- #1

DryRun

Gold Member

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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.