Question

在三角形ABC中,AC=2.BC=1cosC=3/4.求sin(2A+C)

Answer 1

楼上的好像解错了吧……
根据余弦定理：
c^2=1^2+2^2-2*1*2*3/4
=2
∴c=√2
∴cosA=(b^2+c^2-a^2)/2bc
=(4+2-1)/4√2
=5/(4√2)
∵(sinA)^2+(cosA)^2=1
∵sinA>0
∴sinA=√7/(4√2)
∵cosC=3/4
∴sinC=√7/4
sin2A=2sinAcosA
=5√7/16
cos2A=(cosA)^2-(sinA)^2
=9/16
∴sin(2A+C)=sin2AcosC+cos2AsinC
=3√7/8

Answer 2

c^2=a^2+b^2-2abcosC=1+4-2*2*3/4=2
c=√2
cosC=3/4,sinC=√5/2
sinA=asinC/c=1*(√5/2)/√2=√10/4
cosA=√6/4
sin2A=2sinAcosA=2*√60/16=√15/4
cos2A=-1/4

sin(2A+C)
=sin2AcosC+cos2AsinC
=√10/4*3/4-1/4*√5/2
=(3√10-2√5)/16

Answer 3

AB^2=AC^2+BC^2-2ACBCcosC
=4+1-4*3/4=2
AB=√2
sinC=√7/4
sinA=BC*sinC/AB=√14/8
cosA=5√2/8
sin2A=2sinAcosA=5√7/16
cos2A=9/16
sin(2A+C)=sin2AcosC+cos2AsinC
=5√7/16*3/4+9/16*√7/4
=3√7/8