Question

Prove the identity.

16. \(\cosh \left( {x + y} \right) = \cosh x\cosh y + \sinh x\sinh y\)

Short answer

It is proved that \(\cosh \left( {x + y} \right) = \cosh x\cosh y + \sinh x\sinh y\).

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

Prove the identity as shown below:

\(\begin{aligned}\cosh x\cosh y + \sinh x\sinh y &= \left( {\frac{{{e^x} + {e^{ - x}}}}{2}} \right)\left( {\frac{{{e^y} + {e^{ - y}}}}{2}} \right) + \left( {\frac{{{e^x} - {e^{ - x}}}}{2}} \right)\left( {\frac{{{e^y} - {e^{ - y}}}}{2}} \right)\\ &= \frac{{{e^{x + y}} + {e^{x - y}} + {e^{ - x + y}} + {e^{ - x - y}}}}{4} + \frac{\begin{aligned}{l}{e^{x + y}} - {e^{x - y}} - {e^{ - x + y}}\\ - {e^{ - x - y}}\end{aligned}}{4}\\ &= \frac{1}{4}\left( {\left( {{e^{x + y}} + {e^{x - y}} + {e^{ - x + y}} + {e^{ - x - y}}} \right) + \left( \begin{aligned}{l}{e^{x + y}} - {e^{x - y}}\\ - {e^{ - x + y}} - {e^{ - x - y}}\end{aligned} \right)} \right)\\ &= \frac{1}{4}\left( {2{e^{x + y}} + 2{e^{ - x - y}}} \right)\\ &= \frac{2}{4}\left( {{e^{x + y}} + {e^{ - x - y}}} \right)\\ &= \frac{1}{2}\left( {{e^{x + y}} + {e^{ - \left( {x + y} \right)}}} \right)\\ &= \cosh \left( {x + y} \right)\end{aligned}\)

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