Question

在三角形ABC中，sin(C-A)=1,sinB=1/3。(1)求sinA的值；（2）设AC=根号6，求三角形ABC的面积。

Answer 1

1、sin(C-A)=1,
c-A=90°,
C=90°+A,
sinB=sin(180°-A-C)=sin(A+C)=sin(90°+2A)=sin(180°-90°-2A)
=sin(90°-2A)=cos2A=1/3,
sinA=√[(1-cos2A)/2]=√3/3.
2、sinC=sin(90°+A)=cosA=√[1-(sinA)^2]=√6/3,
根据正弦定理，c/sinC=b/sinB,
c=[√6/(1/3)]√6/3=6,
S△ABC=AB*AC*sinA/2=6*√6*√3/3/2=3√2.

Answer 2

（1）sinB=sin(A+C)=1/3
sin(C-A)=1
C-A=π/2
C=π/2+A
sin(A+C)=sin(A+A+π/2)=sin(2A+π/2)=cos2A=1-2sin²A=1/3
sinA=根号3/3
（2）sin(A+C)-sin(C-A)=2sinAcosC=-2/3
cosC=-根号3/3
∵0<角C<π
∴cosC=根号3/3
a/sinA=b/sinB=3倍根号6
a=9倍根号2
S△ABC=1/2*a*b*cosC=9