Cho $\tan \alpha=-\sqrt{5}\left(\frac{\pi}{2}\lt \alpha\lt \pi\right)$.Tính $\cos \alpha$ và $\sin 2 \alpha$.

Q: Cho $\tan \alpha=-\sqrt{5}\left(\frac{\pi}{2}\lt \alpha\lt \pi\right)$.Tính $\cos \alpha$ và $\sin 2 \alpha$.

Do $\frac{\pi}{2}\lt \alpha\lt \pi \Rightarrow \cos \alpha\lt 0$Ta có: $\frac{1}{\cos ^{2} \alpha}=1+\tan ^{2} \alpha=6 \Rightarrow \cos \alpha=\frac{-\sqrt{6}}{6}$ $\sin \alpha=\cos \alpha \cdot \tan \alpha=\frac{\sqrt{30}}{6} \Rightarrow \sin 2 \alpha=2 \sin \alpha \cdot \cos \alpha=\frac{-\sqrt{5}}{3}$

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Accepted Answer

Do $\frac{\pi}{2}\lt \alpha\lt \pi \Rightarrow \cos \alpha\lt 0$

Ta có: $\frac{1}{\cos ^{2} \alpha}=1+\tan ^{2} \alpha=6 \Rightarrow \cos \alpha=\frac{-\sqrt{6}}{6}$

$\sin \alpha=\cos \alpha \cdot \tan \alpha=\frac{\sqrt{30}}{6} \Rightarrow \sin 2 \alpha=2 \sin \alpha \cdot \cos \alpha=\frac{-\sqrt{5}}{3}$

THPT M.V.Lômônôxốp - Đề thi cuối kì 2 (CT)18-19 - Hà Nội - MĐ 6609