Giải thích các bước giải:
\(\begin{array}{l}
a,\\
\frac{{{{\sin }^4}x - {{\cos }^4}x + {{\cos }^2}x}}{{2\left( {1 - \cos x} \right)}}\\
= \frac{{\left( {{{\sin }^2}x - {{\cos }^2}x} \right)\left( {{{\sin }^2}x + {{\cos }^2}x} \right) + {{\cos }^2}x}}{{2\left( {1 - \cos x} \right)}}\\
= \frac{{{{\sin }^2}x - {{\cos }^2}x + {{\cos }^2}x}}{{2.2.{{\sin }^2}\frac{x}{2}}}\\
= \frac{{{{\sin }^2}x}}{{4.{{\sin }^2}\frac{x}{2}}} = \frac{{{{\left( {2\sin \frac{x}{2}.\cos \frac{x}{2}} \right)}^2}}}{{4.{{\sin }^2}\frac{x}{2}}} = {\cos ^2}\frac{x}{2}\\
b,\\
\sin x.{\cos ^5}x - {\sin ^5}x.\cos x\\
= \sin x\cos x\left( {{{\cos }^4}x - {{\sin }^4}x} \right)\\
= \frac{1}{2}\sin 2x.\left( {{{\cos }^2}x - {{\sin }^2}x} \right).\left( {{{\cos }^2}x + {{\sin }^2}x} \right)\\
= \frac{1}{2}\sin 2x.\left( {{{\cos }^2}x - {{\sin }^2}x} \right)\\
= \frac{1}{2}\sin 2x.\cos 2x\\
= \frac{1}{4}\sin 4x
\end{array}\)
