sina+cosb+cosc=0,cosa+sinb+sic=0 --->sina+cosb=-cosc,cosa+sinb=-sinc 两边同时平方以后相加得 (sina)^2+(cosb)^2+2sinacosb+(cosa)^2+(sinb)^2+2cosasinb =(cosc)^2+(sinc)^2 --->1+1+2(sinacosb+cosasinb)=1 --->2+2sin(a+b)=1 --->sin(a+b)=-1/3.
答：sinα+sinβ+sinｒ=0 sinα+sinβ=-sinｒ 平方sin^α+sin^β+2sinαsinβ=sin^r cosα+cosβ+cosｒ=...详情>>
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