Question

高一数学解三角形

1.在△ABC中，若a/cos(A/2)=b/cos(B/2)=c/cos(C/2),那么△ABC是什么三角形
2.在三角形ABC中cos^2(A/2)=(a+b)/2c,则三角形ABC的形状是？
3.在△ABC中，若若lgsinA-lgcosB-lgsinC=lg2,则三角形ABC的形状是？

Answer 1

1
a/sinA=b/sinB=c/sinC
a/cos(A/2)=b/(cosB/2)=C/cos(C/2)
a/b=sinA/sinB=cos(A/2)/cos(B/2)
sinA/cos(A/2)=sinB/cos(B/2)
sin(A/2)=sin(B/2)
A=B
同理，a/c=sinA/sinC=cos(A/2)/cos(C/2)
sin(A/2)=sin(C/2)
A=C
等边三角
2
(1+cosA)/2=(a+b)/2c
(1+cosA)=(a+b)/c
a/sinA=b/sinB=c/sinC=k
(1+cosA)=(sinA+sinB)/sinC
sinC+cosAsinC=sinA+sinB
sinC=sinA+(sin(B+C)-cosAsinC)
sinC=sinA+sinAcosC
sinC/(1+cosC)=sinA
tan(C/2)=sinA
a^2=b^2+c^2-2bccosA=(b+c)^2-2bc(1+cosA)
(1+cosA)=[(b+c)^2-a^2]/2bc=(a+b)/c
(b+c)^2-a^2=2ab+2b^2
2bc+c^2=a^2+2ab+b^2
c^2=(a+b)^2-2bc
c^2=(a+b)^2-2ab(1+cosC)
2bc=2ab(1+cosC)
c/b=(1+cosC)
sinC/sinB=(1+cosC)
sinB=tan(C/2)
sinA=sinB=tan(C/2)
A=B等腰三角
3
sinA=2sinCcosB
sinA=sin(B+C)=sinCcosB+sinBcosC
sinBcosC-sinCcosB=0
sin(B-C)=0
B=C
b=c

Answer 2

解：（1）可化为sinA/cos(A/2)=sinB/cos(B/20)=sinC/cos(C/2)
∴sin(A/2)=sin(B/2)=sin(C/2) (利用倍角公式：sin2A=2sinAcosA)
∴0.5A=0.5B=0.5C
A=B=C所以是正三角形