Biết \(\frac{{\cos 4x + \cos 2x + 1}}{{\sin 4x + \sin 2x}} = m\cot 2x\). Khẳng định nào dưới đây là đúng.

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.

\[\tan \left( {\alpha - \frac{\pi }{4}} \right) = \frac{{\tan \alpha - \tan \frac{\pi }{4}}}{{1 + \tan \alpha .\tan \frac{\pi }{4}}}\]\[ = \frac{{\tan \alpha - 1}}{{1 + \tan \alpha }}\]\[ = \frac{{2 - 1}}{{1 + 2}} = \frac{1}{3}\]. Chọn D.