Prove hyperbolic function

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  • #1
DryRun
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Homework Statement
If [itex]\sinh^{-1}x=2\cosh^{-1}y[/itex], prove that [itex]x^2=4y^2(y^2-1)[/itex]

The attempt at a solution
I re-wrote [itex]\sinh^{-1}x[/itex] and [itex]2\cosh^{-1}y[/itex] in terms of x and y.
[tex]\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
[/tex]
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 [itex]x^2[/itex] on the L.H.S. I tried and squared both sides, but the expressions just expand even more.
 

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  • #2
LCKurtz
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You might have better luck if you begin by taking ##sinh## of both sides and think about identities.
 
  • #3
DryRun
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Hi LCKurtz

OK, i multiplied by [itex]\sinh[/itex] on both sides:

[tex]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}
[/tex]
And then i don't know where it's going...
 
  • #4
LCKurtz
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Hi LCKurtz

OK, i multiplied by [itex]\sinh[/itex] on both sides:

[tex]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}
[/tex]
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...
 
  • #5
DryRun
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[itex]\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}
[/itex]
Then, i expand the R.H.S over a common denominator and end up with a complicated form.
 
Last edited:
  • #6
chiro
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[itex]\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}
[/itex]
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.
 
  • #7
DryRun
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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.

[itex]\sinh 2A=2\sinh A\cosh A[/itex]
Let [itex]A=\cosh^{-1}y
\\\sinh (2\cosh^{-1}y)=2\sinh(\cosh^{-1}y)\cosh(\cosh^{-1}y)[/itex]
Then,
[tex]\sinh(\sinh^{-1}x)=x[/tex]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:
[tex]x=2\sinh(\cosh^{-1}y)\cosh(\cosh^{-1}y)[/tex]
Or, did you mean, to expand the L.H.S:
[tex]\sinh (\sinh^{-1}x)=2\sinh (\frac{\sinh^{-1}x}{2})\cosh(\frac{\sinh^{-1}x)}{2})[/tex]OK, i got it!

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


But i'm wondering how to prove this expression?
[tex]\sinh(\sinh^{-1}x)=x[/tex]
 
Last edited:
  • #8
Curious3141
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But i'm wondering how to prove this?
[tex]\sinh(\sinh^{-1}x)=x[/tex]

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

[itex]f^{-1}(f(x)) = x[/itex]
 
  • #9
DryRun
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It makes perfect sense, but what if i actually worked it out? Will it be too complicated?

Here is a trial...
[tex]\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
[/tex]
 
Last edited:

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