Question

已知在△ABC中，内角A、B、C所对的边分别为a、b、c，且sin（A＋π6）＝2cosA．

Answer 1

sin(A+π/6)=2cosA

√3/2sinA+1/2cosA=2cosA
√3/2sinA=3/2cosA
√3sinA=3cosA
tanA=√3
A=π/3
cosC=√6/3 ,sinC=√[1-(cosC)^2]=√3/3 ,/sinA=√3/2
a/sinA=c/sinC
asinC=csinA
a*√3/3=c*√3/2
2a=3c
2a-3c=0
2)B(0,π/3),cos(A-B)=4/5,
sin(A-B)=3/5,
sinB=sin[A-(A-B)]=sinAcos(A-B)-cosAsin(A-B)
=√3/2*4/5-1/2*3/5
=(4√3-3)/10
sinB=(4√3-3)/10