4.3　单位圆与正弦函数、余弦函数的基本性质

4.4　单位圆的对称性与诱导公式

A组　基础巩固

1.若函数y=2sin x+a的最大值为-2,则a的值等于0(　　)

A.2 B.-2

C.0 D.-4

解析由已知得2+a=-2,所以a=-4.

答案D

2.化简(sin"(" π+α")" cos"(" 2π"-" α")" )/cos(π/2+α) 所得的结果是(　　)

A.sin α B.-sin α

C.cos α D.-cos α

解析原式=("-" sinαcosα)/("-" sinα)=cos α.

答案C

3.已知sin(α"-" π/12)=1/3,则cos(α+17π/12)的值等于(　　)

A.1/3 B.(2√2)/3

C.-1/3 D.-(2√2)/3

解析由sin(α"-" π/12)=1/3,

　　则cos(α+17π/12)=cos(α+3π/2 "-" π/12)

　　=sin(α"-" π/12)=1/3.故选A.

答案A

4.下列四个等式:

①sin(360°+300°)=sin 300°;

②cos(180°-300°)=cos 300°;

③sin(180°+300°)=-sin 300°;

④cos(±300°)=cos 300°.

其中正确的有(　　)

A.1个 B.2个

C.3个 D.4个

解析①③④均正确,②中cos(180°-300°)=-cos 300°.

答案C

5.设函数f(x)(x∈R)满足f(x+π)=f(x)+sin x.当0≤x<π时,f(x)=0,则f23π/6=(　　)

A.1/2 B.√3/2

C.0 D.-1/2

解析反复利用f(x+π)=f(x)+sin x,将f23π/6进行转化,再利用诱导公式求值.f23π/6=f17π/6+sin 17π/6=f11π/6+sin 17π/6+sin 11π/6=f5π/6+sin 17π/6+sin 11π/6+sin 5π/6=2sin 5π/6+sin-π/6=1/2.

答案A

6.已知sin(π/2 "-" α)=2/3,则cos(π+α)=　　　　.

解析cos(π+α)=-cos α=-sin(π/2 "-" α)=-2/3.