已知sina+sinb+sinc=0,cosa+cosb+cosc=0,则cos(a-b)的值是?答:因为(sinc)^2=(sina+sinb)^2=(sina)^2+2sina*sinb+(sinb)^2,同理(cosc)^2=(cosa)^2+2cosa*cosb+(cosb)^2,所以相加得1=1+2(cosa*cosb+sina*sinb)+1,所以cos(a-b)=-1/2.我想知道所
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![已知sina+sinb+sinc=0,cosa+cosb+cosc=0,则cos(a-b)的值是?答:因为(sinc)^2=(sina+sinb)^2=(sina)^2+2sina*sinb+(sinb)^2,同理(cosc)^2=(cosa)^2+2cosa*cosb+(cosb)^2,所以相加得1=1+2(cosa*cosb+sina*sinb)+1,所以cos(a-b)=-1/2.我想知道所](/uploads/image/z/1069306-34-6.jpg?t=%E5%B7%B2%E7%9F%A5sina%2Bsinb%2Bsinc%3D0%2Ccosa%2Bcosb%2Bcosc%3D0%2C%E5%88%99cos%28a-b%29%E7%9A%84%E5%80%BC%E6%98%AF%3F%E7%AD%94%EF%BC%9A%E5%9B%A0%E4%B8%BA%28sinc%29%5E2%3D%28sina%2Bsinb%29%5E2%3D%28sina%29%5E2%2B2sina%2Asinb%2B%28sinb%29%5E2%2C%E5%90%8C%E7%90%86%28cosc%29%5E2%3D%28cosa%29%5E2%2B2cosa%2Acosb%2B%28cosb%29%5E2%2C%E6%89%80%E4%BB%A5%E7%9B%B8%E5%8A%A0%E5%BE%971%3D1%2B2%28cosa%2Acosb%2Bsina%2Asinb%29%2B1%2C%E6%89%80%E4%BB%A5cos%28a-b%29%3D-1%2F2.%E6%88%91%E6%83%B3%E7%9F%A5%E9%81%93%E6%89%80)
已知sina+sinb+sinc=0,cosa+cosb+cosc=0,则cos(a-b)的值是?答:因为(sinc)^2=(sina+sinb)^2=(sina)^2+2sina*sinb+(sinb)^2,同理(cosc)^2=(cosa)^2+2cosa*cosb+(cosb)^2,所以相加得1=1+2(cosa*cosb+sina*sinb)+1,所以cos(a-b)=-1/2.我想知道所
已知sina+sinb+sinc=0,cosa+cosb+cosc=0,则cos(a-b)的值是?
答:
因为(sinc)^2=(sina+sinb)^2=(sina)^2+2sina*sinb+(sinb)^2,
同理(cosc)^2=(cosa)^2+2cosa*cosb+(cosb)^2,
所以相加得1=1+2(cosa*cosb+sina*sinb)+1,
所以cos(a-b)=-1/2.
我想知道所以cos(a-b)=-1/2的细微过程
已知sina+sinb+sinc=0,cosa+cosb+cosc=0,则cos(a-b)的值是?答:因为(sinc)^2=(sina+sinb)^2=(sina)^2+2sina*sinb+(sinb)^2,同理(cosc)^2=(cosa)^2+2cosa*cosb+(cosb)^2,所以相加得1=1+2(cosa*cosb+sina*sinb)+1,所以cos(a-b)=-1/2.我想知道所
为(sinc)^2=(sina+sinb)^2=(sina)^2+2sina*sinb+(sinb)^2,
同理(cosc)^2=(cosa)^2+2cosa*cosb+(cosb)^2,
所以相加得1=1+2(cosa*cosb+sina*sinb)+1,
公式cos(a-b)=cosa*cosb+sina*sinb
∴所以相加得1=1+2cos(a-b)+1,
2cos(a-b)=-1
所以cos(a-b)=-1/2.