问题
填空题
设-
|
答案
∵y=log2(1+sinx)+log2(1-sinx)
=log2[(1+sinx)(1-sinx)]=log2(1-sin2x)=log2cosx2x=2log2cosx
∵-
≤x≤π 6
∴π 4
≤cosx≤1∴-1≤2log2cosx≤02 2
故答案为:0,-1
设-
|
∵y=log2(1+sinx)+log2(1-sinx)
=log2[(1+sinx)(1-sinx)]=log2(1-sin2x)=log2cosx2x=2log2cosx
∵-
≤x≤π 6
∴π 4
≤cosx≤1∴-1≤2log2cosx≤02 2
故答案为:0,-1