Question

在三角形ABC中,AC=2,BC=1,CosC=3/4,求Sin(2A+C)=

Answer 1

余弦定理
AB²=AC²+BC²-2AC*BC*cosC=4+1-2*2*1*3/4=2
AB=√2
cosA=(AB²+AC²-BC²)/(2AC*AB)=5√2/8
A锐角
sinA=√14/8
sin2A=2sinAcosA=5√7/16
cos2A=cos²A-sin²A=9/16
cosC=3/4
C为锐角
sinC=√7/4
sin(2A+C)=sin2AcosC+cos2AsinC=5√7/16*3/4+9/16*√7/4=3√7/8

Answer 2

AB²=AC²+BC²-2AC*BC*cosC=4+1-2*2*1*3/4=2
AB=√2
cosA=(AB²+AC²-BC²)/(2AC*AB)=5√2/8
A锐角
sinA=√14/8
sin2A=2sinAcosA=5√7/16
cos2A=cos²A-sin²A=9/16