[210429] 2021년 4월 학평 미적분 29번

$\angle BAC=\dfrac{2}{3}\pi$이고 $\overline{AB}>\overline{AC}$인 삼각형 $ABC$가 있다. $\overline{BD}=\overline{CD}$인 선분 $AB$ 위의 점 $D$에 대하여 $\angle CBD= \alpha$, $\angle ACD = \beta$라 하자. $\cos^2\alpha=\dfrac{7+\sqrt{21}}{14}$일 때, $54\sqrt3\times\beta$의 값을 구하시오.

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$\angle DCB = \angle DBC = \alpha$이고 $\angle ADC = \angle DBC + \angle DCB = 2\alpha$

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

$\cos2\alpha=2\cos^2\alpha-1=\dfrac{7+\sqrt{21}}{7}-1=\dfrac{\sqrt{21}}{7}$

$\sin2\alpha=\sqrt{1-\cos^22\alpha}=\dfrac{2\sqrt7}{7}$

$\tan2\alpha=\dfrac{\sin2\alpha}{\cos2\alpha}=\dfrac{2}{\sqrt3}$

$\tan\beta=\tan(\dfrac{\pi}{3}-2\alpha)=\dfrac{\tan\dfrac{\pi}{3}-\tan2\alpha}{1+\tan\dfrac{\pi}{3}\tan2\alpha}=\dfrac{\sqrt3-\dfrac{2}{3}\sqrt3}{1+\sqrt3\cdot\dfrac{2}{3}\sqrt3}=\dfrac{\sqrt3}{9}$

$\therefore 54\sqrt3\tan\beta=54\sqrt3\cdot\dfrac{\sqrt3}{9}=18$