
这道题怎么做,sin(a+b)sin(a-b)怎么化成 负二分之一(cos2a-cos2b) 给个详细的步骤
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cos2a=cos[(a+b)+(a-b)]=cos(a+b)cos(a-b)-sin(a+b)sin(a-b)
cos2b=cos[(a+b)-(a-b)]=cos(a+b)cos(a-b)+sin(a+b)sin(a-b)
cos2a-cos2b=cos(a+b)cos(a-b)-sin(a+b)sin(a-b)-(cos(a+b)cos(a-b)+sin(a+b)sin(a-b))
=-2sin(a+b)sin(a-b)
所以
sin(a+b)sin(a-b)=-1/2(cos2a-cos2b)
cos2b=cos[(a+b)-(a-b)]=cos(a+b)cos(a-b)+sin(a+b)sin(a-b)
cos2a-cos2b=cos(a+b)cos(a-b)-sin(a+b)sin(a-b)-(cos(a+b)cos(a-b)+sin(a+b)sin(a-b))
=-2sin(a+b)sin(a-b)
所以
sin(a+b)sin(a-b)=-1/2(cos2a-cos2b)
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