Question

[Sách Giáo Khoa]

Tính các giá trị lượng giác của góc \(\alpha\), nếu :

a) \(\cos\alpha=-\dfrac{1}{4},\pi< \alpha< \dfrac{3\pi}{2}\)

b) \(\sin\alpha=\dfrac{2}{3},\dfrac{\pi}{2}< \alpha< \pi\)

c) \(\tan\alpha=\dfrac{7}{3},0< \alpha< \dfrac{\pi}{2}\)

d) \(\cot\alpha=-\dfrac{14}{9},\dfrac{3\pi}{2}< \alpha< 2\pi\)

Answer 1

[Bùi Thị Vân]

a) Do \(\pi< \alpha< \dfrac{3\pi}{2}\) nên \(sin\alpha< 0;cot\alpha>0;tan\alpha>0\).
Vì vậy: \(sin\alpha=-\sqrt{1-cos^2\alpha}=\dfrac{-\sqrt{15}}{4}\).
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{-\sqrt{15}}{4}:\dfrac{-1}{4}=\sqrt{15}\).
\(cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{1}{\sqrt{15}}\).

Answer 2

[Bùi Thị Vân]

b) Do \(\dfrac{\pi}{2}< \alpha< \pi\) nên \(cos\alpha< 0;tan\alpha< 0;cot\alpha< 0\).
\(cos\alpha=-\sqrt{1-sin^2\alpha}=-\dfrac{\sqrt{5}}{3}\);
\(tan\alpha=\dfrac{2}{3}:\dfrac{-\sqrt{5}}{3}=\dfrac{-2}{\sqrt{5}}\); \(cot\alpha=1:tan\alpha=\dfrac{-\sqrt{5}}{2}\).