Question

9.设函数f(x)=cos(2x+π/3)+sinx^2 (1)求函数f(x)的最小正周期

（2）设△ABC的三个内角A、B、C的对应边分别是a、b、c，若c=根号6，cosB=1/3，f(C/2)=-1/4,求b.

Answer 1

解： f(x)=cos2xcosπ/3-sin2xsinπ/3+(1-cos2x)/2.
=(1/2)cos2x-(√3/2)sin2x+1/2-(1/2)cos2x.
=-√3/2sin2x+1/2.
(1).∴T=2π/2=π.

(2) f(C/2)=-(√3/2)sin(2*C/2)+1/2=-1/4.
=-(√3/2)sinC+1/2=-1/4.
-(√3/2)sinC=-3/4.
sinC=√3/2.
∴∠C=60°, 或∠C=120°.
sinB=√(1-cos^2B)=2√2/3.
由正弦定理得：
b/sinB=c/sinC.
b=csinB/sinC.
=(√6*2√2/3)/(√3/2.).
∴b =8/3.

Answer 2

π