Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cos 2x = 1 - 2{\sin ^2}x \Rightarrow {\sin ^2}x = \dfrac{{1 - \cos 2x}}{2}\\
\cos x.\cos y = \dfrac{1}{2}\left[ {\cos \left( {x + y} \right) + \cos \left( {x - y} \right)} \right]\\
a,\\
{\sin ^4}\alpha + {\cos ^4}\alpha = {\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)^2} - 2{\sin ^2}\alpha .{\cos ^2}\alpha \\
= {1^2} - \dfrac{1}{2}.{\left( {2\sin \alpha .\cos \alpha } \right)^2} = 1 - \dfrac{1}{2}{\sin ^2}2\alpha \\
= 1 - \dfrac{1}{2}.\dfrac{{1 - \cos 4\alpha }}{2} = \dfrac{3}{4} + \dfrac{1}{4}\cos 4\alpha \\
b,\\
{\sin ^6}\alpha + {\cos ^6}\alpha \\
= {\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)^3} - 3{\sin ^2}\alpha .{\cos ^2}\alpha .\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)\\
= {1^3} - 3.{\sin ^2}\alpha .{\cos ^2}\alpha .1\\
= 1 - \dfrac{3}{4}.{\left( {2\sin \alpha .\cos \alpha } \right)^2}\\
= 1 - \dfrac{3}{4}.{\sin ^2}2\alpha \\
= 1 - \dfrac{3}{4}.\dfrac{{1 - \cos 4\alpha }}{2}\\
= \dfrac{5}{8} + \dfrac{3}{8}.\cos 4\alpha \\
c,\\
\cos \alpha .\cos \left( {\dfrac{\pi }{3} - \alpha } \right).\cos \left( {\dfrac{\pi }{3} + \alpha } \right)\\
= \cos \alpha .\dfrac{1}{2}.\left[ {\cos \left( {\dfrac{\pi }{3} - \alpha + \dfrac{\pi }{3} + \alpha } \right) + \cos \left( {\dfrac{\pi }{3} - \alpha - \dfrac{\pi }{3} - \alpha } \right)} \right]\\
= \dfrac{1}{2}\cos \alpha .\left[ {\cos \dfrac{{2\pi }}{3} + \cos \left( { - 2\alpha } \right)} \right]\\
= \dfrac{1}{2}\cos \alpha .\left[ { - \dfrac{1}{2} + \cos 2\alpha } \right]\\
= - \dfrac{1}{4}\cos \alpha + \dfrac{1}{2}\cos \alpha .\cos 2\alpha \\
= - \dfrac{1}{4}\cos \alpha + \dfrac{1}{2}.\dfrac{1}{2}.\left[ {\cos \left( {\alpha + 2\alpha } \right) + \cos \left( {\alpha - 2\alpha } \right)} \right]\\
= - \dfrac{1}{4}\cos \alpha + \dfrac{1}{4}\left( {\cos 3\alpha + \cos \left( { - \alpha } \right)} \right)\\
= - \dfrac{1}{4}\cos \alpha + \dfrac{1}{4}.\left( {\cos 3\alpha + \cos \alpha } \right)\\
= \dfrac{1}{4}\cos 3\alpha
\end{array}\)
