Question

若函数f(x)=根号3/2sin2x-1/2(cos^2*x-sin^2*x)-1,将函数f(x)向左平移π/6个单位后得到函数g（x）

①若c=根号7，f(C)=0,sinB=3sinA,求a，b的值？
②设△ABC的三个角A、B、C的对边分别为a，b，c。求：若g（B）=0且向量m=（cosA,cosB）,向量n=(1,sinA-cosAtanB),求向量m*向量n的取值范围？

Answer 1

f(X)=√3/2sin2x-1/2cos2x-1=sin(2x-π/6）-1
g(x)=f(x+π/6)=sin(2x+π/6)-1
①f(C)=sin(2C-π/6）-1=0
2C-π/6=π/2
C=π/3
sinB=3sinA=sin(A+C) 5/2sinA=√3/2cosA sinA=√21/14 sinB=3√21/14
有正弦定理得
a=3/4 b=9/4
②g(B)=0 B=π/6
m*n=cosA+√3/2(sinA-√3/3cosA)=√3/2sinA+1/2sinA=sin(A+π/6)∈ (0,1)

Answer 2

(1).f(x)=(√3/2)sin2x-(1/2)cos2x-1=sin(2x-π/6)-1
左移π/6,
g(x)=sin2x-1
由题知f(C)=0
f(C)=sin(2C-π/6）-1=0
2C-π/6=π/2
C=π/3
sinB=3sinA=sin(A+C) 5/2sinA=√3/2cosA sinA=√21/14 sinB=3√21/14
有正弦定理得：a=3/4 b=9/4

(2).由（1）知g(x)=sin2x-1且由题知g(B)=0
B=π/4 tanB=1 cosB=√2/2
m*n=cosA+cosB*(sinA-cosA)
=cosBsinA+(1-cosB)cosA
=√[(cosB)^2+(1-cosB)^2] sin(A+u) cosu=cosB/√[(cosB)^2+(1-cosB)^2]=√2/2√(2-√2)
sinu=(1-cosB)/√[(cosB)^2+(1-cosB)^2]=(2-√2)/2√(2-√2)
=√[(1/2)+1+1/2-√2]sin(A+u)
=√(2-√2)sin(A-u)
A>0 sin(A-u)>sin(-u)=(√2-2)/[2√(2-√2)]
(2√2-3)/√(2-√2) < m*n<=√(2-√2)