Cho \(\cos \alpha = \frac{1}{3};\alpha \in \left( {\pi ;2\pi } \right)\) tính giá trị\(\sin \alpha = \)? Và \(A = \frac{{1 - 4{{\sin }^2}x + {{\cos }^2}x}}{{3 + {{\sin }^2}x - 4\cos 2x}}.\)
A.\(\sin \alpha = - \frac{2\sqrt{2}}{3} \,\,;\,\, A = - \frac{4}{11}\)
B.\(\sin \alpha = \frac{2\sqrt{2}}{3} \,\,;\,\, A = \frac{4}{11}\)
C.\(\sin \alpha = - \frac{\sqrt{2}}{3} \,\,;\,\, A = - \frac{2}{11}\)
D.\(\sin \alpha = \frac{\sqrt{2}}{3} \,\,;\,\, A = \frac{2}{11}\)
A.\(\sin \alpha = - \frac{2\sqrt{2}}{3} \,\,;\,\, A = - \frac{4}{11}\)
B.\(\sin \alpha = \frac{2\sqrt{2}}{3} \,\,;\,\, A = \frac{4}{11}\)
C.\(\sin \alpha = - \frac{\sqrt{2}}{3} \,\,;\,\, A = - \frac{2}{11}\)
D.\(\sin \alpha = \frac{\sqrt{2}}{3} \,\,;\,\, A = \frac{2}{11}\)
Loga
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