using logarithm expression of sinh^-1

Hi all,

I'm really stuck on a question and hoping someone can point me in the right direction.

By using the logarithm expression of sinh^{-1}(y), show that sinh^{-1}(-y)=-sinh^{-1}y

So far im at

sinh^{-1}(-y)=ln((-y)+sqrt((-y)^{2}+1))

Thanks :)

Re: using logarithm expression of sinh^-1

So, you have:

$\displaystyle \sinh^{-1}(-y)=\ln(\sqrt{y^2+1}-y)$

My next step would be:

$\displaystyle \sinh^{-1}(-y)=-\ln\left(\frac{1}{\sqrt{y^2+1}-y} \right)$

Now, rationalize the denominator of the argument of the log function on the right.

Re: using logarithm expression of sinh^-1

Finished it, thanks. That was a big help :)

Re: using logarithm expression of sinh^-1

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