Question

# Derive $\cosh ^{2}(x)+\sinh ^{2}(x)=\cosh (2 x)$ from the definition.

Solution

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$\begin{gathered} {\cosh ^2}\left( x \right) + {\sin ^2}\left( x \right) = \cosh \left( {2x} \right) \\ \textcolor{#4257b2}{{\text{Use the definitions of The Hyperbolic Functions}}} \\ \sinh u = \frac{{{e^u} - {e^{ - u}}}}{2}{\text{ and }}\cosh u = \frac{{{e^u} + {e^{ - u}}}}{2} \\ \textcolor{#4257b2}{{\text{Therefore}}{\text{,}}} \\ {\left( {\frac{{{e^x} + {e^{ - x}}}}{2}} \right)^2} + {\left( {\frac{{{e^x} - {e^{ - x}}}}{2}} \right)^2} = \frac{{{e^{2x}} + {e^{ - 2x}}}}{2} \\ \textcolor{#4257b2}{{\text{Simplify}}} \\ \frac{{{e^{2x}} + 2{e^x}{e^{ - x}} + {e^{ - 2x}}}}{4} + \frac{{{e^{2x}} - 2{e^x}{e^{ - x}} + {e^{ - 2x}}}}{4} = \frac{{{e^{2x}} + {e^{ - 2x}}}}{2} \\ \frac{{{e^{2x}} + 2 + {e^{ - 2x}}}}{4} + \frac{{{e^{2x}} - 2 + {e^{ - 2x}}}}{4} = \frac{{{e^{2x}} + {e^{ - 2x}}}}{2} \\ \textcolor{#4257b2}{ {\text{Recall that }}\frac{a}{b} + \frac{c}{b} = \frac{{a + c}}{b}} \\ \frac{{{e^{2x}} + 2 + {e^{ - 2x}} + {e^{2x}} - 2 + {e^{ - 2x}}}}{4} = \frac{{{e^{2x}} + {e^{ - 2x}}}}{2} \\ \frac{{{e^{2x}} + {e^{ - 2x}} + {e^{2x}} + {e^{ - 2x}}}}{4} = \frac{{{e^{2x}} + {e^{ - 2x}}}}{2} \\ \frac{{2{e^{2x}} + 2{e^{ - 2x}}}}{4} = \frac{{{e^{2x}} + {e^{ - 2x}}}}{2} \\ \frac{{{e^{2x}} + {e^{ - 2x}}}}{2} = \frac{{{e^{2x}} + {e^{ - 2x}}}}{2} \\ \textcolor{#4257b2}{{\text{Verified}}} \\ \end{gathered}$

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