证明:
\[
\begin{aligned}
\sinh(x+y)=\sinh(x)\cosh(y)+\cosh(x)\sinh(y)\\
\sinh(x-y)=\sinh(x)\cosh(y)-\cosh(x)\sinh(y)\\
\cosh(x+y)=\cosh(x)\cosh(y)+\sinh(x)\sinh(y)\\
\cosh(x-y)=\cosh(x)\cosh(y)-\sinh(x)\sinh(y)\\
\end{aligned}
\]
1
证明: 我们只需证明第一式和第三式. 第二式只需注意到 $\sinh(x-y)=\sinh(x+(-y))$ 并利用第一式即可.
(1)
\[
\sinh(x+y)=\frac{e^{x+y}-e^{-x-y}}{2}
\]
\[
\begin{split}
\sinh(x)\cosh(y)&=\frac{e^{x}-e^{-x}}{2}\cdot\frac{e^{y}+e^{-y}}{2}\\
&=\frac{1}{4}\Bigl[e^{x+y}+e^{x-y}-e^{-x+y}-e^{-x-y}\Bigr]
\end{split}
\]
\[
\begin{split}
\cosh(x)\sinh(y)&=\frac{e^{x}+e^{-x}}{2}\cdot\frac{e^{y}-e^{-y}}{2}\\
&=\frac{1}{4}\Bigl[e^{x+y}-e^{x-y}+e^{-x+y}-e^{-x-y}\Bigr]
\end{split}
\]
因此,
\[
\begin{split}
\sinh(x)\cosh(y)+\cosh(x)\sinh(y)&=\frac{1}{2}\Bigl[e^{x+y}-e^{-x-y}\Bigr]\\
&=\sinh(x+y).
\end{split}
\]
(3)
\[
\cosh(x+y)=\frac{e^{x+y}+e^{-x-y}}{2}
\]
\[
\begin{split}
\cosh(x)\cosh(y)&=\frac{e^{x}+e^{-x}}{2}\cdot\frac{e^{y}+e^{-y}}{2}\\
&=\frac{1}{4}\Bigl[e^{x+y}+e^{x-y}+e^{-x+y}+e^{-x-y}\Bigr]
\end{split}
\]
\[
\begin{split}
\sinh(x)\sinh(y)&=\frac{e^{x}-e^{-x}}{2}\cdot\frac{e^{y}-e^{-y}}{2}\\
&=\frac{1}{4}\Bigl[e^{x+y}-e^{x-y}-e^{-x+y}+e^{-x-y}\Bigr]
\end{split}
\]
因此,
\[
\begin{split}
\cosh(x)\cosh(y)+\sinh(x)\sinh(y)&=\frac{1}{2}\Bigl[e^{x+y}+e^{-x-y}\Bigr]\\
&=\cosh(x+y).
\end{split}
\]