Question

已知函数:f(x)=根号下3sinx×cosx–cos²x–½ (x∈R) 1，

已知函数:f(x)=根号下3sinx×cosx–cos²x–½ (x∈R) 1，求函数f(x)最大值和最小正周期 2，设△ABC的内角A,B,C的对边分别是a,b,c且c=3,f(C)=0若，sin(A+C)=2sinA求a,b的值

Answer 1

f(x)=根号下3sinx×cosx–cos²x–½
=（根3）/2*sin2x-1/2*(cos2x+1)-1/2
=（根3）/2*sin2x-1/2*cos2x-1
=sin(2x-pi/6)-1
<=1-1
=0
最大值是0，最小正周期为pi.

f(C)=sin(2C-pi/6)-1=0
sin(2C-pi/6)=1
2C-pi/6=pi/2
C=pi/3

sin(A+C)=2sinA
sinB=2sinA
b=2a
a^2+b^2-2abcosC=c^2
a^2+4a^2-4a^2*cospi/3=3^2
a^2*(5-2)=9
a^2=3
a=根3
b=2根3

Answer 2

1.
f(x)=√3sinxcosx-cos²x -1/2
=(√3/2)sin(2x)-[1+cos(2x)]/2 -1/2
=(√3/2)sin(2x)-(1/2)cos(2x) -1
=sin(2x-π/6) -1
当sin(2x-π/6)=1时，f(x)有最大值[f(x)]max=1-1=0
最小正周期Tmin=2π/2=π
2.
f(C)=0
sin(2C-π/6) -1=0
sin(2C-π/6)=1
0<C<π -π/6<2C-π/6<11π/6
2C-π/6=π/2
C=π/3
sin(A+C)=2sinA
sinB=2sinA
由正弦定理得b=2a
由余弦定理得
c²=a²+b²-2abcosC
a²+(2a)²-2a(2a)cos(π/3)=c²
c=3 cos(π/3)=1/2代入，整理，得
3a²=9
a²=3
a=√3 b=2a=2√3