证明：左边=2[sin(π/4)cosa+cos(π/4)sina][sin(π/4)cosa-cos(π/4)sina]=2[cosa/√2+sina/√2][cosa/√2-sina/√2]=(cosa+sina)(cosa-sina)=cos²a-sin²a=cos(2a)=右边故原命题成立,证毕....