Question

在三角形ABC中角ABC的对边为abc,A=45°,bsin(45°+C)-csin(45°+B)=a若a=根号2求△ABC的面积

Answer 1

bsin(45°+C)-csin(45°+B)=a
bsin(A+C)-csin(A+B)=a
bsinB-csinC=a
sin²B-sin²C=sinA (正弦定理)
2sin²B-2sin²C=2sinA
(1-cos2B)-(1-cos2C)=2sinA
cos2C-cos2B=2sinA
cos[(C+B)+(C-B)]-cos[(C+B)-(C-B)]=2sinA
-2sin(C+B)sin(C-B)=2sinA
-2sinAsin(C-B)=2sinA
所以sin(C-B)=-1，那么C-B=-90°，即B-C=90°
而B+C=180°-A=135°，所以B=112.5°，C=22.5°
那么sinB=sin(90°+C)=cosC
根据正弦定理，得：b/sinB=c/sinC=a/sinA=√2/(√2/2)=2
所以b=2sinB=2cosC，c=2sinC
所以S△ABC=1/2*bcsinA=1/2*(2cosC)*(2sinC)*sinA
=sin2CsinA
=sin45°*sin45°
=1/2