# Prove hyperbolic function

Gold Member
Homework Statement
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.

LCKurtz
Homework Helper
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You might have better luck if you begin by taking ##sinh## of both sides and think about identities.

Gold Member
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...

LCKurtz
Homework Helper
Gold Member
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...

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...

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

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

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.

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

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?
$$\sinh(\sinh^{-1}x)=x$$

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Curious3141
Homework Helper
But i'm wondering how to prove this?
$$\sinh(\sinh^{-1}x)=x$$

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

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

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

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

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