Đề bài

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

a, 💛\(cos2\alpha  = \frac{2}{5}, - \frac{\pi }{2} < \alpha  < 0\)

b, 🙈\(\sin 2\alpha  =  - \frac{4}{9},\frac{\pi }{2} < \alpha  < \frac{{3\pi }}{4}\)

Lời giải chi tiết

a, Ta có:

\(\begin{array}{l}cos2\alpha  = 2{\cos ^2}\alpha  - 1 = \frac{2}{5}\\ \Leftrightarrow {\cos ^2}\alpha  = \frac{7}{{10}} \Rightarrow \cos \alpha  =  \pm \frac{{\sqrt {70} }}{{10}}\end{array}\)Vì \( - \frac{\pi }{2} < \alpha  < 0 \Rightarrow \cos \alpha  = \frac{{\sqrt {70} }}{{10}}\)Lại có:\(\begin{array}{l}{\sin ^2}\alpha  + {\cos ^2}\alpha  = 1\\ \Rightarrow {\sin ^2}\alpha  = 1 - \frac{7}{{10}} = \frac{3}{{10}}\\ \Rightarrow \sin \alpha  =  \pm \frac{{\sqrt {30} }}{{10}}\end{array}\)\( - \frac{\pi }{2} < \alpha  < 0 \Rightarrow \sin \alpha  =  - \frac{{\sqrt {30} }}{{10}}\)\(\begin{array}{l}\tan \alpha  = \frac{{\sin \alpha }}{{cos\alpha }} = \frac{{ - \frac{{\sqrt {30} }}{{10}}}}{{\frac{{\sqrt {70} }}{{10}}}} =  - \frac{{\sqrt {21} }}{7}\\\cot \alpha  = \frac{1}{{\tan \alpha }} = \frac{1}{{ - \frac{{\sqrt {21} }}{3}}} =  - \frac{{\sqrt {21} }}{{3 }}\end{array}\)

b, Ta có:

\(\begin{array}{l}{\sin ^2}2\alpha  + {\cos ^2}2\alpha  = 1\\ \Rightarrow \cos 2\alpha  = \sqrt {1 - {{\left( { - \frac{4}{9}} \right)}^2}}  =  \pm \frac{{\sqrt {65} }}{9}\end{array}\) Vì \(\frac{\pi }{2} < \alpha  < \frac{{3\pi }}{4} \Rightarrow \pi  < 2\alpha  < \frac{{3\pi }}{2} \Rightarrow cos2\alpha  =  - \frac{{\sqrt {65} }}{9}\)\(\begin{array}{l}cos2\alpha  = 2{\cos ^2}\alpha  - 1 =  - \frac{{\sqrt {65} }}{9}\\ \Leftrightarrow {\cos ^2}\alpha  = \frac{{9 - \sqrt {65} }}{{18}} \Rightarrow \cos \alpha  =  \pm \sqrt {\frac{{9 - \sqrt {65} }}{{18}}} \end{array}\)Vì \(\frac{\pi }{2} < \alpha  < \frac{{3\pi }}{4} \Rightarrow \cos \alpha  =  - \sqrt {\frac{{9 - \sqrt {65} }}{{18}}} \)Lại có:\(\begin{array}{l}{\sin ^2}\alpha  + {\cos ^2}\alpha  = 1\\ \Rightarrow {\sin ^2}\alpha  = 1 - \frac{{9 - \sqrt {65} }}{{18}} = \frac{{9 + \sqrt {65} }}{{18}}\\ \Rightarrow \sin \alpha  =  \pm \sqrt {\frac{{9 + \sqrt {65} }}{{18}}} \end{array}\)Vì Vì \(\frac{\pi }{2} < \alpha  < \frac{{3\pi }}{4} \Rightarrow \sin \alpha  = \sqrt {\frac{{9 + \sqrt {65} }}{{18}}} \)\(\begin{array}{l}\tan \alpha  = \frac{{\sin \alpha }}{{cos\alpha }} = \frac{{\sqrt {\frac{{9 + \sqrt {65} }}{{18}}} }}{{ - \sqrt {\frac{{9 - \sqrt {65} }}{{18}}} }} \approx  - 4,266\\\cot \alpha  = \frac{1}{{\tan \alpha }} \approx  - 0,234\end{array}\)