Trong bốn khẳng định sau, có bao nhiêu khẳng định đúng?

(1): \(\cos 4a = {\cos ^2}2a - {\sin ^2}2a.\)

(2): \(\cos 4a = 1 - 2{\sin ^2}2a.\)

(3): \(\cos 4a = 2{\cos ^2}2a - 1.\)

(4): \(\cos 4a = 1 - 2{\sin ^2}4a.\)

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.

\(A = \cos \left( {\frac{\pi }{3} - x} \right)\cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{6} + x} \right)\cos \left( {\frac{{3\pi }}{4} + x} \right)\)

\( = \frac{1}{2}\left[ {\cos \left( {\frac{\pi }{{12}} - 2x} \right) + \cos \frac{{7\pi }}{{12}}} \right] + \frac{1}{2}\left[ {\cos \left( { - \frac{{7\pi }}{{12}}} \right) + \cos \left( {\frac{{11\pi }}{{12}} + 2x} \right)} \right]\)

\( = \frac{1}{2}\left[ {\cos \left( {\frac{\pi }{{12}} - 2x} \right) + \cos \left( {\frac{{11\pi }}{{12}} + 2x} \right) + \cos \frac{{7\pi }}{{12}} + \cos \left( { - \frac{{7\pi }}{{12}}} \right)} \right]\)

\( = \frac{1}{2}\left( {0 + 2\cos \frac{{7\pi }}{{12}}} \right)\) (do \(\frac{{11\pi }}{{12}} + 2x + \frac{\pi }{{12}} - 2x = \pi \))

\( = \cos \frac{{7\pi }}{{12}}\).

Vậy \(\frac{a}{b} = \frac{7}{{12}} \approx 0,58\).

Trả lời: 0,58.