Rút gọn các biểu thức
\(\begin{array}{l}
{\rm{a)}}\frac{{2\sin 2\alpha - \sin 4\alpha }}{{2\sin 2\alpha + \sin 4\alpha }}\\
{\rm{b)tan}}\left( {\frac{{1 + {{\cos }^2}\alpha }}{{\sin \alpha }} - \sin \alpha } \right)\\
{\rm{c)}}\frac{{\sin \left( {\frac{\pi }{4} - \alpha } \right) + \cos \left( {\frac{\pi }{4} - \alpha } \right)}}{{\sin \left( {\frac{\pi }{4} - \alpha } \right) - {\rm{cos}}\left( {\frac{\pi }{4} - \alpha } \right)}}\\
{\rm{d)}}\frac{{\sin 5\alpha - \sin 3\alpha }}{{2\cos 4\alpha }}
\end{array}\)
\(a) \begin{array}{l}
2\sin 2a - \sin 4a = 2\sin 2a - 2\sin 2a.\cos 2a\\
= 2\sin 2a\left( {1 - \cos 2a} \right)\\
2\sin 2a + \sin 4a = 2\sin 2a + 2\sin 2a.\cos 2a\\
= 2\sin 2a\left( {1 + \cos 2a} \right)\\
\frac{{2\sin 2a - \sin 4a}}{{2\sin 2a + \sin 4a}} = \frac{{1 - \cos 2a}}{{1 + \cos 2a}}
\end{array}\)
\(\begin{array}{l}
{\rm{b)}}\frac{{1 + {{\cos }^2}a}}{{\sin a}} - \sin a = \frac{{1 + o{s^2}a - {{\sin }^2}a}}{{\sin a}}\\
= \frac{{{{\cos }^2}a + {{\sin }^2}a + {{\cos }^2}a - {{\sin }^2}a}}{{\sin a}} = \frac{{2{{\cos }^2}a}}{{\sin a}}\\
\tan {\rm{ }}a\left( {\frac{{1 + {{\cos }^2}a}}{{\sin a}} - \sin a} \right) = \frac{{\sin a}}{{\cos a}}.\frac{{2{{\cos }^2}a}}{{\sin a}} = 2\cos a
\end{array}\)
\(\begin{array}{l}
c)\sin \left( {\frac{\pi }{4} - a} \right) + \cos \left( {\frac{\pi }{4} - a} \right) = \sin \frac{\pi }{4}\cos a - \sin a\cos \frac{\pi }{4} + \cos a\cos \frac{\pi }{4} + \sin a\sin \frac{\pi }{4}\\
= \frac{{\sqrt 2 }}{2}\left( {\cos a - \sin a} \right) + \frac{{\sqrt 2 }}{2}\left( {\cos a + \sin a} \right) = \sqrt 2 \cos a\\
\sin \left( {\frac{\pi }{4} - a} \right) - \cos \left( {\frac{\pi }{4} - a} \right) = \sin \frac{\pi }{4}\cos a - \sin a\cos \frac{\pi }{4} - \cos a\cos \frac{\pi }{4} - \sin a\sin \frac{\pi }{4}\\
\frac{{\sqrt 2 }}{2}\left( {\cos a - \sin a} \right) - \frac{{\sqrt 2 }}{2}\left( {\cos a + \sin a} \right) = - \sqrt 2 \sin a
\end{array}\)
Suy ra, ta có: \(\frac{{\sin \left( {\frac{\pi }{4} - a} \right) + \cos \left( {\frac{\pi }{4} - a} \right)}}{{\sin \left( {\frac{\pi }{4} - a} \right) - \cos \left( {\frac{\pi }{4} - a} \right)}} = - \cot a\)
\({\rm{d)}}\sin 5a - \sin 3a = 2{\rm{os}}\frac{{5a + 3a}}{2}.\sin \frac{{5a - 3a}}{2} = 2\cos 4a.\sin a\)
Suy ra \(\frac{{\sin 5a - \sin 3a}}{{2\cos 4a}} = \frac{{2\cos 4a.\sin a}}{{2\cos 4a}} = \sin a\)
-- Mod Toán 10