Hello,
How do you cancel this equation down further?
Thanks
$\displaystyle ((exp(x)exp(-x))/4) + ((exp(x)exp(-x))/4) + ((exp(-2x))/4) + ((exp(2x))/4)$
Hello, MattWT!
Are you working with hyperbolic functions?
How do you cancel this equation down further?
. . $\displaystyle \dfrac{e^x\cdot e^{-x}}{4} + \dfrac{e^x\cdot e^{-x}}{4} + \dfrac{e^{-2x}}{4} + \dfrac{e^{2x}}{4}$
Since .$\displaystyle e^x\cdot e^{-x} \:=\:e^0 \:=\:1$
. . we have: .$\displaystyle \displaystyle \frac{1}{4} + \frac{1}{4} + \frac{e^{-2x}}{4} + \frac{e^{2x}}{4} \;=\;\frac{e^{2x}}{4} + \frac{1}{2} + \frac{e^{-2x}}{4} $
. . . . . . . . . . $\displaystyle \displaystyle =\;\frac{e^{2x} + 2 + e^{-2x}}{4} \;=\;\frac{(e^x + e^{-x})^2}{4}$
. . . . . . . . . . $\displaystyle =\;\left(\dfrac{e^x + e^{-x}}{2}\right)^2 \;=\;\cosh^2\!x$