tanA*tanB=1除以根号3,求(2-cos2A)(2-cos2B)
热心网友
=[1+2sinA^2]*[1+2sinB^2]=1+2(sinA^2+sinB^2)+4sinA^2*sinB^2=1+2[sinA^2/(sinA^2+cosA^2)+sinB^2/(sinB^2+cosB^2)]+4sinA^2*sinB^2/ [(sinA^2+cosA^2)*(sinB^2+cosB^2)]=1+2[tanA^2/(tanA^2+1)+tanB^2/(tanB^2+1)]+4tanA^2*tanB^2/{(tanA^2+1)(tanB^2+1)]=1+[2(tanA^2+tanB^2)+8tanA^2*tanB^2]/[(tanA^2+tanB^2)+tanA^2*tanB^2+1]=1+[2(tanA^2+tanB^2)+8/3]/[(tanA^2+tanB^2)+4/3]=1+2=3