Video hướng dẫn giải
- LG a
- LG b
- LG c
- LG d
Rút gọn biểu thức
LG a
\[\displaystyle {{2\sin 2\alpha - \sin 4\alpha } \over {2\sin 2\alpha + \sin 4\alpha }}\]
Phương pháp giải:
Áp dụng các công thức:
\[\begin{array}{l}
+ ]\cos2\alpha = 1 - 2{\sin ^2}\alpha = 2{\cos ^2}\alpha - 1.\\
+ ]\tan\alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }}.\\
+ ]\tan\alpha .\cot\alpha = 1.
\end{array}\]
Lời giải chi tiết:
\[\eqalign{ & {{2\sin 2\alpha - \sin 4\alpha } \over {2\sin 2\alpha + \sin 4\alpha }} \cr&= \frac{{2\sin 2\alpha - \sin \left[ {2.2\alpha } \right]}}{{2\sin 2\alpha + \sin \left[ {2.2\alpha } \right]}}\cr&= {{2\sin 2\alpha - 2\sin 2\alpha .cos2\alpha } \over {2\sin 2\alpha + 2\sin 2\alpha .cos2\alpha }} \cr & = \frac{{2\sin 2\alpha \left[ {1 - \cos 2\alpha } \right]}}{{2\sin 2\alpha \left[ {1 + \cos 2\alpha } \right]}}\cr &= {{1 - \cos 2\alpha } \over {1 + \cos 2\alpha }}= \frac{{1 - \left[ {1 - 2{{\sin }^2}\alpha } \right]}}{{1 + \left[ {2{{\cos }^2}\alpha - 1} \right]}}\cr&= {{2{{\sin }^2}\alpha } \over {2{{\cos }^2}\alpha }} = \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }}\cr&={\left[ {\frac{{\sin \alpha }}{{\cos \alpha }}} \right]^2}=\tan^2\alpha.\cr} \]
LG b
\[\tan \alpha [{{1 + {{\cos }^2}\alpha } \over {\sin \alpha }} - \sin \alpha ]\]
Lời giải chi tiết:
\[\eqalign{& \tan \alpha \left[{{1 + {{\cos }^2}\alpha } \over {\sin \alpha }} - \sin \alpha\right ] \cr&= {{\sin \alpha } \over {\cos \alpha }}\left[{{1 + {{\cos }^2}\alpha - {{\sin }^2}\alpha } \over {\sin \alpha }}\right] \cr & = \frac{{\sin \alpha }}{{\cos \alpha }}.\frac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha + {{\cos }^2}\alpha - {{\sin }^2}\alpha }}{{\sin \alpha }}\cr &= {{\sin \alpha } \over {\cos \alpha }}.{{2{{\cos }^2}\alpha } \over {\sin \alpha }} = 2\cos \alpha. \cr} \]
LG c
\[\displaystyle {{\sin [{\pi \over 4} - \alpha ] + \cos [{\pi \over 4} - \alpha ]} \over {\sin [{\pi \over 4} - \alpha ] - \cos [{\pi \over 4} - \alpha ]}}\]
Lời giải chi tiết:
\[\displaystyle {{\sin [{\pi \over 4} - \alpha ] + \cos [{\pi \over 4} - \alpha ]} \over {\sin [{\pi \over 4} - \alpha ] - \cos [{\pi \over 4} - \alpha ]}}\]
\[\begin{array}{l}
= \dfrac{{\cos \left[ {\frac{\pi }{4} - \alpha } \right]\left[ {\frac{{\sin \left[ {\frac{\pi }{4} - \alpha } \right]}}{{\cos \left[ {\frac{\pi }{4} - \alpha } \right]}} + 1} \right]}}{{\cos \left[ {\frac{\pi }{4} - \alpha } \right]\left[ {\frac{{\sin \left[ {\frac{\pi }{4} - \alpha } \right]}}{{\cos \left[ {\frac{\pi }{4} - \alpha } \right]}} - 1} \right]}}\\
= \dfrac{{\cos \left[ {\frac{\pi }{4} - \alpha } \right]\left[ {\tan \left[ {\frac{\pi }{4} - \alpha } \right] + 1} \right]}}{{\cos \left[ {\frac{\pi }{4} - \alpha } \right]\left[ {\tan \left[ {\frac{\pi }{4} - \alpha } \right] - 1} \right]}}\\
= \dfrac{{\tan \left[ {\frac{\pi }{4} - \alpha } \right] + 1}}{{\tan \left[ {\frac{\pi }{4} - \alpha } \right] - 1}}\\
= \left[ {\tan \left[ {\frac{\pi }{4} - \alpha } \right] + 1} \right]:\left[ {\tan \left[ {\frac{\pi }{4} - \alpha } \right] - 1} \right]\\
= \left[ {\frac{{\tan \frac{\pi }{4} - \tan \alpha }}{{1 + \tan \frac{\pi }{4}.\tan \alpha }} + 1} \right]
:\left[ {\frac{{\tan \frac{\pi }{4} - \tan \alpha }}{{1 + \tan \frac{\pi }{4}.\tan \alpha }} - 1} \right]\\
= \left[ {\frac{{1 - \tan \alpha }}{{1 + \tan \alpha }} + 1} \right]:\left[ {\frac{{1 - \tan \alpha }}{{1 + \tan \alpha }} - 1} \right]\\
= \frac{{1 - \tan \alpha + 1 + \tan \alpha }}{{1 + \tan \alpha }}:\frac{{1 - \tan \alpha - 1 - \tan \alpha }}{{1 + \tan \alpha }}\\
= \frac{2}{{1 + \tan \alpha }}:\frac{{ - 2\tan \alpha }}{{1 + \tan \alpha }}\\
= \frac{2}{{1 + \tan \alpha }}.\frac{{1 + \tan \alpha }}{{ - 2\tan \alpha }}\\
= - \frac{1}{{\tan \alpha }} = - \cot \alpha
\end{array}\]
Cách khác:
LG d
\[\displaystyle {{\sin 5\alpha - \sin 3\alpha } \over {2\cos 4\alpha }}\]
Lời giải chi tiết:
\[\displaystyle {{\sin 5\alpha - \sin 3\alpha } \over {2\cos 4\alpha }} \] \[\displaystyle = {{2\cos {{5\alpha + 3\alpha } \over 2}\sin {{5\alpha - 3\alpha } \over 2}} \over {2\cos 4\alpha }} \] \[\displaystyle = \frac{{2\cos 4\alpha \sin \alpha }}{{2\cos 4\alpha }}\]
\[\displaystyle = \sin \alpha \]