2020-2021学年高中数学人教B版（2019）必修第三册课后习题：7-2-4　诱导公式 WORD版含解析.docx

1、7.2.4诱导公式课后篇巩固提升基础达标练1.(多选)(2020山东日照高一检测)下列函数值中符号为负的是()A.sin (-1 000)B.cos103C.tan 2D.sin 5解析-1000=-3360+80,-1000是第一象限角,sin(-1000)0;103=2+43,103是第三象限角,cos1030;22,2rad是第二象限角,tan20;3252,5rad是第四象限角,sin5acB.abcC.bcaD.acb解析由题可得,a=-33,b=22,c=-22,所以bac.答案A3.(多选)化简1+2sin(-2)cos(-2)的结果是()A.sin 2-cos 2B.|cos

2、2-sin 2|C.(cos 2-sin 2)D.无法确定解析原式=1-2sin2cos2=(sin2-cos2)2=|sin2-cos2|=sin2-cos2.答案AB4.已知tan-6=12,且0,2,则cos23-=()A.-55B.55C.-255D.255解析由题意知,tan(-6)=sin(-6)cos(-6)=12,sin2(-6)+cos2(-6)=1,解得sin-6=55.由tan-6=120,且0,2,可得0-63,则sin-6=55,故cos23-=cos2-6=sin-6=55.答案B5.设tan(5+)=m,则sin(-3)+cos(-)sin(-)-cos(+)的值

3、为.解析由题意知tan=m,原式=-sin-cos-sin+cos=-tan-1-tan+1=m+1m-1.答案m+1m-16.化简:(1)1+cos2+sin2-tan(+);(2)sin(2-)cos(+)cos2+cos112-cos(-)sin(3-)sin(-)sin92+.解(1)原式=1+(-sin)costan=1-sin2=cos2.(2)原式=(-sin)(-cos)(-sin)cos5+2-(-cos)sin(-)-sin(+)sin4+2+=-sin2cos-cos2-(-cos)sin-(-sin)sin2+=sin2cossin-cossin2cos=-sincos

4、=-tan.7.已知sin(5+)=lg1310,求cos(2+)的值.解sin(5+)=sin(+)=-sin,lg1310=lg10-13=-13,sin =13,cos(2+)=cos =1-sin2=1-132=223.能力提升练1.记cos(-80)=k,则tan 100等于()A.1-k2kB.-1-k2kC.k1-k2D.-k1-k2解析cos(-80)=cos80=k,sin80=1-cos280=1-k2,tan100=-tan80=-1-k2k.故选B.答案B2.(多选)已知A=sin(k+)sin+cos(k+)cos(kZ),则A的值是()A.-1B.-2C.1D.2解

5、析当k为偶数时,A=sinsin+coscos=2;当k为奇数时,A=-sinsin-coscos=-2.故选BD.答案BD3.已知为锐角,2tan(-)-3cos2+5=0,tan(+)+6sin(+)-1=0,则sin 的值是()A.355B.377C.31010D.13解析由题可知-2tan+3sin+5=0,tan-6sin-1=0,所以tan=3.又tan=sincos,且sin2+cos2=1,所以9=sin2cos2=sin21-sin2.所以sin2=910.因为为锐角,所以sin=31010.答案C4.已知sin(2+)tan(+)tan(3-)cos2-tan(-)=1,则

6、3sin2+3sincos+2cos2的值是()A.1B.2C.3D.6解析因为原式=sintan(-tan)-sintan=tan=1,所以3sin2+3cos2sin2+3sincos+2cos2=3tan2+3tan2+3tan+2=1.答案A5.已知tan(-)=3,则sin(2+)-cos(-)sin(2-)-sin(-)=()A.-1B.-12C.1D.12解析由tan(-)=3,得-tan=3,即tan=-3,则sin(2+)-cos(-)sin(2-)-sin(-)=cos+coscos-sin=2coscos-sin=21-tan=12.答案D6.在ABC中,3sin2-A=

7、3sin(-A),且cos A=-3cos(-B),则C=.解析由题意得3cosA=3sinA,cosA=3cosB,由得tanA=33,所以A=6.由得cosB=cos63=12,所以B=3,则C=2.答案27.已知是第三象限的角,且f()=sin(-)cos(2-)tan-+32cot(-)sin(-).(1)化简f();(2)若cos-32=15,求f()的值.解(1)f()=sin(-)cos(2-)tan32-cot(-)sin(-)=sincoscot-cotsin=-cos.(2)cos-32=-sin,sin=-15.又是第三象限的角,cos=-1-sin2=-256,f()=

8、256.素养培优练(1)已知sin 是方程5x2-7x-6=0的根,求cos(+2)cos(4+)tan2(2+)tan(6+)sin(2+)sin(8+)的值.(2)已知sin(4+)=2sin ,3cos(6+)=2cos(2+),且0,0,求和的值.解(1)因为方程5x2-7x-6=0的两根为2和-35,所以sin=-35.由sin2+cos2=1,得cos=1-sin2=45.当cos=45时,tan=-34;当cos=-45时,tan=34.所以原式=coscostan2tansinsin=tan=34.(2)因为sin(4+)=2sin,所以sin=2sin.因为3cos(6+)=2cos(2+),所以3cos=2cos.2+2,得sin2+3cos2=2(sin2+cos2)=2,所以cos2=12,即cos=22.又0,所以=4或=34.又0,当=4时,由得=6;当=34时,由得=56.所以=4,=6或=34,=56.