证明(sinα)^2+(sinβ)^2+2sinαsinβcos(α+β)(sinα)^2+(sinβ)^2+2sinαsinβc_数学解答

证明(sinα)^2+(sinβ)^2+2sinαsinβcos(α+β)
(sinα)^2+(sinβ)^2+2sinαsinβcos(α+β)=0

人气：265 ℃　时间：2020-05-20 04:33:39

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