Question

在三角形ABC中,sin^2C=sin(A+B)*sin(A-B),

Answer 1

sin^2C=sin(A+B)*sin(A-B)

1.

sin^2[180-(A+B)]=sin(A+B)*sin(A-B)

sin^2(A+B)=sin(A+B)*sin(A-B)

sin(A+B)=sin(A-B)

sin(A+B)=sinAcosB+cosAsinB(1)

sin(A-B)=sinAcosB-cosAsinB(2)

(1)=(2)

cosAsinB=0

cosA=0(A=90°) 或sinB=0(舍去)

2.

sin(A+B)=sin(A-B)

可得到B=45°

b=c=3√2

S=1/2*3√2*3√2=r*3√2+1/2*6*r

r=9/(3+3√2)

=3√2-3