1.函数f(x)=sin(π/4-x),x属于【-π,0】的单调递减区间是
f(x)=sin[-(x-π/4)]=-sin(x-π/4)；x∈[-π,0]时的单调递减区间为[-π/4,0]
∵f(-π)=-sin(-π-π/4)=sin(π+π/4)=-sin(π/4)=-√2/2；f(-π/4)=-sin(-π/4-π/4)=sin(π/2)=1；
f(0)=-sin(-π/4)=sin(π/4)=√2/2.
2.已知函数f(x)=2sin(π-x)cosx（1）求f(x)的最小正周期（2）求f(x)在区间【-π/6,π/2】上的最大值和最小值
f(x)=2sin(π-x)cosx=2sinxcosx=sin2x,故最小正周期T=2π/2=π；
区间[-π/6,π/2]的长度=π/2-(-π/6)=π/2+π/6=2π/3