Cho \(\cos \alpha = \frac{3}{4},\,\sin \alpha > 0;\,\,\sin \beta = \frac{3}{5};\,\beta \in \left( {\frac{{9\pi }}{2};5\pi } \right)\).
Hãy tính \(\cos 2\alpha ,\,\,\sin 2\alpha ,\,\,\cos 2\beta ,\,\,\sin 2\beta ,\,\,\cos (\alpha + \beta ),\,\,\sin (\alpha - \beta )\).
Áp dụng công thức góc nhân đôi, công thức cơ bản, công thức cộng:
\(\cos 2\alpha = 2{\cos ^2}\alpha - 1\);
\(\sin 2\alpha = 2\sin \alpha \cos \alpha \);
\({\sin ^2}\alpha + {\cos ^2}\alpha = 1\);
\(\cos (\alpha + \beta ) = \cos \alpha \cos \beta - \sin \alpha .\sin \beta \);
\(\sin (\alpha - \beta ) = \sin \alpha \cos \beta - \cos \alpha .\sin \beta \).
Ta có \(\cos 2\alpha = 2{\cos ^2}\alpha - 1 = 2.\frac{9}{{16}} - 1 = \frac{1}{8}.\)
Ta có \({\sin ^2}\alpha = 1 - {\cos ^2}\alpha = 1 - {\left( {\frac{3}{4}} \right)^2} = \frac{7}{{16}}\). Lại do \(\sin \alpha > 0\) nên \(\sin \alpha = \frac{{\sqrt 7 }}{4}\).
Suy ra \(\sin 2\alpha = 2\sin \alpha \cos \alpha = 2.\frac{{\sqrt 7 }}{4}.\frac{3}{4} = \frac{{3\sqrt 7 }}{8}\).
Ta có \(\cos 2\beta = 1 - 2{\sin ^2}\beta = 1 - 2.\frac{9}{{25}} = \frac{7}{{25}}\).
Ta có \({\cos ^2}\beta = 1 - {\sin ^2}\beta = 1 - {\left( {\frac{3}{5}} \right)^2} = \frac{{16}}{{25}}\).
Lại do \(\beta \in \left( {\frac{{9\pi }}{2};5\pi } \right)\) nên \(\cos \beta
\(\sin 2\beta = 2\sin \beta \cos \beta = 2.\frac{3}{5}.\left( { - \frac{4}{5}} \right) = - \frac{{24}}{{25}}\)
Ta có
\(\cos (\alpha + \beta ) = \cos \alpha \cos \beta - \sin \alpha .\sin \beta = \frac{3}{4}.\left( { - \frac{4}{5}} \right) - \frac{{\sqrt 7 }}{4}.\frac{3}{5} = \frac{{ - 12 - 3\sqrt 7 }}{{20}}.\)
\(\sin (\alpha - \beta ) = \sin \alpha \cos \beta - \cos \alpha .\sin \beta = \frac{{\sqrt 7 }}{4}.\left( { - \frac{4}{5}} \right) - \frac{3}{4}.\frac{3}{5} = \frac{{ - 9 - 4\sqrt 7 }}{{20}}.\)