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Cho \(\sin \alpha = m\)
a) Hãy tính\(\cos 2\alpha ;{\sin ^2}2\alpha ;{\tan ^2}2\alpha \) theo \(m\) (giả sử \(\tan 2\alpha \) xác định)
b) Hỏi \(\sin 2\alpha ;\tan 2\alpha \) có xác định duy nhất bởi \(m\) hay không?
a)
\(\cos 2\alpha = 1 – 2{\sin ^2}\alpha = 1 – 2{m^2};\)
\(\begin{array}{l}{\sin ^2}2\alpha = 4{\sin ^2}\alpha {\cos ^2}\alpha \\ = 4{\sin ^2}\alpha \left( {1 – {{\sin }^2}\alpha } \right)\\ = 4{m^2}\left( {1 – {m^2}} \right);\end{array}\)
\({\tan ^2}2\alpha = \dfrac{{{{\sin }^2}2\alpha }}{{{{\cos }^2}2\alpha }} = \dfrac{{4{m^2}\left( {1 – {m^2}} \right)}}{{{{\left( {1 – 2{m^2}} \right)}^2}}}.\)
b) Không, chẳng hạn \(\sin \dfrac{\pi }{3} = \sin \left( {\dfrac{{2\pi }}{3}} \right) = \dfrac{{\sqrt 3 }}{2},\)
nhưng \(\sin \dfrac{{2\pi }}{3} = \dfrac{{\sqrt 3 }}{2},\sin \left( {2.\dfrac{{2\pi }}{3}} \right) = – \dfrac{{\sqrt 3 }}{2},\)
\(\tan \dfrac{{2\pi }}{3} = – \sqrt 3 ,\tan \left( {2.\dfrac{{2\pi }}{3}} \right) = \sqrt 3 .\)