Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
80,\\
A = \dfrac{{3 - 4\cos 2a + \cos 4a}}{{3 + 4\cos 2a + \cos 4a}}\\
= \dfrac{{3 - 4\cos 4a + \left( {2{{\cos }^2}2a - 1} \right)}}{{3 + 4\cos 2a + \left( {2{{\cos }^2}2a - 1} \right)}}\\
= \dfrac{{2.\left( {{{\cos }^2}2a - 2\cos 2a + 1} \right)}}{{2.\left( {{{\cos }^2}2a + 2\cos 2a + 1} \right)}}\\
= \dfrac{{2.{{\left( {\cos 2a - 1} \right)}^2}}}{{2.{{\left( {\cos 2a + 1} \right)}^2}}} = \dfrac{{{{\left( {\left( {1 - 2{{\sin }^2}a} \right) - 1} \right)}^2}}}{{{{\left( {\left( {2{{\cos }^2}a - 1} \right) + 1} \right)}^2}}}\\
= \dfrac{{4{{\sin }^4}a}}{{4{{\cos }^4}a}} = {\tan ^4}a\\
81,\\
A = \dfrac{{{{\sin }^2}2a + 4{{\sin }^4}a - 4{{\sin }^2}a.{{\cos }^2}a}}{{4 - {{\sin }^2}2a - 4{{\sin }^2}a}}\\
= \dfrac{{{{\left( {2\sin a.\cos a} \right)}^2} + 4{{\sin }^4}a - 4{{\sin }^2}a.{{\cos }^2}a}}{{4.\left( {1 - {{\sin }^2}a} \right) - {{\left( {2\sin a.\cos a} \right)}^2}}}\\
= \dfrac{{4{{\sin }^2}a.{{\cos }^2}a + 4{{\sin }^4}a - 4{{\sin }^2}a.{{\cos }^2}a}}{{4.{{\cos }^2}a - 4{{\sin }^2}a.co{s^2}a}}\\
= \dfrac{{4{{\sin }^4}a}}{{4{{\cos }^2}a.\left( {1 - {{\sin }^2}a} \right)}} = \dfrac{{{{\sin }^4}a}}{{{{\cos }^2}a.{{\cos }^2}a}} = {\tan ^4}a = \dfrac{1}{9}\\
85,\\
A = \sin a.{\cos ^5}a - {\sin ^5}a.\cos a\\
= \sin a.\cos a.\left( {{{\cos }^4}a - {{\sin }^4}a} \right)\\
= \dfrac{1}{2}\sin 2a.\left( {{{\cos }^2}a - {{\sin }^2}a} \right).\left( {{{\cos }^2}a + {{\sin }^2}a} \right)\\
= \dfrac{1}{2}\sin 2a.\cos 2a.1\\
= \dfrac{1}{2}.\dfrac{1}{2}\sin 4a = \dfrac{1}{4}\sin 4a
\end{array}\)
