Mollweide公式

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在$\triangle ABE$使用正弦定理\[\frac{\sin \frac{1}{2}(\alpha - \beta)}{\cos \frac{1}{2}\gamma}=\frac{a - b}{c}\]
Proof without Words: Mollweide's Equation
Rex H. Wu
The Story of Mollweide and Some Trigonometric Identities

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Mollweide's formula in teaching trigonometry
Mollweide 公式如下：
\begin{eqnarray}
  \frac{\sin \frac{1}{2}(\alpha - \beta)}{\cos \frac{1}{2}\gamma} &=& \frac{a - b}{c} \label{1}\\
  \frac{\cos \frac{1}{2}(\alpha - \beta)}{\sin \frac{1}{2}\gamma} &=& \frac{a + b}{c}. \label{2}
\end{eqnarray}在整篇文章中，采用以下约定。在 $\triangle ABC$, $\alpha = \measuredangle BAC$, $\beta = \measuredangle ABC$ 和 $\gamma = \measuredangle  ACB$，因此 $\alpha + \beta + \gamma = \pi$。此外，$a = BC$、$b = AC$ 和 $c = AB$。

式 \eqref{1} 的证明见 1#，下面推导式\eqref{2}。首先，考虑上图中的 $\triangle ABC$。 取角 $C$ 的平分线，将 $\gamma$ 切成两半。构造 $\triangle ABF$ 使得 $CD \perp AF$ 和 $BF \perp AF$。因此，$\triangle ACE$ 是一个等腰三角形，其中 $E$ 是边 $BC$ 和 $AF$ 的交点。因此，$2\measuredangle CAD + \gamma = \pi = \alpha + \beta + \gamma$ 因此 $\measuredangle CAD = (\alpha + \beta)/2 = \measuredangle CED$。此外，$\measuredangle BAD + \measuredangle CAD = \alpha$，所以 $\measuredangle BAD = (\alpha - \beta)/2$。考虑直角三角形 $BEF$。我们有 $\measuredangle BEF = \measuredangle CED$ 因为它们与顶点 $E$ 相对。所以，$\measuredangle EBF = \pi/2 - (\alpha + \beta)/2 = \gamma/2$。

考虑 $\triangle ABF$，我们知道
\begin{equation}
  \cos \frac{1}{2}(\alpha - \beta) = \frac{AF}{AB} = \frac{AF}{c}. \label{3}
\end{equation}
此外，分别考虑 $\triangle ACD$、$\triangle DCE$ 和 $\triangle EBF$。我们现在有以下关系：
\begin{align*}
  \sin \frac{1}{2} \gamma &= \frac{AD}{AC} = \frac{AD}{b}                  &\Longrightarrow&  &AD& = b \sin \frac{1}{2} \gamma \\
                                                  &= \frac{DE}{CE} = \frac{DE}{b}                   &\Longrightarrow&  &DE& = b \sin \frac{1}{2} \gamma \\
                                                  &= \frac{EF}{BE} = \frac{EF}{a - b}    &\Longrightarrow&  &EF& = (a - b) \sin \frac{1}{2} \gamma.
\end{align*}
此外，$AF = AD + DE + EF = (a + b) \sin \frac{1}{2} \gamma$。将其代入 \eqref{3}，得到式 \eqref{2}。

根据对称性，我们将 Mollweide 公式 \eqref{1} 和 \eqref{2} 表示如下：
\begin{eqnarray}
  \frac{\sin \frac{1}{2}(\alpha - \beta)}{\cos \frac{1}{2}\gamma} = \frac{a - b}{c} &\qquad&
  \frac{\sin \frac{1}{2}(\beta - \gamma)}{\cos \frac{1}{2}\alpha} = \frac{b - c}{a} \qquad \quad
  \frac{\sin \frac{1}{2}(\gamma - \alpha)}{\cos \frac{1}{2}\beta} = \frac{c - a}{b} \\
  \frac{\cos \frac{1}{2}(\alpha - \beta)}{\sin \frac{1}{2}\gamma} = \frac{a + b}{c} &\qquad&
  \frac{\cos \frac{1}{2}(\beta - \gamma)}{\sin \frac{1}{2}\alpha} = \frac{b + c}{a} \qquad \quad
  \frac{\cos \frac{1}{2}(\gamma - \alpha)}{\sin \frac{1}{2}\beta} = \frac{c + a}{b}.
\end{eqnarray}

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Proof using the law of cotangents
Mollweide's first formula. From the addition formula and the law of cotangents we have
\[\frac {\sin \left( \tfrac{\alpha}{2}-\tfrac{\beta}{2} \right) }{\sin \left( \tfrac{\alpha}{2}+\tfrac{\beta}{2} \right) } = \frac {\cot \left( \tfrac{\beta}{2} \right) -\cot \left( \tfrac{\alpha}{2} \right) }{\cot \left( \tfrac{\beta}{2} \right) +\cot \left( \tfrac{\alpha}{2} \right) }=  \frac {a-b}{2s-a-b}.\]
This gives the result
\[\dfrac {a-b}{c}=\dfrac {\sin \left( \tfrac{\alpha}{2}-\tfrac{\beta}{2} \right)}{\cos \left( \tfrac{\gamma}{2} \right) }\]as required.
Mollweide's second formula. From the addition formula and the law of cotangents we have
\begin{align*}
& \frac {\cos\left( \tfrac{\alpha}{2}-\tfrac{\beta}{2} \right) }{\cos\left( \tfrac{\alpha}{2}+\tfrac{\beta}{2} \right) } =
\frac {\cot\left( \tfrac{\alpha}{2} \right) \cot\left( \tfrac{\beta}{2} \right) +1}{\cot\left( \tfrac{\alpha}{2} \right) \cot \left( \tfrac{\beta}{2} \right) -1} \\[6pt]
= {} & \frac {\cot \left( \tfrac{\alpha}{2} \right) +\cot \left( \tfrac{\beta}{2} \right) +2\cot \left( \tfrac{\gamma}{2} \right) }{\cot \left( \tfrac{\alpha}{2} \right) +\cot \left( \tfrac{\beta}{2} \right) } =
\frac {4s-a-b-2c}{2s-a-b}.
\end{align*}
Here, an extra step is required to transform a product into a sum, according to the sum/product formula.
This gives the result
\[\dfrac {b+a}{c} = \dfrac{\cos \left( \tfrac{\alpha}{2} - \tfrac{\beta}{2} \right) }{\sin \left( \tfrac{\gamma}{2} \right) }\]
as required.
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