# Prove hyperbolic function

1. Apr 14, 2012

### sharks

The problem statement, all variables and given/known data
If $\sinh^{-1}x=2\cosh^{-1}y$, prove that $x^2=4y^2(y^2-1)$

The attempt at a solution
I re-wrote $\sinh^{-1}x$ and $2\cosh^{-1}y$ in terms of x and y.
$$\sinh^{-1}x=\ln(x+\sqrt{x^2+1}) \\2\cosh^{-1}y=2\ln(y+\sqrt{y^2-1})=\ln(y+\sqrt{y^2-1})^2 \\\ln(x+\sqrt{x^2+1})=\ln(y+\sqrt{y^2-1})^2 \\x+\sqrt{x^2+1}=(y+\sqrt{y^2-1})^2 \\x+\sqrt{x^2+1}=2y^2+2y\sqrt{y^2-1}-1$$
From this point onwards, i know that i have to manipulate that equation to get the proof. But, i don't know how to get only $x^2$ on the L.H.S. I tried and squared both sides, but the expressions just expand even more.

2. Apr 14, 2012

### LCKurtz

You might have better luck if you begin by taking $sinh$ of both sides and think about identities.

3. Apr 14, 2012

### sharks

Hi LCKurtz

OK, i multiplied by $\sinh$ on both sides:

$$x=\sinh (2\cosh^{-1}y) \\x=\frac{e^{2\cosh^{-1}y}-e^{-2\cosh^{-1}y}}{2} \\2x=(y+\sqrt{y^2-1})^2-(y+\sqrt{y^2-1})^{-2} \\2x=\frac{(y+\sqrt{y^2-1})^4-1}{(y+\sqrt{y^2-1})^2}$$
And then i don't know where it's going...

4. Apr 14, 2012

### LCKurtz

You don't "multiply" both sides by $sinh$. You take the $sinh$ of both sides. You are too anxious to plug in all those exponentials and square roots. At that very first step you have the $sinh$ of a double "angle" on the right. My hint mentioned identities...

5. Apr 15, 2012

### sharks

$\sinh (\sinh^{-1}x)=\sinh (2\cosh^{-1}y) \\x=2\sinh(\cosh^{-1}y)\cosh(\cosh^{-1}y) \\x=2\sinh(\cosh^{-1}y)(y) \\x=(e^{\cosh^{-1}y}-e^{-\cosh^{-1}y})(y) \\\frac{x}{y}=(y+\sqrt{y^2-1})-(y+\sqrt{y^2-1})^{-1}$
Then, i expand the R.H.S over a common denominator and end up with a complicated form.

Last edited: Apr 15, 2012
6. Apr 15, 2012

### chiro

Following on from LCKurtz, can you find an identity for the double angle formula and then for cosh^-1(sinh(x)) and sinh^-1(cosh(x))?

Think about how to write sinh(2x) in terms of sinh(x) and cosh(x) for the first part.

7. Apr 15, 2012

### sharks

That's exactly what i've done in line 2.

$\sinh 2A=2\sinh A\cosh A$
Let $A=\cosh^{-1}y \\\sinh (2\cosh^{-1}y)=2\sinh(\cosh^{-1}y)\cosh(\cosh^{-1}y)$
Then,
$$\sinh(\sinh^{-1}x)=x$$I've tried to prove the above, but i can't figure it out, however according to my calculator (using random values), it's correct.

Re-writing it here:
$$x=2\sinh(\cosh^{-1}y)\cosh(\cosh^{-1}y)$$
Or, did you mean, to expand the L.H.S:
$$\sinh (\sinh^{-1}x)=2\sinh (\frac{\sinh^{-1}x}{2})\cosh(\frac{\sinh^{-1}x)}{2})$$OK, i got it!

From,
$$x=2\sinh(\cosh^{-1}y)\cosh(\cosh^{-1}y)=2\sinh(\cosh^{-1}y)(y)$$
$$\frac{x}{2y}=\sinh(\cosh^{-1}y)$$
Squaring both sides:
$$\frac{x^2}{4y^2}=\sinh^2(\cosh^{-1}y)$$
$$\frac{x^2}{4y^2}=\cosh^2(\cosh^{-1}y)-1$$
$$\frac{x^2}{4y^2}=y^2-1$$
Therefore,
$$x^2=4y^2(y^2-1)$$

But i'm wondering how to prove this expression?
$$\sinh(\sinh^{-1}x)=x$$

Last edited: Apr 15, 2012
8. Apr 15, 2012

### Curious3141

That's simply the inverse function acting on the output of a function to return the original value.

$f^{-1}(f(x)) = x$

9. Apr 15, 2012

### sharks

It makes perfect sense, but what if i actually worked it out? Will it be too complicated?

Here is a trial...
$$\sinh(\sinh^{-1}x) \\=\sinh(\ln (x +\sqrt{x^2+1})) \\=\frac { e^{\ln (x +\sqrt {x^2+1})}-e^{-\ln (x +\sqrt {x^2+1})} } {2} \\=\frac { (x +\sqrt {x^2+1})-(x +\sqrt {x^2+1})^{-1} } {2} \\=\frac { \frac{(2x^2 +2x\sqrt {x^2+1})}{(x +\sqrt {x^2+1})} } {2} \\=\frac { 2x\frac{(x +\sqrt {x^2+1})}{(x +\sqrt {x^2+1})} } {2} \\=\frac{2x}{2} \\=x$$

Last edited: Apr 15, 2012