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....亲我的回答你还满意吧如果满意给个好评吧