Question

Write \(8\sinh x + 5\cosh x\) in terms of \({e^x}\), and\({e^{ - x}}\).

Short answer

The expression \(8\sinh x + 5\cosh x\) in terms of \({e^x}\) and \({e^{ - x}}\) is \(\frac{{13}}{2}{e^x} - \frac{{13}}{2}{e^{ - x}}\).

Step-by-step solution

Definition of Hyperbolic function

The formulas for the hyperbolic function as shown below:

\(\begin{aligned}\sinh x = \frac{{{e^x} - {e^{ - x}}}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\mathop{\rm csch}\nolimits} x = \frac{1}{{\sinh x}}\\\cosh x = \frac{{{e^x} + {e^{ - x}}}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,{\mathop{\rm sech}\nolimits} x = \frac{1}{{\cosh x}}\\\tanh x = \frac{{\sinh x}}{{\cosh x}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\coth x = \frac{{\cosh x}}{{\sinh x}}\end{aligned}\)

Write the expression in terms of \({e^x}\) and \({e^{ - x}}\)

Write the expression in terms of \({e^x}\) and \({e^{ - x}}\) as shown below:

\(\begin{aligned}8\sinh x + 5\cosh x = 8\left( {\frac{{{e^x} - {e^{ - x}}}}{2}} \right) + 5\left( {\frac{{{e^x} + {e^{ - x}}}}{2}} \right)\\ &= \frac{8}{2}\left( {\frac{{{e^x}}}{2} - \frac{{{e^{ - x}}}}{2}} \right) + 5\left( {\frac{{{e^x}}}{2} + \frac{{{e^{ - x}}}}{2}} \right)\\ &= \frac{8}{2}{e^x} - \frac{8}{2}{e^{ - x}} + \frac{5}{2}{e^x} + \frac{5}{2}{e^{ - x}}\\ &= \frac{{13}}{2}{e^x} - \frac{{13}}{2}{e^{ - x}}\end{aligned}\)

Thus, the expression in terms of \({e^x}\) and \({e^{ - x}}\) is \(\frac{{13}}{2}{e^x} - \frac{{13}}{2}{e^{ - x}}\).