Đáp án:
$ a) \cos\alpha=\dfrac{\sqrt{5}}{3}; \tan \alpha=\dfrac{2\sqrt{5}}{5}; \cot \alpha=\dfrac{\sqrt{5}}{2}\\ b)\sin\alpha=\dfrac{2\sqrt{6}}{5}; \tan \alpha=2\sqrt{6}; \cot \alpha=\dfrac{\sqrt{6}}{12}\\ c)\cot \alpha=\dfrac{1}{2}; \cos \alpha=\dfrac{\sqrt{5}}{5};\sin\alpha=\dfrac{2\sqrt{5}}{5}$
Giải thích các bước giải:
$0^\circ < \alpha < 90^\circ\\ \Rightarrow \sin \alpha; \cos \alpha; \tan \alpha; \cot \alpha>0\\ a)\sin^2\alpha+\cos^2\alpha=1\\ \Rightarrow \cos\alpha=\sqrt{1-\sin^2\alpha}=\dfrac{\sqrt{5}}{3}\\ \tan \alpha=\dfrac{\sin \alpha}{\cos \alpha}=\dfrac{2\sqrt{5}}{5}\\ \cot \alpha=\dfrac{\cos \alpha}{\sin \alpha}=\dfrac{\sqrt{5}}{2}\\ b)\sin^2\alpha+\cos^2\alpha=1\\ \Rightarrow \sin\alpha=\sqrt{1-\cos^2\alpha}=\dfrac{2\sqrt{6}}{5}\\ \tan \alpha=\dfrac{\sin \alpha}{\cos \alpha}=2\sqrt{6}\\ \cot \alpha=\dfrac{\cos \alpha}{\sin \alpha}=\dfrac{\sqrt{6}}{12}\\ c)\tan \alpha=2\\\cot \alpha=\dfrac{1}{\tan \alpha}=\dfrac{1}{2}\\ \tan^2\alpha+1=\dfrac{\sin^2\alpha}{\cos^2\alpha}+1=\dfrac{\sin^2\alpha+\cos^2\alpha}{\cos^2\alpha}=\dfrac{1}{\cos^2\alpha}\\ \Rightarrow \cos^2\alpha=\dfrac{1}{\tan^2\alpha+1}\\\Rightarrow \cos \alpha=\sqrt{\dfrac{1}{\tan^2\alpha+1}}=\dfrac{\sqrt{5}}{5}\\ \sin^2\alpha+\cos^2\alpha=1\\ \Rightarrow \sin\alpha=\sqrt{1-\cos^2\alpha}=\dfrac{2\sqrt{5}}{5}$
