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Câu hỏi: Tính các giá trị lượng giác của góc 2\(\alpha \), biết:
a, \(\sin \alpha = \frac{{\sqrt 3 }}{3},0 < \alpha < \frac{\pi }{2}\)
b, \(\sin \frac{\alpha }{2} = \frac{3}{4},\pi < \alpha < 2\pi \)
a, \(\sin \alpha = \frac{{\sqrt 3 }}{3},0 < \alpha < \frac{\pi }{2}\)
b, \(\sin \frac{\alpha }{2} = \frac{3}{4},\pi < \alpha < 2\pi \)
Phương pháp giải
Áp dụng công thức:
\({\sin ^2}\alpha + {\cos ^2}\alpha = 1\)
\(\begin{array}{l}\sin 2a = 2\sin a\cos a\\\cos 2a = {\cos ^2}a - {\sin ^2}a = 2{\cos ^2}a - 1 = 1 - 2{\sin ^2}a\\\tan 2a = \frac{{2\tan a}}{{1 - {{\tan }^2}a}}\end{array}\)
Lời giải chi tiết
a, Ta có:
\(\begin{array}{l}{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\ \Rightarrow \cos \alpha = \pm \sqrt {1 - {{\sin }^2}\alpha } = \pm \sqrt {1 - {{\left( {\frac{{\sqrt 3 }}{3}} \right)}^2}} = \pm \frac{{\sqrt 6 }}{3}\end{array}\)
Vì \(0 < \alpha < \frac{\pi }{2} \Rightarrow \cos \alpha = \frac{{\sqrt 6 }}{3}\)
Khi đó:
\(\begin{array}{l}\sin 2\alpha = 2\sin \alpha .cos\alpha = 2.\frac{{\sqrt 3 }}{3}.\frac{{\sqrt 6 }}{3} = \frac{{2\sqrt 2 }}{3}\\cos2\alpha = 2{\cos ^2}\alpha - 1 = 2.{\left( {\frac{{\sqrt 6 }}{3}} \right)^2} - 1 = \frac{1}{3}\\\tan 2\alpha = \frac{{\sin 2\alpha }}{{cos2\alpha }} = \frac{{\frac{{2\sqrt 2 }}{3}}}{{\frac{1}{3}}} = 2\sqrt 2 \\\cot 2\alpha = \frac{1}{{\tan 2\alpha }} = \frac{1}{{2\sqrt 2 }} = \frac{{\sqrt 2 }}{4}\end{array}\)
b,
Ta có:
\(\begin{array}{l}{\sin ^2}\frac{\alpha }{2} + {\cos ^2}\frac{\alpha }{2} = 1\\ \Rightarrow \cos \frac{\alpha }{2} = \pm \sqrt {1 - {{\sin }^2}\frac{\alpha }{2}} = \pm \sqrt {1 - {{\left( {\frac{3}{4}} \right)}^2}} = \pm \frac{{\sqrt 7 }}{4}\end{array}\)
Vì \(\pi < \alpha < 2\pi \Rightarrow \frac{\pi }{2} < \frac{\alpha }{2} < \pi \Rightarrow cos\alpha = - \frac{{\sqrt 7 }}{4}\)
Khi đó
\(\begin{array}{l}\sin \alpha = 2\sin \frac{\alpha }{2}.cos\frac{\alpha }{2} = 2.\frac{3}{4}.\left( { - \frac{{\sqrt 7 }}{4}} \right) = - \frac{{3\sqrt 7 }}{8}\\cos\alpha = 2{\cos ^2}\frac{\alpha }{2} - 1 = 2.{\left( { - \frac{{\sqrt 7 }}{4}} \right)^2} - 1 = - \frac{1}{8}\\\sin 2\alpha = 2\sin \alpha .cos\alpha = 2.\left( { - \frac{{3\sqrt 7 }}{8}} \right).\left( { - \frac{1}{8}} \right) = \frac{{3\sqrt 7 }}{{32}}\\cos2\alpha = 2{\cos ^2}\alpha - 1 = 2.{\left( { - \frac{1}{8}} \right)^2} - 1 = - \frac{{31}}{{32}}\\\tan 2\alpha = \frac{{\sin 2\alpha }}{{cos2\alpha }} = \frac{{\frac{{3\sqrt 7 }}{{32}}}}{{ - \frac{{31}}{{32}}}} = - \frac{{3\sqrt 7 }}{{31}}\\\cot 2\alpha = \frac{1}{{\tan 2\alpha }} = \frac{1}{{ - \frac{{3\sqrt 7 }}{{31}}}} = - \frac{{31\sqrt 7 }}{{21}}\end{array}\)
Áp dụng công thức:
\({\sin ^2}\alpha + {\cos ^2}\alpha = 1\)
\(\begin{array}{l}\sin 2a = 2\sin a\cos a\\\cos 2a = {\cos ^2}a - {\sin ^2}a = 2{\cos ^2}a - 1 = 1 - 2{\sin ^2}a\\\tan 2a = \frac{{2\tan a}}{{1 - {{\tan }^2}a}}\end{array}\)
Lời giải chi tiết
a, Ta có:
\(\begin{array}{l}{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\ \Rightarrow \cos \alpha = \pm \sqrt {1 - {{\sin }^2}\alpha } = \pm \sqrt {1 - {{\left( {\frac{{\sqrt 3 }}{3}} \right)}^2}} = \pm \frac{{\sqrt 6 }}{3}\end{array}\)
Vì \(0 < \alpha < \frac{\pi }{2} \Rightarrow \cos \alpha = \frac{{\sqrt 6 }}{3}\)
Khi đó:
\(\begin{array}{l}\sin 2\alpha = 2\sin \alpha .cos\alpha = 2.\frac{{\sqrt 3 }}{3}.\frac{{\sqrt 6 }}{3} = \frac{{2\sqrt 2 }}{3}\\cos2\alpha = 2{\cos ^2}\alpha - 1 = 2.{\left( {\frac{{\sqrt 6 }}{3}} \right)^2} - 1 = \frac{1}{3}\\\tan 2\alpha = \frac{{\sin 2\alpha }}{{cos2\alpha }} = \frac{{\frac{{2\sqrt 2 }}{3}}}{{\frac{1}{3}}} = 2\sqrt 2 \\\cot 2\alpha = \frac{1}{{\tan 2\alpha }} = \frac{1}{{2\sqrt 2 }} = \frac{{\sqrt 2 }}{4}\end{array}\)
b,
Ta có:
\(\begin{array}{l}{\sin ^2}\frac{\alpha }{2} + {\cos ^2}\frac{\alpha }{2} = 1\\ \Rightarrow \cos \frac{\alpha }{2} = \pm \sqrt {1 - {{\sin }^2}\frac{\alpha }{2}} = \pm \sqrt {1 - {{\left( {\frac{3}{4}} \right)}^2}} = \pm \frac{{\sqrt 7 }}{4}\end{array}\)
Vì \(\pi < \alpha < 2\pi \Rightarrow \frac{\pi }{2} < \frac{\alpha }{2} < \pi \Rightarrow cos\alpha = - \frac{{\sqrt 7 }}{4}\)
Khi đó
\(\begin{array}{l}\sin \alpha = 2\sin \frac{\alpha }{2}.cos\frac{\alpha }{2} = 2.\frac{3}{4}.\left( { - \frac{{\sqrt 7 }}{4}} \right) = - \frac{{3\sqrt 7 }}{8}\\cos\alpha = 2{\cos ^2}\frac{\alpha }{2} - 1 = 2.{\left( { - \frac{{\sqrt 7 }}{4}} \right)^2} - 1 = - \frac{1}{8}\\\sin 2\alpha = 2\sin \alpha .cos\alpha = 2.\left( { - \frac{{3\sqrt 7 }}{8}} \right).\left( { - \frac{1}{8}} \right) = \frac{{3\sqrt 7 }}{{32}}\\cos2\alpha = 2{\cos ^2}\alpha - 1 = 2.{\left( { - \frac{1}{8}} \right)^2} - 1 = - \frac{{31}}{{32}}\\\tan 2\alpha = \frac{{\sin 2\alpha }}{{cos2\alpha }} = \frac{{\frac{{3\sqrt 7 }}{{32}}}}{{ - \frac{{31}}{{32}}}} = - \frac{{3\sqrt 7 }}{{31}}\\\cot 2\alpha = \frac{1}{{\tan 2\alpha }} = \frac{1}{{ - \frac{{3\sqrt 7 }}{{31}}}} = - \frac{{31\sqrt 7 }}{{21}}\end{array}\)