Question

在ΔABC中，角A,B,C所对应的边分别是a,b,c，且满足2cos(B-C)=4sinBsinC-1

1、 若（b+c+a）*(b+c-a)=k*bc,求实数k的值.
2、 若a=3,b=2，求sin(C-π/6)的值

Answer 1

解：2cos(B-C)=4sinBsinC-1
2（cosBcosC+sinBsinC）=4sinBsinC-1
2cosBcosC+2sinBsinC=4sinBsinC-1
2cosBcosC-2sinBsinC= -1
2cos(B+C)= -1
cos[π-(B+C)]=-1/2
cosA=1/2 A=π/3
1.（b+c+a）*(b+c-a)=k*bc
k=[(b+c)^2-a^2]/bc
={[(b^2+c^2-a^2]/2bc}x2+2
=cosAX2+2
=3
2.sin(C-π/6)=sin(π-(A+B)-π/6)
=sin(π-π/3+B-π/6)
=sin(π/2+B)
=cosB
a/sinA=b/sinB
sinB=2/3X√3/2=√3/3
cosB=√6/3
sin(C-π/6)=√6/3

Answer 2

(1)
2cos(B-C)=4sinBsinC-1=2cosBcosC+2sinBsinC
2cosBcosC-2sinBsinC=-1
2cos(B+C)=-1
B+C=120
A=60
(b+c+a)(b+c-a)=(b+c)^2-a^2=b^2+c^2-a^2+2bc=kbc
b^2+c^2-a^2=(k-2)bc=2bccosA
k-2=2cosA=2cos60=1
k=3
(2) cosA=(b^2+c^2-a^2)/2bc
=(c^2-5)/4c=1/2
2c=c^2-5
c=1+根号6
a/sinA=2R=2*根号3=c/sinC
sinC=..............