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Rút gọn biểu thức

Rút gọn biểu thức

a) \({{2\sin 2\alpha  - \sin 4\alpha } \over {2\sin 2\alpha  + \sin 4\alpha }}\)

b) \(\tan \alpha ({{1 + {{\cos }^2}\alpha } \over {\sin \alpha }} - \sin \alpha )\)

c) \({{\sin ({\pi  \over 4} - \alpha ) + \cos ({\pi  \over 4} - \alpha )} \over {\sin ({\pi  \over 4} - \alpha ) - \cos ({\pi  \over 4} - \alpha )}}\)

d) \({{\sin 5\alpha  - \sin 3\alpha } \over {2\cos 4\alpha }}\)

Trả lời:

a)

\(\eqalign{
& {{2\sin 2\alpha - \sin 4\alpha } \over {2\sin 2\alpha + \sin 4\alpha }} = {{2\sin 2\alpha - 2\sin 2\alpha .cos2\alpha } \over {2\sin 2\alpha + 22\sin 2\alpha .cos2\alpha }} \cr
& = {{1 - \cos 2\alpha } \over {1 + \cos 2\alpha }} = {{2{{\sin }^2}\alpha } \over {2{{\cos }^2}\alpha }} \cr} \)

b)

\(\eqalign{
& \tan \alpha ({{1 + {{\cos }^2}\alpha } \over {\sin \alpha }} - \sin \alpha ) = {{\sin \alpha } \over {\cos \alpha }}({{1 + {{\cos }^2}\alpha - {{\sin }^2}\alpha } \over {\sin \alpha }}) \cr
& = {{\sin \alpha } \over {\cos \alpha }}.{{2{{\cos }^2}\alpha } \over {\sin \alpha }} = 2\cos \alpha \cr} \)

c)

\(\eqalign{
& = {{\tan ({\pi \over 4} - \alpha ) + 1} \over {\tan ({\pi \over 4} - \alpha ) - 1}} = ({{\tan {\pi \over 4} - \tan \alpha } \over {1 + \tan {\pi \over 4}.\tan \alpha }} + 1):({{\tan {\pi \over 4} - \tan \alpha } \over {1 + \tan {\pi \over 4}.\tan \alpha }} - 1) \cr
& = ({{1 - \tan \alpha + 1 + \tan \alpha } \over {1 + \tan \alpha }}):({{1 - \tan \alpha - 1 - \tan \alpha } \over {1 + \tan \alpha }}) \cr
& = {{ - 1} \over {\tan \alpha }} = - \cot \alpha \cr} \) 

d) 

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