Question

已知函数f(x)=根号3sinxcosx-1/2cos2x-1 1.求函数f(x)的最小值

2.设△ABC的内角A,B,C所对的边分别是a,b,c，且c=根号7，f(C)=0，sinB=3sinA求△ABC面积

Answer 1

f(x)=√3sinxcosx-1/2cos2x-1
=(√3/2)sin2x-(1/2)cos2x-1
=cosπ/6sin2x-sinπ/6cos2x-1
=sin(2x-π/6)-1
1.-1≤sin(2x-π/6)≤1
f(x)min=-1-1=-2
2.f(C)=sin(2C-π/6)-1=0
2C-π/6=π/2
C=π/3
在△ABC中
b/sinB=a/sinA
b=asinB/sinA=3a
cosC=(a²+b²-c²)/2ab
=(10a²-7)/6a²
=cosπ/3=1/2
a=1(a>0)
b=3
面积 S=(1/2)absinC=(3/2)(√3/2)=3√3/4

Answer 2

f(x)=(√3/2)sin2x－(1/2)cos2x-1=sin(2x－π/6)-1

故函数的最小值是:-1-1=-2
(2)f(C)=sin(2C-Pai/6)-1=0,即2C-Pai/6=Pai/2,C=Pai/3.
∵sinB=3sinA
由正弦定理可知
b=3a
由余弦定理可知
cosC=(a^2+b^2-c^2)/2ab
=(10a^2-7)/6a^2
=cosπ/3=1/2
∴10a^2-7=3a^2
a^2=1
a=1
∴b=3a=3
S=1/2absinC=1/2*1*3*根号3/2=3根号3/4