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Câu hỏi: Tính các giá trị lượng giác của góc \(\alpha \), biết:
a, \(cos2\alpha = \frac{2}{5}, - \frac{\pi }{2} < \alpha < 0\)
b, \(\sin 2\alpha = - \frac{4}{9},\frac{\pi }{2} < \alpha < \frac{{3\pi }}{4}\)
a, \(cos2\alpha = \frac{2}{5}, - \frac{\pi }{2} < \alpha < 0\)
b, \(\sin 2\alpha = - \frac{4}{9},\frac{\pi }{2} < \alpha < \frac{{3\pi }}{4}\)
Phương pháp giải
Sử dụng các công thức lượng giác để tính toán
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:
\(\begin{array}{l}{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\ \Rightarrow {\sin ^2}\alpha = 1 - \frac{7}{{10}} = \frac{3}{{10}}\\ \Rightarrow \sin \alpha = \pm \frac{{\sqrt {30} }}{{100}}\end{array}\)
\( - \frac{\pi }{2} < \alpha < 0 \Rightarrow \sin \alpha = - \frac{{\sqrt {30} }}{{100}}\)
\(\begin{array}{l}\tan \alpha = \frac{{\sin \alpha }}{{cos\alpha }} = \frac{{ - \frac{{\sqrt {30} }}{{100}}}}{{\frac{{\sqrt {70} }}{{10}}}} = - \frac{{\sqrt 3 }}{{\sqrt 7 }}\\\cot \alpha = \frac{1}{{\tan \alpha }} = \frac{1}{{ - \frac{{\sqrt 3 }}{{\sqrt 7 }}}} = - \frac{{\sqrt 7 }}{{\sqrt 3 }}\end{array}\)
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 \(\frac{\pi }{2} < \alpha < \frac{{3\pi }}{4} \Rightarrow \sin \alpha = \sqrt {\frac{{9 + \sqrt {65} }}{{18}}} \)
\(\begin{array}{l}\tan \alpha = \frac{{\sin \alpha }}{{cos\alpha }} = \frac{{\sqrt {\frac{{9 + \sqrt {65} }}{{18}}} }}{{ - \sqrt {\frac{{9 - \sqrt {65} }}{{18}}} }} \approx - 4,266\\\cot \alpha = \frac{1}{{\tan \alpha }} \approx - 0,234\end{array}\)
Sử dụng các công thức lượng giác để tính toán
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:
\(\begin{array}{l}{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\ \Rightarrow {\sin ^2}\alpha = 1 - \frac{7}{{10}} = \frac{3}{{10}}\\ \Rightarrow \sin \alpha = \pm \frac{{\sqrt {30} }}{{100}}\end{array}\)
\( - \frac{\pi }{2} < \alpha < 0 \Rightarrow \sin \alpha = - \frac{{\sqrt {30} }}{{100}}\)
\(\begin{array}{l}\tan \alpha = \frac{{\sin \alpha }}{{cos\alpha }} = \frac{{ - \frac{{\sqrt {30} }}{{100}}}}{{\frac{{\sqrt {70} }}{{10}}}} = - \frac{{\sqrt 3 }}{{\sqrt 7 }}\\\cot \alpha = \frac{1}{{\tan \alpha }} = \frac{1}{{ - \frac{{\sqrt 3 }}{{\sqrt 7 }}}} = - \frac{{\sqrt 7 }}{{\sqrt 3 }}\end{array}\)
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 \(\frac{\pi }{2} < \alpha < \frac{{3\pi }}{4} \Rightarrow \sin \alpha = \sqrt {\frac{{9 + \sqrt {65} }}{{18}}} \)
\(\begin{array}{l}\tan \alpha = \frac{{\sin \alpha }}{{cos\alpha }} = \frac{{\sqrt {\frac{{9 + \sqrt {65} }}{{18}}} }}{{ - \sqrt {\frac{{9 - \sqrt {65} }}{{18}}} }} \approx - 4,266\\\cot \alpha = \frac{1}{{\tan \alpha }} \approx - 0,234\end{array}\)