Nếu \(\sin \alpha = \frac{1}{{\sqrt 3 }}\) với \(0 < \alpha < \frac{\pi }{2}\) thì giá trị của \(\cos \left( {\alpha + \frac{\pi }{3}} \right)\) bằng:

A. \(\frac{{\sqrt 6 }}{6} - \frac{1}{2}\).

B. \(\sqrt 6 - 3\).

C. \(\frac{{\sqrt 6 }}{6} - 3\).

D. \(\sqrt 6 - \frac{1}{2}\).

Đáp án đúng là: B

Ta có sin⁴ x + cos⁴ x = 1 – 2sin² x cos² x (theo Bài 9a)

= 1 – 2 (sin x cos x)² = \(1 - 2{\left( {\frac{{\sin 2x}}{2}} \right)^2} = 1 - 2.\frac{{{{\sin }^2}2x}}{4} = 1 - \frac{{2\left( {1 - {{\cos }^2}2x} \right)}}{4}\)

\( = 1 - \frac{{2 - 2{{\cos }^2}2x}}{4} = \frac{{4 - 2 + 2{{\cos }^2}2x}}{4}\)\( = \frac{{3 + \left( {2{{\cos }^2}2x - 1} \right)}}{4} = \frac{{3 + \cos 4x}}{4}\).

Vậy \({\sin ^4}x + {\cos ^4}x = \frac{{3 + \cos 4x}}{4}\).

Ta có \(\frac{{A + B + C}}{2} = \frac{\pi }{2}\), suy ra \(\frac{A}{2} + \frac{B}{2} = \frac{\pi }{2} - \frac{C}{2}\) nên \(\tan \left( {\frac{A}{2} + \frac{B}{2}} \right) = \cot \frac{C}{2}\)

\( \Leftrightarrow \frac{{\tan \frac{A}{2} + \tan \frac{B}{2}}}{{1 - \tan \frac{A}{2}.\tan \frac{B}{2}}} = \frac{1}{{\tan \frac{C}{2}}}\)

\( \Leftrightarrow \left( {\tan \frac{A}{2} + \tan \frac{B}{2}} \right)\tan \frac{C}{2} = 1 - \tan \frac{A}{2}.\tan \frac{B}{2}\)

\( \Leftrightarrow \tan \frac{A}{2}.\tan \frac{C}{2} + \tan \frac{B}{2}.\tan \frac{C}{2} + \tan \frac{A}{2}.\tan \frac{B}{2} = 1\)

\( \Leftrightarrow \tan \frac{A}{2}.\tan \frac{B}{2} + \tan \frac{B}{2}.\tan \frac{C}{2} + \tan \frac{C}{2}.\tan \frac{A}{2} = 1\).