quote:

Prove: $\text{tanh}^{-1}\left(\frac{x^2-a^2}{x^2+a^2}\right) \:= \:\ln\left(\frac{x}{a}\right)$

Let $u \:=\:\text{tanh}^{-1}\left(\frac{x^2-a^2}{x^2+a^2}\right) \quad\Rightarrow\quad \text{tanh }u \:=\:\frac{x^2-a^2}{x^2+a^2}$

From the definition of $\text{tanh }u$, we have: . $\frac{e^u - e^{-u}}{e^u+ e^{-u}} \:=\:\frac{x^2-a^2}{x^2+a^2}$

On the left side, multiply top and bottom by $e^u\!:\;\;\frac{e^{2u} - 1}{e^{2u} + 1} \:=\:\frac{x^2-a^2}{x^2+a^2}$

Then: . . . . $(x^2+a^2)(e^{2u} - 1) \:=\:(x^2-a^2)(e^{2u} + 1)$

. . . . $(x^2 + a^2)e^{2u} - (x^2+a^2) \:=\:(x^2-a^2)e^{2u} + (x^2 - a^2)$

. . $(x^2+a^2)e^{2u} - (x^2-a^2)e^{2u} \:=\:(x^2-a^2) + (x^2 + a^2)$

. . . . . . . . . . . . . . . $2a^2e^{2u} \:=\:2x^2$

. . . . . . . . . . . . . . . . . $e^{2u} \:=\:\frac{x^2}{a^2} \:=\:\left(\frac{x}{a}\right)^2$

. . . . . . . . . . . . . . . . . $2u \:=\:\ln\left(\frac{x}{a}\right)^2\:=\:2\ln\left(\ frac{x}{a}\right)$

. . . . . . . . . . . . . . . . . . $u \:=\:\ln\left(\frac{x}{a}\right)$

Therefore: . $\text{tanh}^{-1}\left(\frac{x^2-a^2}{x^2+a^2}\right)\;=\;\ln\left(\frac{x}{a}\righ t)$