Question

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

Answer 1

C=90+A
sinB=sin(A+C) = sinAcosC+cosAsinC = -sin^2A + cos^2A = 1-2sin^2A
sinA = sqrt(3)/3
AC=b = sqrt(6)
a /sinA = b/sinB
a = sinA/升袜冲明sinB*b
S=1/2 ab sinC = 1/2 b^2 sinAsinC/吵判激sinB = sqrt(3)*sqrt(6)/3*3 = 3sqrt(2)

追问

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?

追答

cosC = -sinA

追问

对啊，那sinB=sin(A+C) = sinAcosC+cosAsinC = -sin^2A + cos^2A = 1-2sin^2A
是怎么来的

追答

cosC = -sinA
sinAcosC = -sin^2 A

追问

额，不好意思啊，看错了，看成2倍的sinA了