Question

已知sinA+sinB+sinC=0,cosA+cosB+cosC=0,求cos(A－B)的值

Answer 1

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.