Chapter 3 - Section 3.11 - Hyperbolic Functions - 3.11 Exercises - Page 266: 5

(a) $sech~0 = 1$ (b) $cosh^{-1}~1 = 0$

(a) $sech~0 = \frac{1}{cosh~0}$ $sech~0 = \frac{2}{e^0+e^{-0}}$ $sech~0 = \frac{2}{1+1}$ $sech~0 = 1$ (b) $cosh^{-1}~x = ln(x+\sqrt{x^2-1})$ $cosh^{-1}~1 = ln(1+\sqrt{1^2-1})$ $cosh^{-1}~1 = ln(1+\sqrt{0})$ $cosh^{-1}~1 = ln(1+0)$ $cosh^{-1}~1 = ln(1)$ $cosh^{-1}~1 = 0$