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\[\frac{AD}{\sin x}=\frac{CD}{\sin30^{\circ}}, \frac{AD}{\sin30^{\circ}}=\frac{BD}{\sin65^{\circ}}, \frac{CD}{\sin40^{\circ}}=\frac{BD}{\sin(x+75^{\circ})}\]
\[\frac{\sin x}{\sin30^{\circ}}=\frac{\sin30^{\circ}\sin(x+75^{\circ})}{\sin65^{\circ}\sin40^{\circ}}\]
\[\sin(x+75^{\circ})=4\sin65^{\circ}\sin40^{\circ}\sin x=2(\cos25^{\circ}-\cos105^{\circ})\sin x=2(\sin65^{\circ}+\cos75^{\circ})\sin x\]
\[\sin x\cos75^{\circ}+\cos x\sin75^{\circ}=2(\sin65^{\circ}+\cos75^{\circ})\sin x\]
\[\cot x=\frac{2\sin65^{\circ}+\cos75^{\circ}}{\sin75^{\circ}}\]
因为$\cot x$在$[0^{\circ},180^{\circ}]$中是单调的，所以它只能有一个解，而画图可知那个解就是$x=25^{\circ}$，所以只要验证$x=25^{\circ}$满足方程就行了，就是证明$\frac{\cos25^{\circ}}{\sin25^{\circ}}=\frac{2\sin65^{\circ}+\cos75^{\circ}}{\sin75^{\circ}}$，而
\begin{align*}
\cos25^{\circ}\sin75^{\circ}-(2\sin65^{\circ}+\cos75^{\circ})\sin25^{\circ}
&=\sin65^{\circ}\sin75^{\circ}-2\sin65^{\circ}\sin25^{\circ}-\sin15^{\circ}\sin25^{\circ}\\
&=\frac{1}{2}[\cos140^{\circ}-\cos10^{\circ}-2\cos90^{\circ}+2\cos40^{\circ}-\cos40^{\circ}+\cos10^{\circ}]\\
&=0
\end{align*}
所以这是成立的，$x=25^{\circ}$就是唯一的解。 |
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abababa
posted 2022-5-19 17:24
kuing
posted 2022-5-20 13:33
