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isee
Posted 2021-11-26 22:26
Last edited by isee 2021-11-26 22:46回复 5# player1703
记 $\angle CAB=x$，易知 $x<60^\circ$，在 $\triangle ABD$ 中由 Ceva 定理角元形式得
\[\frac {\sin 15^\circ}{\sin x}\cdot \frac {\sin \left(90^\circ-\frac x2\right)}{\sin x}\cdot \frac {\sin \left(90^\circ -\frac {3x}2\right)}{\sin (x+15^\circ)}=1,\]
开始三角化简
\begin{align*}
\sin 15^\circ\cos \left(\frac x2\right)\cos\left(\frac {3x}2\right)&=\sin^2 x\sin (x+15^\circ)\\[1em]
\sin 15^\circ\cdot \frac 12(\cos 2x+\cos x)&=(1-\cos^2x)\sin (x+15^\circ)\\[1em]
\sin 15^\circ(\cos x+1)(2\cos x-1)&=2(1-\cos x)(1+\cos x)\sin (x+15^\circ)\\[1em]
\sin 15^\circ(2\cos x-1)-2(1-\cos x)\sin (x+15^\circ)&=0
\end{align*}
函数$$f(x)=\sin 15^\circ(2\cos x-1)-2(1-\cos x)\sin (x+15^\circ),x\in (0,60^\circ),$$单调递减，且$$f(30^\circ)=0.$$ |
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