# using logarithm expression of sinh^-1

• Nov 18th 2012, 11:52 AM
carla1985
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-1y

So far im at
sinh-1(-y)=ln((-y)+sqrt((-y)2+1))

Thanks :)
• Nov 18th 2012, 03:36 PM
MarkFL
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.
• Nov 19th 2012, 06:04 AM
carla1985
Re: using logarithm expression of sinh^-1
Finished it, thanks. That was a big help :)
• Nov 26th 2012, 10:05 PM
happysoso
Re: using logarithm expression of sinh^-1
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