设-π6≤x≤π4,函数y=log2(1+sinx)+log2(1-sinx)的最大值是_,最小值是_.

问题描述:

设-

π
6
≤x≤
π
4
,函数y=log2(1+sinx)+log2(1-sinx)的最大值是______,最小值是______.

∵y=log2(1+sinx)+log2(1-sinx)
=log2[(1+sinx)(1-sinx)]=log2(1-sin2x)=log2cosx2x=2log2cosx
∵-

π
6
≤x≤
π
4
2
2
≤cosx≤1∴-1≤2log2cosx≤0
故答案为:0,-1