因为sin(a+b)=1,所以cos(a+b)=0, 所以cos(a+2b)+sin(2a+b)=cos(a+b+b)+sin(a+b+a)= cos(a+b)coab-sin(a+b)sinb+sin(a+b)cosa+cos(a+b)sina= -sinb+cosa,又因为sin(a+b)=1,所以a+b=90+360k, 所以a=90+360k-b,所以-sinb+cosa=-sinb+cos(90+360k-b)= -sinb+cos(90-b)=-sinb+sinb=0
答：1)|AB|^2=(cosa-cosb)^2+(sina-cosb)^2 =1+1-2(cosacosb+sinasinb) =2-2cos(a-b)=<2 -...详情>>
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