Cho \(\sin \alpha = \frac{2}{3}\) với \(0 < \alpha < \frac{\pi }{2}\).

a) \(\sin 2\alpha = \frac{{4\sqrt 5 }}{9}\).

b) \(\cos \left( {\alpha + \frac{{3\pi }}{2}} \right) = \frac{2}{3}\).

c) \(\sqrt 2 \cos \left( {\alpha + \frac{\pi }{4}} \right) = \frac{{2 + \sqrt 5 }}{3}\).

d) \(D = \frac{{\cot \alpha + \tan \alpha }}{{\cot \alpha - \tan \alpha }} = \frac{1}{9}\).

\[T = \sin \left( {\frac{\pi }{3} + x} \right) - \sin \left( {\frac{\pi }{3} - x} \right)\]

\[ = 2\cos \frac{{\left( {\frac{\pi }{3} + x} \right) + \left( {\frac{\pi }{3} - x} \right)}}{2}\sin \frac{{\left( {\frac{\pi }{3} + x} \right) - \left( {\frac{\pi }{3} - x} \right)}}{2}\]

\[ = 2\cos \frac{\pi }{3}\sin x = \sin x.\] Chọn B.

a) \({\sin ^2}\alpha = \frac{{1 - \cos 2\alpha }}{2}\).

b) \(\cos 2\alpha = - \frac{1}{9}\)\( \Leftrightarrow 2{\cos ^2}\alpha - 1 = - \frac{1}{9}\)\( \Leftrightarrow {\cos ^2}\alpha = \frac{4}{9}\)\( \Leftrightarrow \cos \alpha = \frac{2}{3}\) vì \(\alpha \in \left( { - \frac{\pi }{2};0} \right)\).

c) Ta có \({\sin ^2}2\alpha = 1 - {\cos ^2}2\alpha = 1 - {\left( { - \frac{1}{9}} \right)^2} = \frac{{80}}{{81}}\)\( \Rightarrow \sin 2\alpha = - \frac{{\sqrt {80} }}{9}\).

Có \(\sin 4\alpha = 2\sin 2\alpha \cos 2\alpha = 2.\frac{{ - \sqrt {80} }}{9}.\frac{{ - 1}}{9} = \frac{{2\sqrt {80} }}{{81}}\).

d) Ta có \({\tan ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }} - 1 = \frac{1}{{{{\left( {\frac{2}{3}} \right)}^2}}} - 1 = \frac{5}{4}\).

Vì \(\alpha \in \left( { - \frac{\pi }{2};0} \right)\) nên \(\tan \alpha < 0 \Rightarrow \tan \alpha = - \frac{{\sqrt 5 }}{2}\).

Ta có \(\tan \left( {\alpha + \frac{\pi }{4}} \right) = \frac{{\tan \alpha + \tan \frac{\pi }{4}}}{{1 - \tan \alpha .\tan \frac{\pi }{4}}} = \frac{{\frac{{ - \sqrt 5 }}{2} + 1}}{{1 - \left( { - \frac{{\sqrt 5 }}{2}} \right).1}} = - 9 + 4\sqrt 5 \).

Suy ra a = −9 ; b = 4; c = 5. Do đó a + b + c = 0.

Đáp án: a) Sai;   b) Đúng;   c) Sai;   d) Đúng.