second question:
$\displaystyle cosh^{-1}2=?$
I tried
$\displaystyle \frac{1}{cosh2}=sech2=\frac{2}{e^2+e^{-2}}$
The answer given is $\displaystyle ln(1+\sqrt{3})$
How did they get to a answer w/ natural log?
Unfortunate notation- $\displaystyle f^{-1}(x)$ means the inverse function, not the reciprocal!
Specifically, $\displaystyle cosh^{-1}(2)$ is the value of x such that $\displaystyle cosh(x)= \frac{e^x+ e^{-x}}{2}= 2$. Solve that equation for x: $\displaystyle e^x+ e^{-x}= 4$. Let $\displaystyle y= e^x$ so the equation becomes $\displaystyle y+ y^{-1}= 4$. Multiply both sides by y: $\displaystyle y^2+ 1= 4y$ or $\displaystyle y^2- 4y+ 1= 0$. Solve that equation for y then take the logarithm to find x.