高中4.4 诱导公式与旋转免费一课一练

第一章　三角函数

4.3　诱导公式与对称

4.4　诱导公式与旋转

基础过关练

题组一　知角求值

1.(2020首师大附中高一上期末)sin-=(　　)

A.- B.- C. D.

2.cos 960°=(　　)

A. B. C.- D.-

3.(2020安徽马鞍山二中月考)sin2150°+sin2315°+2sin 210°+cos2225°的值是(　　)

A. B. C. D.

4.的值是　　　　. 

5.计算下列各式的值:

(1)cos+cos +cos+cos;

(2)sin 420°cos 330°+sin(-690°)cos(-660°).

题组二　给值求值

6.(2020辽宁大连第二十四中学期末)若sin(π+α)+cos+α=-m,则cos-α+2sin(6π-α)的值为(　　)

A.-m B.-m C.m D.m

7.(2020湖南长沙高一上期末联考)已知cosα+=,则sinα-=　　　　. 

8.(2020黑龙江东部四校期末联考)若sin-α=,则cos+α=　　　　. 

9.(2020湖北黄冈中学高二下期末)已知sinα-=-,则cos(3π+α)的值为　　　　. 

10.(2020四川成都石室中学高一上期末)已知sin-α=,求cos+α·sin+α的值.

题组三　化简求值

11.(2020安徽合肥第一中学高一上期末)已知角α的顶点在坐标原点,始边与x轴的非负半轴重合,终边经过点P(3,-4).

(1)求sin α-cos α的值;

(2)求的值.

12.(2019华南师范大学附属中学高二下学期期末)化简:(k∈Z).

题组四　诱导公式的综合应用

13.(2020山东省实验中学期末)已知θ是三角形的一个内角,且sin-θ=,则角θ等于(　　)

A. B. C.或 D.

14.(2020山西晋城四校联考)已知函数f(x)=asin(πx+α)+bcos(πx+β),且f(-1)=3,则f(2 020)的值为(　　)

A.-1 B.1 C.3 D.-3

15.(2020江西临川一中高二上学期期末)若f(cos x)=cos 2x,则f(sin 15°)=　　　　. 

16.(2020黑龙江哈尔滨三中高一下期中)在△ABC中,sin=sin ,试判断△ABC的形状.

17.(2018江西南昌三校联考)已知角α的终边过点P(1,).

(1)求:sin(π-α)-sin+α的值;

(2)写出角α的集合S.

18.(2020江西师范大学附中高一上期末)已知函数f(x)=,且f(m)=2,试求f(-m)的值.

答案全解全析

4.3　诱导公式与对称

4.4　诱导公式与旋转

基础过关练

1.A 2.C 3.A 6.B 13.A

14.D    

1.A　sin -2π/3 =-sin 2π/3=-√3/2.故选A.

2.C　cos 960°=cos(720°+240°)=cos 240°=cos(180°+60°)=-cos 60°=-1/2.故选C.

3.A　原式=sin230°+sin245°-2sin 30°+cos245°=1/4+1/2-1+1/2=1/4.

4.答案　√2-2

解析　原式=(cos(360° +225° )/(sin(360° +135°)- sin(210° +360°))

=cos225"°" /(sin135"°-" sin210"°" )

=(cos"(" 180"°" +45"°)" )/(sin"(" 180"°-" 45"°)-" sin"(" 180"°" +30"°)" )

=("-" cos45"°" )/(sin45"°" +sin30"°" )=("-"  √2/2)/(√2/2+1/2)=√2-2.

5.解析　(1)cosπ/5+cos 2π/5+cos3π/5+cos4π/5=cosπ/5+cos2π/5+cos π-2π/5 +cos π-π/5 =cosπ/5+cos2π/5-cos2π/5-cosπ/5=0.

(2)原式=sin(360°+60°)cos(360°-30°)+sin(-2×360°+30°)•cos(-2×360°+60°)

=sin 60°cos 30°+sin 30°cos 60°

=√3/2×√3/2+1/2×1/2=1.

6.B　∵sin(π+α)+cos π/2+α =-m,

即-sin α-sin α=-2sin α=-m,

∴sin α=m/2,

∴cos 3π/2-α +2sin(6π-α)

=-sin α-2sin α=-3sin α=-3m/2.

7.答案　-1/3

解析　∵cos α+π/6 =1/3,

∴sin α-π/3 =sin  α+π/6 -π/2

=-cos α+π/6 =-1/3.

8.答案　1/3

解析　∵sin π/6-α =1/3,

∴cos π/3+α =cos π/2- π/6-α  =sin π/6-α =1/3 .

9.答案　3/5

解析　sin α-7π/2 =-sin 7π/2-α

=-sin -π/2-α

=sin π/2+α =cos α=-3/5,

∴cos(3π+α)=cos(π+α)=-cos α=3/5.

10.解析　cos π/6+α •sin 2π/3+α

=cos π/2- π/3-α  •sin π- π/3-α 

=sin π/3-α •sin π/3-α

=1/2×1/2=1/4.

11.解析　(1)由已知可得|OP|=√(3^2+"(-" 4")" ^2 )=5,

根据三角函数的定义知sin α=-4/5,cos α=3/5,所以sin α-cos α=-4/5-3/5=-7/5.

(2)根据诱导公式知(sin"(" π+α")" +cos(π/2+α))/(cos"(" 2π+α")" +sin"(-" α")" )=("-" sinα"-" sinα)/(cosα"-" sinα)=

("-" 2sinα)/(cosα"-" sinα)=("-" 2×("-"  4/5))/(3/5 "-" ("-"  4/5) )=(8/5)/(7/5)=8/7.

12.解析　当k=2n(n∈Z)时,

原式=

(sin"(" 2nπ"-" α")" cos"[(" 2n"-" 1")" π"-" α"]" )/(sin"[(" 2n+1")" π+α"]" cos"(" 2nπ+α")" )

=(sin"(-" α")•" cos"(-" π"-" α")" )/(sin"(" π+α")•" cosα)=("-" sinα"(-" cosα")" )/("-" sinα"•" cosα)=-1;

当k=2n+1(n∈Z)时,

原式

=(sin"[(" 2n+1")" π"-" α"]•" cos"[(" 2n+1"-" 1")" π"-" α"]" )/(sin"[(" 2n+1+1")" π+α"]•" cos"[(" 2n+1")" π+α"]" )

=(sin"(" π"-" α")•" cosα)/(sinα"•" cos"(" π+α")" )=(sinα"•" cosα)/(sinα"(-" cosα")" )=-1.

综上,原式=-1.

13.A　由已知得sin π/2-θ =cos θ=√2/2,

∵θ是三角形的一个内角,∴θ=π/4.

14.D　∵f(x)=asin(πx+α)+bcos(πx+β),且f(-1)=3,

∴asin(-π+α)+bcos(-π+β)=-asin α-bcos β=3,∴asin α+bcos β=-3.

则f(2 020)=asin(2 020π+α)+b•cos(2 020π+β)=asin α+bcos β=-3,

故选D.

15.答案　-√3/2

解析　f(sin 15°)=f(cos 75°)=cos 150°=cos(180°-30°)=-cos 30°=-√3/2.

16.解析　∵A+B+C=π,

∴A+B-C=π-2C,A-B+C=π-2B.

∵sin (A+B"-" C)/2=sin (A"-" B+C)/2,

∴sin (π"-" 2C)/2=sin (π"-" 2B)/2,

∴sin π/2-C =sin π/2-B ,

∴cos C=cos B,

∴B=C.

∴△ABC为等腰三角形.

17.解析　∵α的终边过点P(1,√3),∴x=1,y=√3,r=√(1+3)=2,

∴sin α=y/r=√3/2,cos α=x/r=1/2.

(1)sin(π-α)-sin π/2+α =sin α-cos α=(√3 "-" 1)/2.

(2)若α∈[0,2π],由sin α=√3/2,cos α=1/2,得α=π/3,故角α的集合S= α α=2kπ+π/3,k∈Z .

18.解析　易得f(x)的定义域关于原点对称.

因为f(x)=

(6cos"(" π+x")" +5sin^2 "(" π"-" x")-" 4)/(cos"(" 2π"-" x")" )

=("-" 6cosx+5sin^2 x"-" 4)/cosx,

所以f(-x)=("-" 6cos"(-" x")" +5sin^2 "(-" x")-" 4)/(cos"(-" x")" )

=("-" 6cosx+5sin^2 x"-" 4)/cosx=f(x),

所以f(x)是偶函数,

所以f(-m)=f(m)=2.