在三角形ABC中,sin(C-A)=1,sinB=1/3,求sinA,若AC=根号下6,求三角型面积

C=90+AsinB=sin(A+C) = sinAcosC+cosAsinC = -sin^2A + cos^2A = 1-2sin^2AsinA = sqrt(3)/3AC=b = sqrt(6)a /sinA = b/sinBa = sinA/sinB*bS=1/2 ab sinC = 1/2 b^2 sinAsinC/sinB = sqrt(3)*sqrt(6)/3*3 = 3sqrt(2...问一下，sinB=sin(A+C) = sinAcosC+cosAsinC = -sin^2A + cos^2A = 1-2sin^2A这一步怎么来的，sinC=sin（90+A)=cosA,那么sinB=sin（A+C)=sinAcosc+cosAsinC不是等于sin^A+cos^A=1么?cosC = -sinA对啊，那sinB=sin(A+C) = sinAcosC+cosAsinC = -sin^2A + cos^2A = 1-2sin^2A是怎么来的cosC = -sinAsinAcosC = -sin^2 A额，不好意思啊，看错了，看成2倍的sinA了