A.π/8 B.π/4 C.3π/8 D.3π/4

解析由题意f(x)=sin 2x+cos 2x=√2sin(2x+π/4),将其图像向右平移φ个单位长度,得函数y=√2sin[2"(" x"-" φ")" +π/4]=√2sin(2x"-" 2φ+π/4)的图像,要使图像关于y轴对称,则π/4-2φ=π/2+kπ,解得φ=-π/8-kπ/2,当k=-1时,φ取最小正值3π/8.

答案C

6.若cos(A-B)=1/3,则(sin A+sin B)2+(cos A+cos B)2=.

解析原式=sin2A+2sin Asin B+sin2B+cos2A+2cos Acos B+cos2B=2+2cos(A-B)=2+2/3=8/3.

答案8/3

7.若sin α-sin β=1-√3/2,cos α-cos β=1/2,则cos(α-β)的值为　　　　　.

答案√3/2

8.函数f(x)=cos x+cos(x+π/3)的对称轴方程为　　　　　　　　　　　　.

解析y=cos x+cos(x+π/3)

　　=cos x+1/2cos x-√3/2sin x

　　=√3 (√3/2 cosx"-" 1/2 sinx)

　　=√3sin(π/3 "-" x)=-√3sin(x"-" π/3),

　　令x-π/3=kπ+π/2(k∈Z),得x=kπ+5π/6(k∈Z),

　　即对称轴方程是x=kπ+5π/6(k∈Z).

答案x=kπ+5π/6(k∈Z)

9.化简:sin(x+π/3)+2sin(x"-" π/3)-√3cos(2π/3 "-" x).

解原式=sin xcosπ/3+cos xsinπ/3+2sin xcosπ/3-2cos xsinπ/3-√3cos2π/3cos x-√3sin2π/3sin x

　　=(cos π/3+2cos π/3 "-" √3 sin 2π/3)sin x+(sin π/3┤-2sinπ/3-├ √3 cos 2π/3)cos x

　　=(1/2+1"-" √3×√3/2)sin x+(√3/2 "-" √3+√3/2)cos x=0.

10.求函数f(x)=sin 2x+cos 2x的递增区间.

解f(x)=sin 2x+cos 2x

　　=√2 (sin2x"·" √2/2+cos2x"·" √2/2)

　　=√2sin(2x+π/4).

　　令2kπ-π/2≤2x+π/4≤2kπ+π/2(k∈Z),

　　解得kπ-3π/8≤x≤kπ+π/8(k∈Z),

　　故f(x)的递增区间是[kπ"-" 3π/8 "," kπ+π/8](k∈Z).

11.导学号93774092已知cos α-cos β=1/2,sin α-sin β=-1/3,求cos(α-β)的值.

解由已知cos α-cos β=1/2,sin α-sin β=-1/3,两式两边平方并相加,得(cos α-cos β)2+(sin α-sin β)2=1/4+1/9,即2-2cos αcos β-2sin αsin β=13/36,

　　∴cos αcos β+sin αsin β=1/2×(2"-" 13/36)=59/72,

　　∴cos(α-β)=59/72.

B组　能力提升

1.sin(θ+75°)+cos(θ+45°)-√3cos(θ+15°)的值等于0(　　)

A.±1 B.1