北师版高中数学必修第二册课后习题第1章 §4 4.3 诱导公式与对称--4.4 诱导公式与旋转.docVIP

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4.3诱导公式与对称

4.4诱导公式与旋转

课后训练巩固提升

1.(多选题)已知x∈R,则下列等式恒成立的是().

A.sin(3π-x)=sinx

B.sinπ-x

C.cos5π2

D.cos3π2

2.已知sin(θ+π)0,cos(θ-π)0,则θ是().

A.第一象限角 B.第二象限角

C.第三象限角 D.第四象限角

3.若sin(9π+α)=-12,则cos7π

A.-12 B.1

C.32 D.-

4.已知sin5π4+α=

A.12 B.-1

C.32 D.-

5.化简sin(

A.sinα B.-sinα

C.cosα D.-cosα

6.已知cosπ12-θ=13

A.13 B.2

C.-13 D.-

7.化简cos(

A.1 B.-1

C.sinαcosα D.-

8.(多选题)已知角α,β,γ,满足α+β+γ=π,则下列结论正确的是().

A.sin(α+β)=sinγ

B.cos(β+γ)=cosα

C.sinα+γ2=sin

D.cosα+β2=sin

9.cos1°+cos2°+cos3°+…+cos360°=().

A.0 B.2

C.-2 D.1

10.已知sinx+π6=14,则sin5π6-x+cos

11.下列三角函数:①sinnπ+4π3;②cos2nπ+π6;③sin2nπ+π3;④cos[(2n+1)π-π6];⑤sin[(2n+1)π-π3],n∈

12.已知cosπ6-θ=a(|a|≤1),则cos(5π6+θ)+sin

13.设f(θ)=2cos

求fπ3

14.已知α是第四象限角,且f(α)=sin(

(1)若cosα-

(2)若α=-1860°,求f(α)的值.

答案：

1.ABsin(3π-x)=sin(π-x)=sinx,sinπ-x2=sinπ2-x2=cosx

2.B由sin(θ+π)=-sinθ0?sinθ0,cos(θ-π)=-cosθ0?cosθ0,由sinθ0,

3.A∵sin(9π+α)=sin(π+α)=-sinα=-12

∴sinα=12

∴cos7π2-α=cos3π

4.Dsin3π4-α=sinπ-3π4-α=sin(

5.C原式=-sinαcosα

6.A∵cosπ12-θ=13,∴sin5π12

7.B原式=-cosαsinαsinα

8.AD因为α+β+γ=π,所以sin(α+β)=sin(π-γ)=sinγ,cos(γ+β)=cos(π-α)=-cosα,

α+β+γ2=π2,sinα+γ2=sinπ2-β2

9.A利用诱导公式:cos(180°+α)=-cosα,可得cos1°+cos2°+cos3°+…+cos360°=(cos1°+cos2°+cos3°+…+cos180°)+(cos181°+cos182°+cos183°+…+cos360°)=(cos1°+cos2°+cos3°+…+cos180°)-(cos1°+cos2°+cos3°+…+cos180°)=0.

10.516∵sinx+π6=14,∴sin5π6-x+cos2π3-x=sin[π-(x+π6)]+cos2

11.②③⑤①sinnπ+

②cos2nπ+π6=cosπ6

③sin2nπ+π3=sin

④cos(2n+1)π-π6=cos2nπ+π-π6=cos

⑤sin(2n+1)π-π3=sin2nπ+π-π3=sin(π-

12.0cos5π6+θ

=cos[π-(π6-θ)]+sin

=-cosπ6-θ+cos(π

13.解因为f(θ)=2cos3

=2co

=cosθ(

所以fπ3=cosπ3=cos673π+π3=-cos

14.解f(α)=sin(

(1)∵cosα-3π2=1

∴cosπ2+α=15

∴f(α)=1sinα

(2)当α=-1860°时,f(α)=1sinα=1