Question

Prove the identity \(cosh(x + y) = coshxcoshy + sinhxsinhy\).

Short answer

The identity \(\cos h(x + y) = \cos hx\cosh y + \sinh x\sinh y\) is proved.

Step-by-step solution

Given identity

The identity is \(\cos h(x + y) = \cos hx\cosh y + \sinh x\sinh y\).

Formula of hyperbolic function and sine cosine functions

Hyperbolic function and sine cosine function:

\(\begin{array}{c}sinhx = \frac{{{e^x} - {e^{ - x}}}}{2}\\\cosh x = \frac{{{e^x} + {e^{ - x}}}}{2}\end{array}\)

Use the formula and substitute the value

The definition of the hyperbolic function is \(\cos hx = \frac{{{e^x} + {e^{ - x}}}}{2}\).

\(\begin{array}{c}\cosh (x + y) = \frac{{{e^{x + y}} + {e^{ - (x + y)}}}}{2}\\ = \frac{{{e^{x + y}} + {e^{ - x - y}}}}{2}\end{array}\)

Multiply the numerator and the denominator by \(2\).

\(\cos h(x + y) = \frac{{2{e^{x + y}} + 2{e^{ - (x - y)}}}}{4}\)

Add and subtract the term \({e^{x - y}}\) and \({e^{y - x}}\) on numerator and obtain the required identity.

\(\begin{array}{c}\cos h(x + y) = \frac{{2{e^{x + y}} + 2{e^{ - (x + y)}} + {e^{x - y}} - {e^{x - y}} + {e^{y - x}} - {e^{y - x}}}}{4}\\ = \frac{{\left( {{e^{x + y}} + {e^{y - x}} + {e^{x - y}} + {e^{ - x - y}}} \right)}}{4} + \frac{{\left( {{e^{x + y}} - {e^{x - y}} - {e^{y - x}} + {e^{ - y - x}}} \right)}}{4}\\ = \frac{{\left( {{e^y}\left( {{e^x} + {e^{ - x}}} \right) + {e^{ - y}}\left( {{e^x} + {e^{ - x}}} \right)} \right)}}{4} + \frac{{\left( {{e^x}\left( {{e^y} - {e^{ - y}}} \right) - {e^{ - x}}\left( {{e^y} - {e^{ - y}}} \right)} \right)}}{4}\\ = \left( {\frac{{{e^x} + {e^{ - x}}}}{2}} \right)\left( {\frac{{{e^y} + {e^{ - y}}}}{2}} \right) + \left( {\frac{{{e^x} - {e^{ - x}}}}{2}} \right)\left( {\frac{{{e^y} - {e^{ - y}}}}{2}} \right)\end{array}\)

Use the formula of the hyperbolic function and sine cosine functions \(\sin hx = \frac{{{e^x} - {e^{ - x}}}}{2}\)and \(\cos hx = \frac{{{e^2} + {e^{ - x}}}}{2}\).

\(\begin{array}{c}\cos h(x + y) = (\cos hx)(\cosh y) + (\sin hx)(\sinh y)\\ = \cosh x\cosh y + \sinh x\sinh y\end{array}\)

Hence, the required identity \(\cos h(x + y) = \cos hx\cosh y + \sinh x{\mathop{\rm sinhy}\nolimits} \) is proved.