Bài 5 trang 24 SGK Toán 11 tập 1 – Chân trời sáng tạo

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 \(\frac{\pi }{2} < \alpha  < \frac{{3\pi }}{4} \Rightarrow \sin \alpha  = \sqrt {\frac{{9 + \sqrt {65} }}{{18}}} \)

\(\begin{array}{l}\tan \alpha  = \frac{{\sin \alpha }}{{cos\alpha }} = \frac{{\sqrt {\frac{{9 + \sqrt {65} }}{{18}}} }}{{ – \sqrt {\frac{{9 – \sqrt {65} }}{{18}}} }} \approx  – 4,266\\\cot \alpha  = \frac{1}{{\tan \alpha }} \approx  – 0,234\end{array}\)