1、第7章三角函数7.2三角函数概念7.2.3三角函数的诱导公式第1课时诱导公式一、二、三、四课后篇巩固提升必备知识基础练1.tan 300+sin 450的值是()A.-1+3B.1+3C.-1-3D.1-3答案D解析原式=tan(360-60)+sin(360+90)=tan(-60)+sin90=-tan60+1=-3+1.2.已知sin(-)=13,则sin(-2 021)的值为()A.223B.-223C.13D.-13答案D解析sin(-2021)=sin(-)-2020=sin(-)=sin-(-)=-sin(-)=-13.3.已知角的终边与单位圆的交点为P-35,45,则cos(+
2、)+sin(-)=()A.-15B.15C.-75D.75答案A解析因为角的终边与单位圆的交点为P-35,45,所以cos=-35,sin=45,则cos(+)+sin(-)=-cos-sin=-15.故选A.4.若P(-4,3)是角终边上一点,则cos(-3)tan(-2)sin2(-)的值为.答案-53解析由题意知sin=35,原式=(-cos)tansin2=-sinsin2=-1sin=-53.5.若cos(+)=-12,322,则sin(-2)=.答案-32解析由cos(+)=-12,得cos=12.又322,故sin(-2)=sin=-1-cos2=-1-(12)2=-32.6.化
3、简下列各式:(1)sin-193cos76;(2)sin(-960)cos 1 470-cos(-240)sin(-210).解(1)sin-193cos76=-sin6+3cos+6=sin3cos6=34.(2)sin(-960)cos1470-cos(-240)sin(-210)=-sin(180+60+2360)cos(30+4360)+cos(180+60)sin(180+30)=sin60cos30+cos60sin30=1.7.已知cos(+)=-12,且是第四象限角,计算:(1)sin(2-);(2)sin+(2n+1)+sin-(2n+1)sin(+2n)cos(-2n)(n
4、Z).解(1)由cos(+)=-12可得cos=12,而sin(2-)=-sin,因为是第四象限角,所以sin=-32,故sin(2-)=32.(2)sin+(2n+1)+sin-(2n+1)sin(+2n)cos(-2n)=(-sin)+(-sin)sincos=-2cos,由(1)得cos=12,所以sin+(2n+1)+sin-(2n+1)sin(+2n)cos(-2n)=-4.关键能力提升练8.(2020江西抚州期末)sin21200等于()A.12B.32C.-32D.32答案D解析sin21200=sin2(3603+120)=sin2120=(32)2=32.9.若sin(-11
5、0)=a,则tan 70=()A.a1-a2B.-a1-a2C.a1+a2D.-a1+a2答案B解析sin(-110)=-sin110=-sin(180-70)=-sin70=a,sin70=-a,cos70=1-(-a)2=1-a2,tan70=sin70cos70=-a1-a2.10.已知tan3-=13,则tan23+=()A.13B.-13C.233D.-233答案B解析因为tan23+=tan-3-=-tan3-,所以tan23+=-13.11.已知函数f(x)=asin(x+)+bcos(x+),且f(4)=3,则f(2 019)的值为()A.-1B.1C.3D.-3答案D解析f(
6、x)=asin(x+)+bcos(x+),f(4)=asin(4+)+bcos(4+)=asin+bcos=3,f(2019)=asin(2019+)+bcos(2019+)=asin(+)+bcos(+)=-asin-bcos=-f(4)=-3.故选D.12.化简1+2sin(-2)cos(-2)的结果是()A.sin 2-cos 2B.|cos 2+sin 2|C.(cos 2-sin 2)D.无法确定答案A解析原式=|sin(-2)+cos(-2)|=|sin2-cos2|=sin2-cos2.13.(多选)已知f(x)=sin x,下列式子中不成立的是()A.f(x+)=sin xB.
7、f(2-x)=sin xC.f(-x)=-sin xD.f(-x)=-f(x)答案ABD解析f(x+)=sin(x+)=-sinx,f(2-x)=sin(2-x)=-sinx,f(-x)=sin(-x)=-sinx,f(-x)=sin(-x)=sinx=f(x).故A,B,D不成立.14.(多选)有下列三角函数式,其中nZ,则函数值与sin3的值相同的是()A.sin2n+34B.cos2n-6C.sin2n+3D.cos(2n+1)-6答案BC解析sin2n+34=sin34sin3,故A错误;cos2n-6=cos6=sin3,故B正确;sin2n+3=sin3,故C正确;cos(2n+1
8、)-6=cos-6=-cos6sin3,故D错误.15.cos(-585)sin495+sin(-570)的值是.答案2-2解析原式=cos(360+225)sin(360+135)-sin(360+210)=cos225sin135-sin210=cos(180+45)sin(180-45)-sin(180+30)=-cos45sin45+sin30=-2222+12=2-2.16.已知f(x)=|sin(3+x)|+4cos 2x+2,则f(x)为函数(选填“奇”或“偶”);f54=.答案偶22解析f(x)=|sin(3+x)|+4cos2x+2=|sinx|+4cos(2x+)=|sin
9、x|-4cos2x,则f(-x)=|sin(-x)|-4cos2(-x)=|sinx|-4cos2x=f(x),f(x)为偶函数.f54=sin54-4cos52=sin4-4cos2=22.17.已知f(x)=cos2(n+x)sin2(n-x)cos2(2n+1)-x(nZ).(1)化简f(x)的表达式;(2)求f20203.解(1)当n为偶数,即n=2k(kZ)时,f(x)=cos2(2k+x)sin2(2k-x)cos2(22k+1)-x=cos2xsin2(-x)cos2(-x)=cos2x(-sinx)2(-cosx)2=sin2x;当n为奇数,即n=2k+1(kZ)时,f(x)=
10、cos2(2k+1)+xsin2(2k+1)-xcos22(2k+1)+1-x=cos2(+x)sin2(-x)cos2(-x)=(-cosx)2sin2x(-cosx)2=sin2x,综上得f(x)=sin2x.(2)由(1)知f20203=sin220203=sin2672+43=sin243=sin2+3=sn23=34.学科素养拔高练18.(多选)(2021江苏连云港赣榆中学月考)在ABC中,给出下列四个式子,其中为常数的是()A.sin(A+B)+sin CB.cos(A+B)+cos CC.sin(2A+2B)+sin 2CD.cos(2A+2B)+cos 2C答案BC解析sin(
11、A+B)+sinC=2sinC,故A错误;cos(A+B)+cosC=-cosC+cosC=0,故B正确;sin(2A+2B)+sin2C=sin2(A+B)+sin2C=sin2(-C)+sin2C=sin(2-2C)+sin2C=-sin2C+sin2C=0,故C正确;cos(2A+2B)+cos2C=cos2(A+B)+cos2C=cos2(-C)+cos2C=cos(2-2C)+cos2C=cos2C+cos2C=2cos2C,故D错误.19.设f(x)=sin6x,则f(1)+f(2)+f(3)+f(13)=.答案12解析f(x)=sin6x,当x=1时,f(1)=sin6=12;当
12、x=2时,f(2)=sin3=32;当x=3时,f(3)=sin2=1;当x=4时,f(4)=sin23=32;当x=5时,f(5)=sin56=12;当x=6时,f(6)=sin=0;当x=7时,f(7)=sin76=-12;当x=8时,f(8)=sin43=-32;当x=9时,f(9)=sin32=-1;当x=10时,f(10)=sin53=-32;当x=11时,f(11)=sin116=-12;当x=12时,f(12)=sin2=0;当x=13时,f(13)=sin136=12,则f(1)+f(2)+f(3)+f(13)=12+32+1+32+12+0-12-32-1-32-12+0+12=12.
