Hi,

question:

$\displaystyle tanh(-2)=?$

I tried simplifying it and got

$\displaystyle \frac{sinh(-2)}{cosh(-2)}=\frac{\frac{e^2}{1}-\frac{1}{e^2}}{\frac{e^2}{1}+\frac{1}{e^2}}=1+e^4-\frac{1}{e^4}-1=e^4-\frac{1}{e^4}$

but this calculates into 54.579 when the answer is -.964

I could use a calculator and figure out the answer from the point

$\displaystyle \frac{e^{-2} - e^2}{e^{-2} + e^2}$

Im trying to figure out where my algebra mistake is b/c my professor forbids calculators on our test which is tomorrow. Help?