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Câu hỏi: Tính các giá trị lượng giác của góc $\alpha,$ nếu
a) $\cos \alpha=-\frac{1}{4}, \pi<\alpha<\frac{3 \pi}{2}$
b) $\sin \alpha=\frac{2}{3}, \frac{\pi}{2}<\alpha<\pi$;
c) $\tan \alpha=\frac{7}{3}, 0<\alpha<\frac{\pi}{2}$
d) $\cot \alpha=-\frac{14}{9}, \frac{3 \pi}{2}<\alpha<2 \pi$.
a) $\cos \alpha=-\frac{1}{4}, \pi<\alpha<\frac{3 \pi}{2}$
b) $\sin \alpha=\frac{2}{3}, \frac{\pi}{2}<\alpha<\pi$;
c) $\tan \alpha=\frac{7}{3}, 0<\alpha<\frac{\pi}{2}$
d) $\cot \alpha=-\frac{14}{9}, \frac{3 \pi}{2}<\alpha<2 \pi$.
a) $\pi<\alpha<\frac{3 \pi}{2} \Rightarrow \sin \alpha<0$
Vậy $\quad \sin \alpha=-\sqrt{1-\cos ^{2} \alpha}=-\sqrt{1-\frac{1}{16}}=-\frac{\sqrt{15}}{4},$
$\tan \alpha=\frac{\sin \alpha}{\cos \alpha}=\sqrt{15}, \cot \alpha=\frac{1}{\sqrt{15}}$
b) $\frac{\pi}{2}<\alpha<\pi \Rightarrow \cos \alpha<0$
Vậy $\quad \cos \alpha=-\sqrt{1-\sin ^{2} \alpha}=-\sqrt{1-\frac{4}{9}}=\frac{-\sqrt{5}}{3},$
$\tan \alpha=\frac{\sin \alpha}{\cos \alpha}=-\frac{2}{\sqrt{5}}, \cot \alpha=-\frac{\sqrt{5}}{2}$
c) $0<\alpha<\frac{\pi}{2} \Rightarrow \cos \alpha>0, \cos ^{2} \alpha=\frac{1}{1+\tan ^{2} \alpha}$
Vậy $\cos \alpha=\frac{1}{\sqrt{1+\frac{49}{9}}}=\frac{3}{\sqrt{\bar{p} 8}}$,
$\sin \alpha=\cos \alpha \tan \alpha=\frac{7}{\sqrt{58}}, \cot \alpha=\frac{3}{7}$
d) $\frac{3 \pi}{2}<\alpha<2 \pi \Rightarrow \sin \alpha<0, \sin ^{2} \alpha=\frac{1}{1+\cot ^{2} \alpha}$.
Vậy $\sin \alpha=-\frac{1}{\sqrt{1+\frac{196}{81}}}=-\frac{9}{\sqrt{277}}$,
$\cos \alpha=\sin \alpha \cot \alpha=\frac{14}{\sqrt{277}}, \quad \tan \alpha=\frac{1}{\cot \alpha}=-\frac{9}{14}$
Vậy $\quad \sin \alpha=-\sqrt{1-\cos ^{2} \alpha}=-\sqrt{1-\frac{1}{16}}=-\frac{\sqrt{15}}{4},$
$\tan \alpha=\frac{\sin \alpha}{\cos \alpha}=\sqrt{15}, \cot \alpha=\frac{1}{\sqrt{15}}$
b) $\frac{\pi}{2}<\alpha<\pi \Rightarrow \cos \alpha<0$
Vậy $\quad \cos \alpha=-\sqrt{1-\sin ^{2} \alpha}=-\sqrt{1-\frac{4}{9}}=\frac{-\sqrt{5}}{3},$
$\tan \alpha=\frac{\sin \alpha}{\cos \alpha}=-\frac{2}{\sqrt{5}}, \cot \alpha=-\frac{\sqrt{5}}{2}$
c) $0<\alpha<\frac{\pi}{2} \Rightarrow \cos \alpha>0, \cos ^{2} \alpha=\frac{1}{1+\tan ^{2} \alpha}$
Vậy $\cos \alpha=\frac{1}{\sqrt{1+\frac{49}{9}}}=\frac{3}{\sqrt{\bar{p} 8}}$,
$\sin \alpha=\cos \alpha \tan \alpha=\frac{7}{\sqrt{58}}, \cot \alpha=\frac{3}{7}$
d) $\frac{3 \pi}{2}<\alpha<2 \pi \Rightarrow \sin \alpha<0, \sin ^{2} \alpha=\frac{1}{1+\cot ^{2} \alpha}$.
Vậy $\sin \alpha=-\frac{1}{\sqrt{1+\frac{196}{81}}}=-\frac{9}{\sqrt{277}}$,
$\cos \alpha=\sin \alpha \cot \alpha=\frac{14}{\sqrt{277}}, \quad \tan \alpha=\frac{1}{\cot \alpha}=-\frac{9}{14}$