# Basic Hyperbolic Simplification

• Sep 30th 2010, 11:11 PM
ugkwan
Basic Hyperbolic Simplification
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?
• Sep 30th 2010, 11:37 PM
Prove It
$\displaystyle \displaystyle{\tanh{x} = \frac{e^{2x} - 1}{e^{2x} + 1}}$.

So what does $\displaystyle \tanh{(-2)}$ equal?
• Oct 1st 2010, 02:14 AM
HallsofIvy
Of course, the definition of "tanh" is $\displaystyle tanh(x)= \frac{sinh(x)}{cosh(x)}= \frac{\frac{e^x- e^{-1}}{2}}{\frac{e^x+ e^{-x}}{2}}= \frac{e^x- e^{-x}}{e^x+ e^{-x}}$ as you have.

What Prove It did was multiply both numerator and denominator by $\displaystyle e^x$ to get $\displaystyle tanh(x)= \frac{e^{2x}- 1}{e^{2x}+1}$

You are (almost) correct to $\displaystyle \frac{\frac{e^2}{1}-\frac{1}{e^2}}{\frac{e^2}{1}+\frac{1}{e^2}}$ (each "2" should be "-2") but I cannot imagine how you then wrote that fraction as four added and subtracted terms If you multiply numerator and denominator of
$\displaystyle \dfrac{\frac{e^{-2}}{1}-\frac{1}{e^{-2}}}{\frac{e^{-2}}{1}+\frac{1}{e^{-2}}}$
by $\displaystyle e^2$ you get
$\displaystyle \dfrac{e^{-4}- 1}{e^{-4}+ 1}$.