Bài 11: Theo giả thiết, ta có: \(sin\alpha=\dfrac{\sqrt{3}}{2}\Rightarrow\alpha=60^o\)
Do đó: \(cos\alpha=cos60^o=\dfrac{1}{2},\) \(tan\alpha=tan60^o=\sqrt{3},\) \(cot\alpha=cot60^o=\dfrac{\sqrt{3}}{3}.\)
Bài 12: Ta có: \(sin\alpha=cos\alpha.tan\alpha\Rightarrow sin\alpha=2cos\alpha\)
Khi đó: \(sin^2\alpha+cos^2\alpha=1\Rightarrow5cos^2\alpha=1\Rightarrow cos^2\alpha=\dfrac{1}{5}\Rightarrow cos\alpha=\dfrac{\sqrt{5}}{5}\Rightarrow sin\alpha=\dfrac{2\sqrt{5}}{5}\)
Theo giả thiết, ta có: \(tan\alpha=2\Rightarrow cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{1}{2}\)
Bài 11:
\(cosa=\sqrt{1-\left(\dfrac{\sqrt{3}}{2}\right)^2}=\dfrac{1}{2}\)
\(tana=\sqrt{3}\)
cot a=1/ căn 3
