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# 和差角幾何解

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$$
已知
\overline{AB}=\overline{AC}=\overline{AD}
$$

## 和角

$$
令\cases{
\angle{\alpha}=\angle{BAC}=\angle{EDH}\\
\angle{\beta}=\angle{CAD}
}
\implies\cases{
\sin\beta=\frac{\overline{ED}}{\overline{AD}}\\
\cos\beta=\frac{\overline{AE}}{\overline{AD}}\\
}
$$

$$\begin{array}l
\sin{(\alpha+\beta)} &= \frac{\overline{HD}}{\overline{AD}}\\
&=\frac{\overline{HG}}{\overline{AD}}+\frac{\overline{GD}}{\overline{AD}}\\
&又 \cases{
\overline{HG} = \overline{FE} =\overline{AE}\sin\alpha\\
\overline{GD} = \overline{ED}\cos\alpha
}\\
&=\frac{\overline{AE}}{\overline{AD}}\sin\alpha + \frac{\overline{ED}}{\overline{AD}}\cos\alpha\\
&=\sin\alpha\cos\beta + \cos\alpha\sin\beta
\end{array}
$$

$$
\begin{array}l
\cos{(\alpha+\beta)} &= \frac{\overline{AH}}{\overline{AD}}\\
&= \frac{\overline{AF}}{\overline{AD}}-\frac{\overline{HF}}{\overline{AD}}\\
&又 \cases{
\overline{HF} = \overline{GE} = \overline{ED}\sin\alpha\\
\overline{AF} = \overline{AE}\cos\alpha\\
}\\
&=\frac{\overline{AE}}{\overline{AD}}\cos\alpha - \frac{\overline{ED}}{\overline{AD}}\sin\alpha\\
&=\cos\alpha\cos\beta - \sin\alpha\sin\beta
\end{array}
$$

## 差角

$$
令\cases{
\angle{\alpha}=\angle{BAD}=\angle{HAD}\\
\angle{\beta}=\angle{BAC}=\angle{EDH}
}
\implies\cases{
\sin\alpha=\frac{\overline{HD}}{\overline{AD}}\\
\cos\alpha=\frac{\overline{AH}}{\overline{AD}}\\
}
$$

$$
\begin{array}l
\sin{(\alpha-\beta)} &= \frac{\overline{ED}}{\overline{AD}}\\
&= \frac{\overline{KD}}{\overline{AD}}-\frac{\overline{KE}}{\overline{AD}}\\
&又\cases{
\overline{KD}=\overline{HD}\cos\beta\\
\overline{KE}=\overline{HM}=\overline{AH}\sin\beta
}\\
&=\frac{\overline{HD}}{{\overline{AD}}}\cos\beta-\frac{\overline{AH}}{\overline{AD}}\sin\beta\\
&=\sin\alpha\cos\beta-\cos\alpha\sin\beta
\end{array}
$$

$$
\begin{array}l
\cos{(\alpha-\beta)}
&=\frac{\overline{AE}}{\overline{AD}}\\
&=\frac{\overline{AM}}{\overline{AD}}+\frac{\overline{ME}}{\overline{AD}}\\
&又\cases{
\overline{AM}=\overline{AH}\cos\beta\\
\overline{ME}=\overline{HK}=\overline{HD}\sin\beta
}\\
&=\frac{\overline{AH}}{\overline{AD}}\cos\beta+\frac{\overline{HD}}{\overline{AD}}\sin\beta\\
&=\cos\alpha\cos\beta+\sin\alpha\sin\beta
\end{array}
$$

###### tags: `math` `geometry`