Question

Prove the identity.

13. \(\cosh x + \sinh x = {e^x}\)

Short answer

It is proved that \(\cosh x + \sinh x = {e^x}\).

Step-by-step solution

Definition of Hyperbolic function

The formulas of thehyperbolic functionas shown below:

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

Prove the identity\(\cosh x + \sinh x = {e^x}\)

Prove the identity as shown below:

\(\begin{aligned}\cosh x + \sinh x &= \frac{{{e^x} + {e^{ - x}}}}{2} + \frac{{{e^x} - {e^{ - x}}}}{2}\\ &= \frac{{{e^x} + {e^{ - x}} + {e^x} - {e^{ - x}}}}{2}\\ &= \frac{{2{e^x}}}{2}\\ &= {e^x}\end{aligned}\)

Thus, it is proved that \(\cosh x + \sinh x = {e^x}\).