b) Do \(0< \alpha< \dfrac{\pi}{2}\) nên các giá trị lượng giác của \(\alpha\) đều dương.
Vì vậy:
\(cos\alpha=\sqrt{1-0,6^2}=\dfrac{4}{5}\).
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=0,6:\dfrac{4}{5}=0,75;cot\alpha=1:tan\alpha=\dfrac{4}{3}\).
Do \(\dfrac{\pi}{2}< \alpha< \pi\) nên \(sin\alpha>0;tan\alpha< 0;cot\alpha< 0\).
\(sin\alpha=\sqrt{1-cos^2\alpha}=\dfrac{\sqrt{51}}{10}\).
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\sqrt{51}}{10}:\left(-0,7\right)=-\dfrac{\sqrt{51}}{7}\).
\(cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{-7}{\sqrt{51}}\).
Do \(\pi< \alpha< \dfrac{3\pi}{2}\) nên \(sin\alpha< 0;cos\alpha< 0;cot\alpha>2\).
\(cos\alpha=-\sqrt{\dfrac{1}{tan^2+1}}=-\dfrac{1}{\sqrt{5}}\);
\(cot\alpha=1:tan\alpha=\dfrac{1}{2}\);
\(sin\alpha=-\sqrt{1-cos^2\alpha}=-\dfrac{2}{\sqrt{5}}\).
d) Do \(\dfrac{3\pi}{2}< \alpha< 2\pi\) nên \(sin\alpha< 0;cos\alpha>0;cot\alpha< 0;tan\alpha< 0\).
\(sin\alpha=-\sqrt{\dfrac{1}{1+cot^2\alpha}}=-\dfrac{1}{\sqrt{10}}\);
\(tan\alpha=1:cot\alpha=\dfrac{-1}{3}\).
\(cos\alpha=\sqrt{1-sin^2\alpha}=\dfrac{3\sqrt{10}}{10}\).