a)\(\sin^2\alpha+\cos^2\alpha=1\Rightarrow\sin^2\alpha=1-\cos^2\alpha\)
\(\Rightarrow1-2^2=-3\) \(\Rightarrow\cos=-\sqrt{3}\left(0< \alpha< \dfrac{\pi}{2}\right)\)
b) \(\tan\alpha\times\cot\alpha=1\Rightarrow\tan\alpha=\dfrac{1}{\cot\alpha}\Rightarrow\tan=\dfrac{1}{4}\)
a)Do \(0< \alpha< \dfrac{\pi}{2}\) nên các giá trị lượng giác của \(\alpha\) đều dương.
\(cos\alpha=2sin\alpha\)(1)
Nếu \(sin\alpha=0\Rightarrow cos\alpha\) (vô lý).
Vì vậy \(sin\alpha\ne0\) . Từ (1) \(\Rightarrow\dfrac{cos\alpha}{sin\alpha}=2\)\(\Leftrightarrow cot\alpha=2\).
Suy ra: \(tan\alpha=\dfrac{1}{2}\).
\(sin\alpha=\sqrt{\dfrac{1}{1+cot^2\alpha}}=\dfrac{1}{\sqrt{3}}\).
\(cos\alpha=\sqrt{1-sin^2\alpha}=\sqrt{\dfrac{2}{3}}\).
b) Do \(\dfrac{\pi}{2}< \alpha< \pi\) nên \(cos\alpha< 0;sin\alpha>0;tan\alpha< 0;cot\alpha< 0\).
\(cot\alpha=4tan\alpha\Leftrightarrow cot^2\alpha=4\) \(\Leftrightarrow cot\alpha=-2\) (do \(cot\alpha< 0\) ).
Vì vậy \(tan\alpha=\dfrac{-1}{2}\).
\(sin\alpha=\sqrt{\dfrac{1}{1+cot^2\alpha}}=\dfrac{1}{\sqrt{5}}\).
\(cos\alpha=-\sqrt{1-\left(\dfrac{1}{\sqrt{5}}\right)^2}=-\dfrac{2}{\sqrt{5}}\).