Question

Prove the identity\(sinh(x + y) = sinhxcoshy + coshxsinhy\).

Short answer

The identity \(\sinh (x + y) = \sinh x\cosh y + \cosh x{\mathop{\rm sinhy}\nolimits} \) is proved.

Step-by-step solution

Given

The identity is \(\sinh (x + y) = \sinh x\cosh y + \cosh x{\mathop{\rm sinhy}\nolimits} \).

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 sine function is \(\sin hx = \frac{{{e^x} - {e^{ - x}}}}{2}\).

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

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

\(\begin{array}{c}\sinh (x + y) = \frac{{2{e^{x + y}} - 2{e^{ - (x - y)}}}}{4}\\ = \frac{{2{e^{x + y}} - 2{e^{ - (s - y)}} + {e^{x - y}} - {e^{x - y}} + {e^{y - x}} - {e^{y - x}}}}{4}\\ = \frac{{\left( {{e^{x + y}} + {e^{x - y}} - {e^{y - x}} - {e^{ - (x + y)}}} \right) + \left( {{e^{x + y}} - {e^{x - y}} + {e^{y - x}} - {e^{ - (sty)}}} \right)}}{4}\\ = \frac{{\left( {{e^{x + y}} + {e^{x - y}} - {e^{\left. {{e^{ - x}} - {e^{ - (x + y)}}} \right)}}} \right.}}{4} + \frac{{\left( {{e^{x + y}} - {e^{x - y}} + {e^{y - x}} - {e^{ - (x + y)}}} \right)}}{4}\end{array}\)

Simplify further to obtain the identity.

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

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

\(\begin{array}{c}\sinh (x + y) = (\cosh y)(\sinh x) + (\cosh x)(\sinh y)\\ = \cosh x\sinh y + \sinh x\cosh y\end{array}\)

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