说一个更简单的办法:对付π/4+x,π/4-x之类问题有奇效!
(π/4+x)+(π/4-x)=π/2,
(π/4+x)=π/2-(π/4-x),
sin(π/4+x)=sin[π/2-(π/4-x)]=cos(π/4-x),
函数f(x)=sin(π/4+x)sin(π/4-x)=cos(π/4-x)sin(π/4-x)
=(1/2)[2cos(π/4-x)sin(π/4-x)] (符合二倍角正弦)
=(1/2)sin[2(π/4-x)]
=(1/2)sin(π/2-2x)
=(1/2)cos2x,
单调递减区间是:2KPai