giúp mình những câu đó đi ạaaaaaaaa

Đáp án:

\(\begin{array}{l}
a,\\
{\sin ^4}\alpha  + {\cos ^4}\alpha  = \dfrac{3}{4} + \dfrac{{\cos 4\alpha }}{4}\\
b,\\
{\sin ^6}\alpha  + {\cos ^6}\alpha  = \dfrac{5}{8} + \dfrac{3}{8}\cos 4\alpha \\
c,\\
\dfrac{{1 - \sin 2\alpha }}{{1 + \sin 2\alpha }} = {\cot ^2}\left( {\alpha  + \dfrac{\pi }{4}} \right)\\
d,\\
\sin 3\alpha  = 3\sin \alpha  - 4{\sin ^3}\alpha \\
 = 4\sin \alpha .\sin \left( {\dfrac{\pi }{3} - \alpha } \right).\sin \left( {\dfrac{\pi }{3} + \alpha } \right)
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
{\sin ^4}\alpha  + {\cos ^4}\alpha \\
 = \left( {{{\sin }^4}\alpha  + 2.{{\sin }^2}\alpha .{{\cos }^2}\alpha  + {{\cos }^4}\alpha } \right) - 2{\sin ^2}\alpha .{\cos ^2}\alpha \\
 = {\left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right)^2} - 2.{\left( {\sin \alpha .\cos \alpha } \right)^2}\\
 = {1^2} - 2.{\left( {\dfrac{1}{2}\sin 2\alpha } \right)^2}\\
 = 1 - 2.\dfrac{1}{4}{\sin ^2}2\alpha \\
 = 1 - \dfrac{1}{2}{\sin ^2}2\alpha \\
 = \dfrac{3}{4} + \left( {\dfrac{1}{4} - \dfrac{1}{2}{{\sin }^2}2\alpha } \right)\\
 = \dfrac{3}{4} + \dfrac{1}{4}.\left( {1 - 2{{\sin }^2}2\alpha } \right)\\
 = \dfrac{3}{4} + \dfrac{1}{4}.\cos \left( {2.2\alpha } \right)\\
 = \dfrac{3}{4} + \dfrac{{\cos 4\alpha }}{4}\\
b,\\
{\sin ^6}\alpha  + {\cos ^6}\alpha \\
 = {\left( {{{\sin }^2}\alpha } \right)^3} + {\left( {{{\cos }^2}\alpha } \right)^3}\\
 = \left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right).\left( {{{\sin }^4}\alpha  - {{\sin }^2}\alpha .{{\cos }^2}\alpha  + {{\cos }^4}\alpha } \right)\\
 = 1.\left[ {\left( {{{\sin }^4}\alpha  + 2{{\sin }^2}\alpha .{{\cos }^2}\alpha  + {{\cos }^4}\alpha } \right) - 3{{\sin }^2}\alpha .{{\cos }^2}\alpha } \right]\\
 = {\left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right)^2} - 3{\sin ^2}\alpha .{\cos ^2}\alpha \\
 = {1^2} - 3.{\left( {\sin \alpha .\cos \alpha } \right)^2}\\
 = 1 - 3.{\left( {\dfrac{1}{2}\sin 2\alpha } \right)^2}\\
 = 1 - \dfrac{3}{4}{\sin ^2}2\alpha \\
 = \dfrac{5}{8} + \left( {\dfrac{3}{8} - \dfrac{3}{4}{{\sin }^2}2\alpha } \right)\\
 = \dfrac{5}{8} + \dfrac{3}{8}.\left( {1 - 2{{\sin }^2}2\alpha } \right)\\
 = \dfrac{5}{8} + \dfrac{3}{8}.\cos \left( {2.2\alpha } \right)\\
 = \dfrac{5}{8} + \dfrac{3}{8}\cos 4\alpha \\
c,\\
\dfrac{{1 - \sin 2\alpha }}{{1 + \sin 2\alpha }}\\
 = \dfrac{{\left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right) - 2.\sin \alpha .\cos \alpha }}{{\left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right) + 2.\sin \alpha .\cos \alpha }}\\
 = \dfrac{{{{\sin }^2}\alpha  - 2\sin \alpha .\cos \alpha  + {{\cos }^2}\alpha }}{{{{\sin }^2}\alpha  + 2\sin \alpha .\cos \alpha  + {{\cos }^2}\alpha }}\\
 = \dfrac{{{{\left( {\sin \alpha  - \cos \alpha } \right)}^2}}}{{{{\left( {\sin \alpha  + \cos \alpha } \right)}^2}}}\\
 = {\left( {\dfrac{{\cos \alpha  - \sin \alpha }}{{\cos \alpha  + \sin \alpha }}} \right)^2}\\
 = {\left( {\dfrac{{\dfrac{{\sqrt 2 }}{2}.\cos \alpha  - \dfrac{{\sqrt 2 }}{2}\sin \alpha }}{{\dfrac{{\sqrt 2 }}{2}.\cos \alpha  + \dfrac{{\sqrt 2 }}{2}\sin \alpha }}} \right)^2}\\
 = {\left( {\dfrac{{\cos \alpha .\cos \dfrac{\pi }{4} - \sin \alpha .\sin \dfrac{\pi }{4}}}{{\cos \alpha .\sin \dfrac{\pi }{4} + \sin \alpha .\cos \dfrac{\pi }{4}}}} \right)^2}\\
 = {\left( {\dfrac{{\cos \left( {\alpha  + \dfrac{\pi }{4}} \right)}}{{\sin \left( {\alpha  + \dfrac{\pi }{4}} \right)}}} \right)^2}\\
 = {\left( {\cot \left( {\alpha  + \dfrac{\pi }{4}} \right)} \right)^2}\\
 = {\cot ^2}\left( {\alpha  + \dfrac{\pi }{4}} \right)\\
d,\\
\sin 3\alpha \\
 = \sin \left( {\alpha  + 2\alpha } \right)\\
 = \sin \alpha .\cos 2\alpha  + \cos \alpha .\sin 2\alpha \\
 = \sin \alpha .\left( {1 - 2{{\sin }^2}\alpha } \right) + \cos \alpha .2\sin \alpha .\cos \alpha \\
 = \sin \alpha  - 2{\sin ^3}\alpha  + 2\sin \alpha .{\cos ^2}\alpha \\
 = \sin \alpha  - 2{\sin ^3}\alpha  + 2\sin \alpha .\left( {1 - {{\sin }^2}\alpha } \right)\\
 = \sin \alpha  - 2{\sin ^3}\alpha  + 2\sin \alpha  - 2{\sin ^3}\alpha \\
 = 3\sin \alpha  - 4{\sin ^3}\alpha \\
 = 4.\sin \alpha .\left( {\dfrac{3}{4} - {{\sin }^2}\alpha } \right)\\
 = 4\sin \alpha .\left[ {\dfrac{3}{4}\left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right) - {{\sin }^2}\alpha } \right]\\
 = 4\sin \alpha .\left( {\dfrac{3}{4}{{\cos }^2}\alpha  - \dfrac{1}{4}{{\sin }^2}\alpha } \right)\\
 = 4\sin \alpha .\left( {\dfrac{{\sqrt 3 }}{2}\cos \alpha  - \dfrac{1}{2}\sin \alpha } \right).\left( {\dfrac{{\sqrt 3 }}{2}\cos \alpha  + \dfrac{1}{2}\sin \alpha } \right)\\
 = 4\sin \alpha .\left( {\sin \dfrac{\pi }{3}.\cos \alpha  - \sin \alpha .\cos \dfrac{\pi }{3}} \right).\left( {\sin \dfrac{\pi }{3}.\cos \alpha  + \cos \dfrac{\pi }{3}.\sin \alpha } \right)\\
 = 4\sin \alpha .\sin \left( {\dfrac{\pi }{3} - \alpha } \right).\sin \left( {\dfrac{\pi }{3} + \alpha } \right)
\end{array}\)