函数y=cosx\(1-sinx)的单调递增区间?求步骤!多谢!

=sinc/(1-cosc)(c=0.5pi-x)
=2sin(c/2)cos0.5c/2sin0.5csin0.5c
=cot0.5(0.5pi-x))
下面的讨论就简单了。根据cot函数的单调区间分段讨论。

cosx=cos(x/2)平方-sin(x/2)平方
1-sinx=cos(x/2)平方 -2sin(x/2)cos(x/2)+sin(x/2)平方
=[cos(x/2)-sin(x/2)]平方
所以y=cos(x/2)平方-sin(x/2)平方 / [cos(x/2)-sin(x/2)]平方
=cos(x/2)+sin(x/2) / cos(x/2)-sin(x/2)
=1+tan(x/2) / 1-tan(x/2)