C.-1 D.0

解析原式=sin[60°+(θ+15°)]+cos(θ+45°)-√3cos(θ+15°)=-√3/2cos(θ+15°)+1/2sin(θ+15°)+cos(θ+45°)=sin(-60°)cos(θ+15°)+cos(-60°)·sin(θ+15°)+cos(θ+45°)=sin(θ-45°)+cos(θ+45°)=0,故选D.

答案D

2.设函数f(x)=sin(ωx+φ)+cos(ωx+φ)(ω>0",|" φ"|" <π/2)的最小正周期为π,且f(-x)=f(x),则(　　)

A.f(x)在(0"," π/2)内是减少的

B.f(x)在(π/4 "," 3π/4)内是减少的

C.f(x)在(0"," π/2)内是增加的

D.f(x)在(π/4 "," 3π/4)内是增加的

解析f(x)=√2sin(ωx+φ+π/4),由T=2π/ω=π,得ω=2,则f(x)=√2sin(2x+φ+π/4).

　　又f(-x)=f(x),则f(x)是偶函数,

　　因为|φ|<π/2,所以φ+π/4=π/2,

　　所以f(x)=√2cos 2x,易知A正确.

答案A

3.若sin 2α=√5/5,sin(β-α)=√10/10,且α∈[π/4 "," π],β∈[π"," 3π/2],则α+β的值是(　　)

A.7π/4 B.9π/4

C.5π/4 或 7π/4 D.5π/4 或 9π/4

解析∵α∈[π/4 "," π],∴2α∈[π/2 "," 2π].

　　∵sin 2α=√5/5,∴2α∈[π/2 "," π],

　　∴α∈[π/4 "," π/2],∴cos 2α=-(2√5)/5.

　　∵β∈[π"," 3π/2],

　　故β+α∈[5π/4 "," 2π],β-α∈[π/2 "," 5π/4],

　　于是cos(β-α)=-(3√10)/10,

　　∴cos(α+β)=cos[2α+(β-α)]

　　=cos 2αcos(β-α)-sin 2αsin(β-α)

　　=-(2√5)/5×("-" (3√10)/10)-√5/5×√10/10=√2/2.

　　又α+β∈[5π/4 "," 2π],故α+β=7π/4.

答案A

4.已知α为锐角,且cos(α+π/4)=3/5,则sin α=　　　　　.

答案√2/10

5.设cos α=-√5/5,tan β=1/3,π<α<3π/2,0<β<π/2,求α-β的值.

解由cos α=-√5/5,π<α<3π/2,得sin α=-(2√5)/5.

　　由tan β=1/3,0<β<π/2,得sin β=1/√10,cos β=3/√10,

　　所以sin(α-β)=sin αcos β-cos αsin β=("-" (2√5)/5)×3/√10-("-" √5/5)×1/√10=-√2/2.

　　由π<α<3π/2,0<β<π/2,得-π/2<-β<0,π/2<α-β<3π/2,因此,α-β=5π/4.

6.导学号93774093如图,设A是单位圆和x轴正半轴的交点,P,Q是单位圆上的两点,O是坐标原点,且∠AOP=π/6,∠AOQ=α,α∈[0,π).