Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
A = \sin x.\cos x.\cos 2x.\cos 4x.\cos 8x\\
 = \dfrac{1}{2}\left( {2\sin x.\cos x} \right).\cos 2x.\cos 4x.\cos 8x\\
 = \dfrac{1}{2}\sin 2x.\cos 2x.\cos 4x.\cos 8x\\
 = \dfrac{1}{2}.\dfrac{1}{2}\sin 4x.\cos 4x.\cos 8x\\
 = \dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.sin8x.\cos 8x\\
 = \dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}\sin 16x\\
 = \dfrac{1}{{16}}\sin 16x\\
b,\\
B = \dfrac{{\sin 5a - \sin 3a}}{{\sin a}} = \dfrac{{2.\cos \dfrac{{5a + 3a}}{2}.\sin \dfrac{{5a - 3a}}{2}}}{{\sin a}} = \dfrac{{2\cos 4a.\sin a}}{{\sin a}} = 2\cos 4a\\
c,\\
C = \dfrac{{\cos 5a + \cos a}}{{\cos 3a}} = \dfrac{{2\cos \dfrac{{5a + a}}{2}.cos\dfrac{{5a - a}}{2}}}{{\cos 3a}} = \dfrac{{2\cos 3a.\cos 2a}}{{\cos 3a}} = 2\cos 2a\\
d,\\
D = \dfrac{{\cos 7a + \sin \left( {\dfrac{\pi }{2} - a} \right)}}{{\cos 3a}} = \dfrac{{\cos 7a + \cos a}}{{\cos 3a}} = \dfrac{{2.\cos \dfrac{{7a + a}}{2}.\cos \dfrac{{7a - a}}{2}}}{{\cos 3a}} = \dfrac{{2.\cos 4a.cos3a}}{{\cos 3a}} = \cos 4a\\
7,\\
a,\\
{\sin ^4}x + {\cos ^4}x\\
 = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x.{\cos ^2}x\\
 = {1^2} - \dfrac{1}{2}.{\left( {2\sin x.\cos x} \right)^2}\\
 = 1 - \dfrac{1}{2}{\sin ^2}2x\\
 = 1 - \dfrac{1}{2}.\dfrac{{1 - \cos 4x}}{2}\\
 = \dfrac{{3 + \cos 4x}}{4}\\
b,\\
{\sin ^6}x + {\cos ^6}x\\
 = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^3} - 3{\sin ^2}x.{\cos ^2}x.\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\\
 = {1^3} - 3{\sin ^2}x.{\cos ^2}x.1\\
 = 1 - \dfrac{3}{4}.{\left( {2\sin x.\cos x} \right)^2}\\
 = 1 - \dfrac{3}{4}{\sin ^2}2x\\
 = 1 - \dfrac{3}{4}.\dfrac{{1 - \cos 4x}}{2}\\
 = \dfrac{{5 + 3\cos 4x}}{8}
\end{array}\)