Question

asina+bsinb等于什么

Answer 1

asinA+bsinB=csinC。
a=2R*sinA,b=2R*sinB,c=2R*sinC,即有
(sinA)^2+(sinB)^2=(sinC)^2,
(sinA)^2=(sinC)^2-(sinB)^2=(sinC+sinB)*(sinC-sinB),
而,sinC+sinB=2sin[(C+B)/2]*cos[(C-B)/2],
sinC-sinB=2cos[(C+B)/2]*sin[(C-B)/2],
(sinA)^2=[2*sin(A/2)*cos(A/2)]^2,
A+B+C=180,A/2=[180-(C+B)]/2=90-(C+B)/2,
即有,sin(A/2)=sin[90-(C+B)/2]=cos[(C+B)/2],
cos(A/2)=cos[(90-(C+B)/2]=sin[(C+B)/2],
可得到,
2cos(A/2)*sin(A/2)=2cos[(C-B)/2]*sin[(C-B)/2],
sinA=sin(C-B),
sinA-sin(C-B)=0,
2cos[(A+C-B)/2]*sin[(A+B-C)/2]=0,
cos[(A+C-B)/2]=0,或sin[(A+B-C)/2]=0,
A+C=180-B,或A+B=180-C,
cos[(A+C-B)/2]=cos(90-B)=sinB=0(不合,舍去).
当sin[(A+B-C)/2]=0,时,
sin(90-C)=cosC=0,即有,
cosC=(a^2+b^2-c^2)/2ab=0,
a^2+b^2=c^2,
C=90度,ABC为直角三角形.