Đề bài

Tính các giá trị lượng giác của góc 2\(\alpha \), biết:

a, ▨\(\sin \alpha  = \frac{{\sqrt 3 }}{3},0 < \alpha  < \frac{\pi }{2}\)

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

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\(\begin{array}{l}\sin \alpha  = 2\sin \frac{\alpha }{2}.cos\frac{\alpha }{2} = 2.\frac{3}{4}.\left( { - \frac{{\sqrt 7 }}{4}} \right) =  - \frac{{3\sqrt 7 }}{8}\\cos\alpha  = 2{\cos ^2}\frac{\alpha }{2} - 1 = 2.{\left( { - \frac{{\sqrt 7 }}{4}} \right)^2} - 1 =  - \frac{1}{8}\\\sin 2\alpha  = 2\sin \alpha .cos\alpha  = 2.\left( { - \frac{{3\sqrt 7 }}{8}} \right).\left( { - \frac{1}{8}} \right) = \frac{{3\sqrt 7 }}{{32}}\\cos2\alpha  = 2{\cos ^2}\alpha  - 1 = 2.{\left( { - \frac{1}{8}} \right)^2} - 1 =  - \frac{{31}}{{32}}\\\tan 2\alpha  = \frac{{\sin 2\alpha }}{{cos2\alpha }} = \frac{{\frac{{3\sqrt 7 }}{{32}}}}{{ - \frac{{31}}{{32}}}} =  - \frac{{3\sqrt 7 }}{{31}}\\\cot 2\alpha  = \frac{1}{{\tan 2\alpha }} = \frac{1}{{ - \frac{{3\sqrt 7 }}{{31}}}} =  - \frac{{31\sqrt 7 }}{{21}}\end{array}\)