即cos[π/2-(a-π/4)]=1/3cos(π/4+a)cos(3π/4-a)=1/3cos(π/4+a)-cos[π-(3π/4-a)]=1/3cos(π/4+a)-cos(π/4+a)=1/3cos(π/4+a)所以cos(π/4+a)=0π/4+a=kπ+π/2a=kπ+π/4