Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A = {\sin ^6}x + 2{\sin ^2}x.{\cos ^4}x + 3{\sin ^4}x.{\cos ^2}x + {\cos ^4}x\\
= \left( {{{\sin }^6}x + 3{{\sin }^4}x.{{\cos }^2}x + 3{{\sin }^2}x.{{\cos }^4}x + {{\cos }^6}x} \right) + {\cos ^4}x - {\cos ^6}x - {\sin ^2}x.{\cos ^4}x\\
= {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^3} + {\cos ^4}x.\left( {1 - {{\sin }^2}x} \right) - {\cos ^6}x\\
= {1^3} + {\cos ^4}x.{\cos ^2}x - {\cos ^6}x\\
= 1\\
b,\\
P = \dfrac{{\sin x + \sin 5x}}{{\cos 5x - \cos x}}\\
= \dfrac{{2.\sin \dfrac{{5x + x}}{2}.\cos \dfrac{{5x - x}}{2}}}{{ - 2.\sin \dfrac{{5x + x}}{2}.\sin \dfrac{{5x - x}}{2}}}\\
= \dfrac{{2.\sin 3x.\cos 2x}}{{ - 2.\sin 3x.\sin 2x}}\\
= - \dfrac{{\cos 2x}}{{\sin 2x}} = - \cot 2x\\
c,\\
P = \dfrac{{\sin 5x + \sin x}}{{\cos 5x + \cos x}}\\
= \dfrac{{2\sin \dfrac{{5x + x}}{2}.\cos \dfrac{{5x - x}}{2}}}{{2.\cos \dfrac{{5x + x}}{2}.\cos \dfrac{{5x - x}}{2}}}\\
= \dfrac{{2.\sin 3x.\cos 2x}}{{2.\cos 3x.\cos 2x}}\\
= \dfrac{{\sin 3x}}{{\cos 3x}} = \tan 3x
\end{array}\)
