Bài 8 trang 161 SGK Đại số 10

Đề bài

Rút gọn các biểu thức sau:

a) \({{1 + \sin 4a - \cos 4a} \over {1 + \cos 4a + \sin 4a}}\)

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

c) \({{\cos 2x - \sin 4x - \cos 6x} \over {\cos 2x + \sin 4x - \cos 6x}}\)

Hướng dẫn giải

\(\eqalign{a) \, \, & {{1 + \sin 4a - \cos 4a} \over {1 + \cos 4a + \sin 4a}} \cr&= {{2{{\sin }^2}2a + 2\sin 2a\cos 2a} \over {2{{\cos }^2}2a + 2\sin 2a\cos 2a}} \cr & = {{2\sin 2a(\sin 2a + \cos 2a)} \over {2\cos 2a(\sin 2a + \cos 2a)}} \cr&= \tan 2a \cr} \) 

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

\(\eqalign{c) \, \, 
& {{\cos 2x - \sin 4x - \cos 6x} \over {\cos 2x + \sin 4x - \cos 6x}} \cr&= {{(cos2x - \cos 6x) - sin4x} \over {(cos2x - \cos 6x) + sin4x}} \cr
& = {{-2\sin {{2x + 6x} \over 2}\sin {{6x - 2x} \over 2} - \sin 4x} \over {-2\sin {{2x + 6x} \over 2}\sin {{2x - 6x} \over 2} + \sin 4x}} \cr
& = {{2\sin 2x - 1} \over {2\sin 2x + 1}} \cr} \)