삼각함수 값

$\displaystyle 4 \sin ^{2} \theta = 5 \sin ^{2} \left( \frac{\pi}{2} - \theta \right)$에서 $\displaystyle \sin \left( \frac{\pi}{2} - \theta \right) = \cos \theta$이므로 $4 \sin ^{2} \theta = 5 \cos ^{2} \theta$ $\cdots$$\cdots$ ㉠ $\sin ^{2} \theta + \cos ^{2} \theta = 1$에서 $\sin ^{2} \theta = 1 - \cos ^{2} \theta$이므로 ㉠의 좌변에 대입하면 $4 \left( 1 - \cos ^{2} \theta \right) = 5 \cos ^{2} \theta$, $9 \cos ^{2} \theta = 4$ $\displaystyle \cos ^{2} \theta = \frac{4}{9}$ $\displaystyle \frac{\pi}{2} < \theta < \pi$에서 $\cos \theta < 0$이므로 $\displaystyle \cos \theta = - \frac{2}{3}$ 또한 ㉠에서 $\displaystyle 4 \sin ^{2} \theta = 5 \cos ^{2} \theta = 5 \times \frac{4}{9} = \frac{20}{9}$이므로 $\displaystyle \sin ^{2} \theta = \frac{5}{9}$ $\displaystyle \frac{\pi}{2} < \theta < \pi$에서 $\sin \theta > 0$이므로 $\displaystyle \sin \theta = \frac{\sqrt{5}}{3}$ $\displaystyle \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{\sqrt{5}}{3}}{- \frac{2}{3}} = - \frac{\sqrt{5}}{2}$ 따라서 구하는 값은 $\displaystyle \sin \theta \times \tan \theta = \frac{\sqrt{5}}{3} \times \left( - \frac{\sqrt{5}}{2} \right) = - \frac{5}{6}$