证明(sinα)^2+(sinβ)^2+2sinαsinβcos(α+β)
证明(sinα)^2+(sinβ)^2+2sinαsinβcos(α+β)
(sinα)^2+(sinβ)^2+2sinαsinβcos(α+β)=0
(sinα)^2+(sinβ)^2+2sinαsinβcos(α+β)=0
数学人气：453 ℃时间：2020-05-20 04:33:39
优质解答
没结论啊 左边＝(sina)^2+(sinb)^2+2sinasinb(cosacosb-sinasinb) =(sina)^2(1-(sinb)^2)+(sinb)^2(1-(sina)^2)+2sinacosasinbcosb =(sina)^2(cosb)^2+(cosa)^2(sinb)^2+2sinacosasinbcosb =(sinacosb+cosasinb)^2=(sin(a+b))^2
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