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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