高中数学问题！求大神！在线等

Question

已知函数f（x）=√3/2sin2-cos^2x-1/2，x属于R.①求函数f（x）的最小值和最小正周期。②设△ABC的内角A，B，C的对边分别为a，b，c且c=√3，f（c）=0，若sinB=2sinA，求a，b的值

Answer 1

（1）原式=f(x)=√3/2 *sin2x-1/2 *cos2x-1=sin(2x-pi/6)-1
所以最小值为0，最小正周期为pi。

（2）由f(C)=0，得C＝pi/3.
由sinB=2sinA,得sin(2pi/3-A)=2sinA,化简得√3/2 *cosA=3/2 *sinA
√3*sin(A-pi/6)=0，得A=pi/6，所以B＝pi/2.
所以b^2=a^2+c^2。且a=b/2。计算得a=1,b=2.

Answer 2

解：f(x)=2sinx/2cosx/2 √3cosx
=sin(x/2 x/2) √3cosx
=sinx √3cosx
=√(1^2 √3^2)sin(x π/3)
=2sin(x π/3)
函数f(x)的最小正周期 T=2π/1=2π
f(x)的值域 [-2,2] 即最小值是-2.

Answer 3

f(x)=√3/2sin2x-(1+cos2x)/2-1/2
=√3/2sin2x-1/2cos2x-1
=sin2xcosπ/6-cos2xsinπ/6-1
=sin(2x-π/6)-1
1.最小值=-1-1=-2, 最小正周期=2π/2=π
2.f(C)=sin(2C-π/6)-1=0, sin(2C-π/6)=1, 2C-π/6=π/2, C=π/3
A+B=π-C=π-π/3=2π/3, B=2π/3-A
sinB=sin(2π/3-A)=sin2π/3cosA-cos2π/3sinA=√3/2cosA+1/2sinA=2sinA
√3cosA=3sinA, tanA=√3/3, A=π/6
B=2π/3-π/6=π/2
c/sinC=a/sinA=b/sinB, √3/(√3/2)=a/(1/2)=b/1
a=1, b=2本回答被提问者采纳

Answer 4

f(x)
=2sinx/2cosx/2 √3cosx
=sin(x/2 x/2) √3cosx
=sinx √3cosx
=√(1^2 √3^2)sin(x π/3)
=2sin(x π/3)
函数f(x)的最小正周期 T=2π/1=2π
f(x)的值域 [-2,2]