Question

已知函数f（x）=√3/2sin2x-cos²x-1/2,x∈R

已知函数f（x）=√3/2sin2x-cos²x-1/2,x∈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2sinA=sinB,求a,b的值

Answer 1

解：先用降幂公式把函数化为：f(x)=√3/2sin2x-1/2cos2x-1=sin(2x-π/6)-1
（1）最小值为-2，最小正周期为π
（2）由f(C)=0知sin(2C-π/6)=1，从而可得C=π/3，再由余弦定理知：c^2=a^2+b^2-2abcosC
3=a^2+4a^2-2a*2acosπ/3，解得a=1，
a/sinA=b/sinB,故b=asinB/sinA=2a=2

Answer 2

1. f（x）=√3/2sin2x-cos²x-1/2
=√3/2sin2x-1/2cos2x-1
=sin(2x-π/6)-1
fmin=-2 最小正周期T=2π/2=π
2. f(c)=sin(2C-π/6)-1=0
2C-π/6=π/2 C=π/3
2sinA=sinB
正弦定理
2a=b
余弦定理
cosC=(a^2+b^2-c^2)/2ab=(a^2+4a^2-3)/4a^2=1/2
a=1
b=2a=2

Answer 3

1. f（x）=√3/2sin2x-cos²x-1/2
=√3/2sin2x-1/2cos2x-1
=sin(2x-π/6)-1
fmin=-2 最小正周期T=2π/2=π
2. f(c)=sin(2C-π/6)-1=0
2C-π/6=π/2 C=π/3
2sinA=sinB
正弦定理
2a=b
余弦定理
cosC=(a^2+b^2-c^2)/2ab=(a^2+4a^2-3)/4a^2=1/2
a=1
b=2a=2