根据sina*sinb=[cos(a-b)-cos(a+b)]/2
sinx·sin(x-π/4)=[cosπ/4-cos(2x-π/4)]/2=[√2/2-cos(2x-π/4)]/2
显然上式在cos(2x-π/4)=1时取最小值为(√2/2-1)/2=√2/4-1/2sina*sinb=[cos(a-b)-cos(a+b)]/2是怎么得出的cos(a-b)=cosacosb+sinasinbcos(a+b)=cosacosb-sinasinb两式相减，除以2即可得到