$a,b,c,A,B,C$ 是任意实数，$A,B,C\in(0,\pi)$，由
$$\cos A= \frac{\cosh b \cosh c -\cosh a}{\sinh a \sinh b}$$
$$\cos B = \frac{\cosh c \cosh a - \cosh b}{\sinh b \sinh c}$$
$$\cos C = \frac{\cosh a \cosh b - \cosh c}{\sinh c \sinh a}$$
怎么推出
$$\cosh a =\frac{\cos B \cos C + \cos A}{\sin B \sin C}$$
直接代入
$$\frac{\cos B \cos C + \cos A}{\sin B \sin C}=\frac{\left(\frac{\cosh c \cosh a - \cosh b}{\sinh b \sinh c}\right) \left(\frac{\cosh a \cosh b - \cosh c}{\sinh c \sinh a}\right) + \frac{\cosh b \cosh c - \cosh a}{\sinh a \sinh b}}{\sqrt{1-\left(\frac{\cosh a \cosh b - \cosh c}{\sinh c \sinh a}\right)^2}\sqrt{1-\left(\frac{\cosh b \cosh c - \cosh a}{\sinh a \sinh b}\right)^2}}$$
这个能化简出$\cos a$吗
由第1式$\cos C = \frac{\cosh a \cosh b - \cosh c}{\sinh c \sinh a}$，得
\begin{aligned}
\frac{\sin ^2 C}{\sinh^2 c} & =\left(1-\frac{\left(\cosh a \cosh b-\cosh^2 c\right)^2}{\sinh^2 a \sinh^2 b}\right) / \sinh^2 c \\
& =\frac{\sinh^2 a \sinh^2 b-\cosh^2 c \cosh^2 b-\cosh^2 c+2 \cosh a \cosh b \cosh c}{\sinh^2 a \sinh^2 b \sinh^2 c} \\
& =\frac{1-\cosh^2 a-\cosh^2 b-\cosh^2 c+2 \cosh a \cosh b \cosh c}{\sinh^2 a \sinh^2 b \sinh^2 c}
\end{aligned}令$F=1-\cosh^2 a-\cosh^2 b-\cosh^2 c+2 \cosh a \cosh b \cosh c$，上式两边开方
\[
\sin C=\sqrt{F} / \sinh a \sinh b .
\]
同样有
\[
\begin{aligned}
& \sin B=\sqrt{F} / \sinh a\sinh c . \\
& \sin A=\sqrt{F} / \sinh b \sinh c .
\end{aligned}
\]
这样, 我们有
\[
\frac{\cos A \cos B+\cos C}{\sin A \sin B}=\frac{1}{F}\left(\sinh a \sinh^2 \sinh^2 c\right)
(\cos A \cos B+\cos C)\]
由第1式
\[
\begin{aligned}
& \cos C=\frac{\cosh a \cosh b-\cosh c}{\sinh a \sinh b}, \\
& \cos B=\frac{\cosh a \cosh c-\cosh b}{\sinh a \sinh c}, \\
& \cos A=\frac{\cosh b \cosh c-\cosh a}{\sinh b \sinh c} .
\end{aligned}
\]
代入得
\[
\begin{aligned}
& \frac{\cos A \cos B+\cos C}{\sin A \sin B}=\frac{1}{F}\left(\sinh a \sinh b \sinh^2 c\right)\times\left[\frac{(\cosh a \cosh c-\cosh b)(\cosh b \cosh c-\cosh a)}{\sinh a \sinh b \sinh^2 c}+\frac{\cosh a \cosh b-\cosh c}{\sinh a \sinh b}\right] \\
& =\frac{1}{F}[(\cosh a \cosh c-\cosh b)(\cosh b \cosh c-\cosh a) +\sinh^2 c(\cosh a \cosh b-\cosh c)]
\end{aligned}
\]
注意到 $\sinh^2 c=\cosh^2 c-1$, 上式右端经过化简后即得 $\cosh c$。这样就证明了第 2 式.
kuing
发表于 2024-4-2 00:30
