2021년 04월 미적분 29번

29. 그림과 같이 $\mathrm{\angle }BAC=\frac{2}{3}\pi$$\angle BAC=\cfrac 23 \pi$이고 $\overline{AB}>\overline{AC}$$\overline{AB}>\overline{AC}$인 삼각형 $ABC$$ABC$가 있다. $\overline{BD}=\overline{CD}$$\overline{BD}=\overline{CD}$인 선분 $AB$$AB$ 위의 점 $D$$D$에 대하여 $\mathrm{\angle }CBD=\alpha$$\angle CBD=\alpha$, $\mathrm{\angle }ACD=\beta$$\angle ACD=\beta$라 하자. ${\cos }^2\alpha =\frac{7+\sqrt{21}}{14}$$\cos^2 \alpha = \cfrac{7+\sqrt{21}}{14}$일 때, $54\sqrt{3}\times \tan \beta$$54\sqrt 3 \times \tan\beta$의 값을 구하시오.

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i. 정리

• $\mathrm{\angle }DCB=\alpha$$\angle DCB=\alpha$
• $\mathrm{\angle }ADC=2\alpha$$\angle ADC=2\alpha$

얼래?

• $\beta =\pi -(\frac{2}{3}\pi +2\alpha )=\frac{\pi }{3}+2\alpha$$\beta =\pi -(\cfrac 23\pi +2\alpha)=\cfrac {\pi}3 +2\alpha$

ii. 계산

$\cos 2\alpha$$\cos 2\alpha$를 찾으면, 대충 삼각형을 그려서 $\tan 2\alpha$$\tan 2\alpha$를 구하면 되겠다.

$\begin{array}{rl}\cos 2\alpha & ={\cos }^2\alpha -{\sin }^2\alpha \\ \\ & ={\cos }^2\alpha -(1-{\sin }^2\alpha )\\ \\ & =2{\cos }^2-1\\ \\ & =\frac{\sqrt{21}}{7}\end{array}$$\begin{aligned} \cos 2\alpha &= \cos ^2 \alpha -\sin^2 \alpha \\ \\ &= \cos^2 \alpha -(1-\sin^2 \alpha ) \\ \\ &=2\cos^2 -1\\ \\ &=\cfrac {\sqrt{21}} 7\end{aligned}$

$\therefore \tan 2\alpha =\frac{2\sqrt{7}}{\sqrt{21}}=\frac{2}{\sqrt{3}}$$\therefore~\tan 2\alpha =\cfrac {2\sqrt 7}{\sqrt{21}}=\cfrac {2}{\sqrt 3}$

$\begin{array}{rl}\tan \beta & =\tan (\frac{\pi }{3}+2\alpha )\\ \\ & =\frac{\tan \frac{\pi }{3}-\tan 2\alpha }{1+\tan \frac{\pi }{3}\cdot \tan 2\alpha }\\ \\ & =\frac{\sqrt{3}-\frac{2}{\sqrt{3}}}{3}\end{array}$$\begin{aligned} \tan \beta &=\tan \big(\cfrac {\pi}3+2\alpha) \\ \\&= \cfrac {\tan \frac{\pi}3 -\tan 2\alpha }{1+\tan \frac{\pi}3\cdot \tan 2\alpha} \\ \\ &=\cfrac {\sqrt 3-\frac 2{\sqrt 3}}3\end{aligned}$

$\therefore 54\sqrt{3}\times \tan \beta =18$$\therefore~54\sqrt 3\times\tan \beta =18$