Homework SolutionProve Hyperbolic Function: Solving for x in Terms of y

[DryRun]

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})<br /> \\2\cosh^{-1}y=2\ln(y+\sqrt{y^2-1})=\ln(y+\sqrt{y^2-1})^2<br /> \\\ln(x+\sqrt{x^2+1})=\ln(y+\sqrt{y^2-1})^2<br /> \\x+\sqrt{x^2+1}=(y+\sqrt{y^2-1})^2<br /> \\x+\sqrt{x^2+1}=2y^2+2y\sqrt{y^2-1}-1<br />
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]

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

 

[DryRun]

Hi LCKurtz

OK, i multiplied by \sinh on both sides:

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

 

[LCKurtz]

Hi LCKurtz

OK, i multiplied by \sinh on both sides:

x=\sinh (2\cosh^{-1}y)<br /> \\x=\frac{e^{2\cosh^{-1}y}-e^{-2\cosh^{-1}y}}{2}<br /> \\2x=(y+\sqrt{y^2-1})^2-(y+\sqrt{y^2-1})^{-2}<br /> \\2x=\frac{(y+\sqrt{y^2-1})^4-1}{(y+\sqrt{y^2-1})^2}<br />
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...

 

[DryRun]

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

 

[chiro]

\sinh (\sinh^{-1}x)=\sinh (2\cosh^{-1}y)<br /> \\x=2\sinh(\cosh^{-1}y)\cosh(\cosh^{-1}y)<br /> \\x=2\sinh(\cosh^{-1}y)(y)<br /> \\x=(e^{\cosh^{-1}y}-e^{-\cosh^{-1}y})(y)<br /> \\\frac{x}{y}=(y+\sqrt{y^2-1})-(y+\sqrt{y^2-1})^{-1} <br />
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.

 

[DryRun]

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<br /> \\\sinh (2\cosh^{-1}y)=2\sinh(\cosh^{-1}y)\cosh(\cosh^{-1}y)
Then,
\sinh(\sinh^{-1}x)=xI'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

 

[Curious3141]

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

 

[DryRun]

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)<br /> \\=\sinh(\ln (x +\sqrt{x^2+1}))<br /> \\=\frac { e^{\ln (x +\sqrt {x^2+1})}-e^{-\ln (x +\sqrt {x^2+1})} } {2}<br /> \\=\frac { (x +\sqrt {x^2+1})-(x +\sqrt {x^2+1})^{-1} } {2}<br /> \\=\frac { \frac{(2x^2 +2x\sqrt {x^2+1})}{(x +\sqrt {x^2+1})} } {2}<br /> \\=\frac { 2x\frac{(x +\sqrt {x^2+1})}{(x +\sqrt {x^2+1})} } {2}<br /> \\=\frac{2x}{2}<br /> \\=x<br />