Question

已知函数fx)=根号3/2 Sin2x -Cos^2 x -1/2

求函数最小正周期。设三角形ABC 内角对边分别为abc ，且c =根号3，f(C)=0，Sin B =2Sin A ，求a，b

Answer 1

f(x)=√3/2sin2x-(1+cos2x)/2-1/2
=√3/2sin2x-1/2cos2x-1
=sin(2x-π/6)-1
则最小正周期T=2π/2=π
f(c)=0,得sin(2c-π/6)=1,即2c-π/6=π/2,得：c=π/3
sinB/b=sinA/a
由sinB=2sinA,得：2sinA/b=sinA/a,得：b=2a
由余弦定理：c²=a²+b²-2abcosC
即3=a²+4a²-2a²
得a=1
故b=2