Nội dung bài giảng
Bài 4. 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)
\(\eqalign{
& {{2\sin 2\alpha - \sin 4\alpha } \over {2\sin 2\alpha + \sin 4\alpha }} = {{2\sin 2\alpha - 2\sin 2\alpha .cos2\alpha } \over {2\sin 2\alpha + 2\sin 2\alpha .cos2\alpha }} \cr
& = {{1 - \cos 2\alpha } \over {1 + \cos 2\alpha }} = {{2{{\sin }^2}\alpha } \over {2{{\cos }^2}\alpha }} =\tan^2\alpha\cr} \)
b)
\(\eqalign{
& \tan \alpha \left({{1 + {{\cos }^2}\alpha } \over {\sin \alpha }} - \sin \alpha\right ) = {{\sin \alpha } \over {\cos \alpha }}\left({{1 + {{\cos }^2}\alpha - {{\sin }^2}\alpha } \over {\sin \alpha }}\right) \cr
& = {{\sin \alpha } \over {\cos \alpha }}.{{2{{\cos }^2}\alpha } \over {\sin \alpha }} = 2\cos \alpha \cr} \)
c)
\(\eqalign{
& = {{\tan \left({\pi \over 4} - \alpha \right) + 1} \over {\tan\left({\pi \over 4} - \alpha \right) - 1}} = \left({{\tan {\pi \over 4} - \tan \alpha } \over {1 + \tan {\pi \over 4}.\tan \alpha }} + 1\right):\left({{\tan {\pi \over 4} - \tan \alpha } \over {1 + \tan {\pi \over 4}.\tan \alpha }} - 1\right) \cr
& = \left({{1 - \tan \alpha + 1 + \tan \alpha } \over {1 + \tan \alpha }} \right):\left({{1 - \tan \alpha - 1 - \tan \alpha } \over {1 + \tan \alpha }} \right) \cr
& = {{ - 1} \over {\tan \alpha }} = - \cot \alpha \cr} \)
d)
\({{\sin 5\alpha - \sin 3\alpha } \over {2\cos 4\alpha }} = {{2\cos {{5\alpha + 3\alpha } \over 2}\sin {{5\alpha - 3\alpha } \over 2}} \over {2\cos 4\alpha }} = \sin \alpha \)