Question

在⊿ABC中,角A、B、C的对边长分别为a、b、c，且A、B、C(1)若向量BA*BC=3/2,b=√3，求a+c的值成等差数列

在⊿ABC中,角A、B、C的对边长分别为a、b、c，且A、B、C(1)若向量BA*BC=3/2,b=√3，求a+c的值成等差数列(2)求2sinA-sinC的取值范围。

Answer 1

1
A、B、C成等差数列
即：2B=A+C
即：3B=π
即：B=π/3
BA·BC=|BA||BC|*cosB
=accos(π/3)
=ac/2=3/2
即：ac=3
b^2=a^2+c^2-2accosB
=a^2+c^2-ac=3
即：(a+c)^2=3+3ac=12
故：a+c=2√3
2
A、B、C成等差数列
1) 如果：A<B<C
故设A=π/3-t，C=π/3+t，0<t<π/3
故：2sinA-sinC
=2sin(π/3-t)-sin(π/3+t)
=2sin(π/3)cost-2cos(π/3)sint-sin(π/3)cost-cos(π/3)sint
=sin(π/3)cost-3cos(π/3)sint
=√3cost/2-3sint/2
=√3cos(t+π/3)
π/3<t+π/3<2π/3
故：cos(t+π/3)∈(-1/2,1/2)
故：2sinA-sinC∈(-√3/2,√3/2)
2) 如果：A>B>C
故设A=π/3+t，C=π/3-t，0<t<π/3
故：2sinA-sinC
=2sin(π/3+t)-sin(π/3-t)
=2sin(π/3)cost+2cos(π/3)sint-sin(π/3)cost+cos(π/3)sint
=sin(π/3)cost+3cos(π/3)sint
=√3cost/2+3sint/2
=√3sin(t+π/6)
π/6<t+π/6<π/2
故：sin(t+π/3)∈(1/2,1)
故：2sinA-sinC∈(√3/2,√3)