Question

已知函数f(x)=二分之根号三Sin2x-cosx的平方-1/2,x属于R

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

【参考答案】

f(C)=(√3 /2)sin2C-cosC^2-(1/2)
=(√3/2)sin2C-[(cos2C+1)/2]-(1/2)
=sin2Ccos(π/6)-sin(π/6)cos2C-1
=sin(2C- π/6)-1
=0
则 sin(2C- π/6)=1, 即C=π/3
由sinB=2sinA得
b/sinB=a/sinA
b/(2sinA)=a/sinA
b/2=a
b=2a
cosC=-1/2=[a^2 +(4a^2)-3]/(2a×2a)
1/2=(5a^2 -3)/(4a^2)
5a^2 -3=2a^2
3a^2=3
a=1,于是 b=2
不理解之处欢迎追问

Answer 2

你好！
先化简：f(x)=√3/2sin2x-cos²x-1/2=√3/2sin2x-（1+cos2x）/2-1/2=√3/2sin2x-1/2cos2x-1=sin(2x-π/6)-1,∵f（C）=0,由于C∈（0，π）解得C=π/3，根据正弦定理得sinB=2sinA，可得b=2a，根据余弦定理得cosC=a²+b²-c²/2ab，即cosπ/3=1/2=a²+（2a）²-（√3）²/2×2a×a，解得a=1，b=2a=2