Bài 4 trang 155 SGK Đại số 10

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:

\(\displaystyle {{\sin 5\alpha  - \sin 3\alpha } \over {2\cos 4\alpha }} \) \(\displaystyle = {{2\cos {{5\alpha  + 3\alpha } \over 2}\sin {{5\alpha  - 3\alpha } \over 2}} \over {2\cos 4\alpha }} \) \(\displaystyle  = \frac{{2\cos 4\alpha \sin \alpha }}{{2\cos 4\alpha }}\)

\(\displaystyle = \sin \alpha \)

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