Question

已知cosB=1/2,求sinA*sinC的取值范围

△ABC中
已知cosB=1/2,求sinA*sinC的取值范围
求解
谢谢

Answer 1

因为cosB=1/2
所以B=60°
所以A+C=120°且0°<=|A-C|<120°
cos(A+C)=-1/2且-1/2<cos(A-C)<=1
sinAsinC-cosAcosC=1/2
且-1/2<cosAcosC+sinAsinC<=1
两式相加，0<2sinAsinC<=3/2
所以0<sinAsinC<=3/4

追问

sinAsinC-cosAcosC=1/2

这步怎么来的？
不应该是余余正正吗

追答

cosAcosC-sinAsinC=cos(A+C)=-1/2
所以sinAsinC-cosAcosC=1/2

追问

请问 为何要两式相加

追答

为了把cosAcosC约掉

Answer 2

cosB=1/2，知∠B=60°
sinA*sinC
=[cos(A-C)-cos(A+C)]/2
=[cos(A-C)-cos120°]/2
=cos(A-C)/2+1/4
∠A+∠B+∠C=180°
∠B=60°
∠A>0°，∠C>0°
故-120°<∠A-∠C<120°
-1/2<cos(A-C)≤1
最终有 0<sinA*sinC≤3/4