[210429] 2021년 4월 학평 미적분 29번

$\mathrm{\angle }BAC={\displaystyle \frac{2}{3}}\pi$이고 $\overline{AB}>\overline{AC}$인 삼각형 $ABC$가 있다. $\overline{BD}=\overline{CD}$인 선분 $AB$ 위의 점 $D$에 대하여 $\mathrm{\angle }CBD=\alpha$, $\mathrm{\angle }ACD=\beta$라 하자. ${\cos }^2\alpha ={\displaystyle \frac{7+\sqrt{21}}{14}}$일 때, $54\sqrt{3}\times \beta$의 값을 구하시오.

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$\mathrm{\angle }DCB=\mathrm{\angle }DBC=\alpha$이고 $\mathrm{\angle }ADC=\mathrm{\angle }DBC+\mathrm{\angle }DCB=2\alpha$

$\beta =2\pi -({\displaystyle \frac{2}{3}}\pi +2\alpha )={\displaystyle \frac{\pi }{3}}-2\alpha$

$\cos 2\alpha =2{\cos }^2\alpha -1={\displaystyle \frac{7+\sqrt{21}}{7}}-1={\displaystyle \frac{\sqrt{21}}{7}}$

$\sin 2\alpha =\sqrt{1-{\cos }^{2}2\alpha }={\displaystyle \frac{2\sqrt{7}}{7}}$

$\tan 2\alpha ={\displaystyle \frac{\sin 2\alpha }{\cos 2\alpha }}={\displaystyle \frac{2}{\sqrt{3}}}$

$\tan \beta =\tan ({\displaystyle \frac{\pi }{3}}-2\alpha )={\displaystyle \frac{\tan {\displaystyle \frac{\pi }{3}}-\tan 2\alpha }{1+\tan {\displaystyle \frac{\pi }{3}}\tan 2\alpha }}={\displaystyle \frac{\sqrt{3}-{\displaystyle \frac{2}{3}}\sqrt{3}}{1+\sqrt{3}\cdot {\displaystyle \frac{2}{3}}\sqrt{3}}}={\displaystyle \frac{\sqrt{3}}{9}}$

$\therefore 54\sqrt{3}\tan \beta =54\sqrt{3}\cdot {\displaystyle \frac{\sqrt{3}}{9}}=18$