cos(x+y)*cos(x-y)=cos*cosx+sin*siny

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cos(x+y)*cos(x-y)=cos*cosx+sin*sinycos(x+y)*cos(x-y)=cos*cosx+sin*sinycos(x+y)*cos(x-y)=cos*cosx+sin

cos(x+y)*cos(x-y)=cos*cosx+sin*siny
cos(x+y)*cos(x-y)=cos*cosx+sin*siny

cos(x+y)*cos(x-y)=cos*cosx+sin*siny
cos(x+y)=cosx*cosy-sinx*siny
cos(x-y)=cosx*cosy+sinx*siny
则cos(x+y)*cos(x-y)
=(cos*cosx)*(cos*cosy) - (sin*sinx)*(sin*siny)
=(cos*cosx)*(1-sin*siny) - (1-cos*cosx)*(sin*siny)
=cos*cosx - (cos*cosx)*(sin*siny) - sin*siny + (cos*cosx)*(sin*siny)
=cos*cosx - sin*siny
这个……我证得应该是相减的关系
是不是题出错了,还是你打错了?
代入x=45°,y=45°就知道应该是相减的关系