Question

Prove the identity

12. \(\cosh \left( { - x} \right) = \cosh x\) (This shows that \(\cosh \) is an even function.)

Short answer

It is proved that \(\cosh \left( { - x} \right) = \cosh x\).

Step-by-step solution

Definition of Hyperbolic function

The formulas of the hyperbolic function as shown below:

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

Prove the identity\(\cosh \left( { - x} \right) = \cosh x\)

Prove the expression as shown below:

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

Thus, it is proved that \(\cosh \left( { - x} \right) = \cosh x\).