## Calculus: Early Transcendentals 8th Edition

$e^{x}$ = 5 sinh x = $\frac{12}{5}$ cosh x = $\frac{13}{5}$
We know that $\frac{1 + tanh x}{1 - tanh x}$ = $e^{2x}$ Substitute tanh x = $\frac{12}{13}$ , To get $\frac{1 + \frac{12}{13}}{1 - \frac{12}{13}}$ = $e^{2x}$ $\frac{13 + 12}{13 - 12}$ = $e^{2x}$ 25 = $e^{2x}$ $e^{x}$ = 5 Thus, we know: sinh = $\frac{e^{x}-e^{-x}}{2}$ = $\frac{5 - \frac{1}{5}}{2}$ = $\frac{\frac{25}{5}-\frac{1}{5}}{2}$ = $\frac{12}{5}$ cosh x = $\frac{e^{x} + e^{-x}}{2}$ = $\frac{5 + \frac{1}{5}}{2}$ = $\frac{\frac{25}{5} + \frac{1}{5}}{2}$ = $\frac{13}{5}$