2022-2023学年苏教版必修第一册 7.2.3　第1课时　诱导公式一、二、三、四 作业

.2.3三角函数的诱导公式第1课时诱导公式一、二、三、四A级必备知识基础练1.cos-11π6=A.12 B.-C.-32 D.2.已知sin(π-α)=13,则sin(α-2021π)的值为(A.223 B.C.13 D.-3.已知角α的终边与单位圆的交点为P-35,45,则cos(π+α)+sin(-α)=(A.-15 B.C.-75 D.4.若P(-4,3)是角α终边上一点,则cos(α-35.若cos(π+α)=-12,32π<α<2π,则sin(α-2π)6.化简下列各式:(1)sin-193πcos76π;(2)sin(-960°)cos1470°-cos(-240°)sin(-210°).7.已知cos(π+α)=-12,且α是第四象限角,计算(1)sin(2π-α);(2)sin(n∈Z).B级关键能力提升练8.sin21200°A.12 B.±C.-32 D.9.(2022广东珠海香洲校级模拟)若cos165°=a,则tan195°=()A.1-a2 BC.1-a2a10.已知tanπ3-α=13,则tan2π3+α=()A.13 B.-C.233 D.11.已知函数f(x)=asin(πx+α)+bcos(πx+β),且f(4)=3,则f(2019)的值为()A.-1 B.1C.3 D.-312.化简1+2sin(π-2A.sin2-cos2B.|cos2+sin2|C.±(cos2-sin2)D.无法确定13.(多选题)已知f(x)=sinx,下列式子中不成立的是()A.f(x+π)=sinxB.f(2π-x)=sinxC.f(-x)=-sinxD.f(π-x)=-f(x)14.(多选题)(2022辽宁抚顺期末)下列化简正确的是()A.tan(π+1)=tan1B.sin(-α)C.sin(π-D.cos(π15.cos(-585°)16.已知f(x)=|sin(3π+x)|+4cos2x+π2,则f(x)为函数(选填“奇”或“偶”);f5π4=.

17.已知f(x)=cos2(nπ+(1)化简f(x)的表达式;(2)求f20203π.C级学科素养创新练18.(多选题)(2022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)+sinCB.cos(A+B)+cosCC.sin(2A+2B)+sin2CD.cos(2A+2B)+cos2C19.设f(x)=sinπ6x,则f(1)+f(2)+f(3)+…+f(13)=.

7.2.3三角函数的诱导公式第1课时诱导公式一、二、三、四1.Dcos-11π6=cos11π6=cos2π-2.Dsin(α-2021π)=sin[(α-π)-2020π]=sin(α-π)=sin[-(π-α)]=-sin(π-α)=-133.A因为角α的终边与单位圆的交点为P-35,45,所以cosα=-35,sinα=45,则cos(π+α)+sin(-α)=-cosα-sinα=-14.-53由题意知sinα=3原式=(-cosα)tanαsin5.-32由cos(π+α)=-12,得cosα=又3π2<α<2π,故sin(α-2π)=sinα=-1-cos6.解(1)sin-193πcos76π=-sin6π+π3cosπ+π6=sinπ3cosπ6=(2)sin(-960°)cos1470°-cos(-240°)sin(-210°)=-sin(180°+60°+2×360°)cos(30°+4×360°)+cos(180°+60°)sin(180°+30°)=sin60°cos30°+cos60°sin30°=1.7.解(1)由cos(π+α)=-12可得cosα=12,而sin(2π-α)=-sinα.因为α是第四象限角,所以sinα=-故sin(2π-α)=32(2)sin=(-sinα)由(1)得cosα=12所以sin[α+8.Dsin9.B∵cos165°=cos(180°-15°)=-cos15°=a,∴cos15°=-a,则sin15°=1-∴tan15°=sin15°cos15°∴tan195°=tan(180°+15°)=tan15°=-1-故选B.10.B因为tan2π3+α=tanπ-π3-α=-tanπ3-α,所以tan2π3+α=-13.11.D∵f(x)=asin(πx+α)+bcos(πx+β),f(4)=asin(4π+α)+bcos(4π+β)=asinα+bcosβ=3,∴f(2019)=asin(2019π+α)+bcos(2019π+β)=asin(π+α)+bcos(π+β)=-asinα-bcosβ=-f(4)=-3.故选D.12.A原式=|sin(π-2)+cos(π-2)|=|sin2-cos2|=sin2-cos2.13.ABDf(x+π)=sin(x+π)=-sinx,f(2π-x)=sin(2π-x)=-sinx,f(-x)=sin(-x)=-sinx,f(π-x)=sin(π-x)=sinx=f(x).故A,B,D不成立.14.AB∵由诱导公式可得tan(π+1)=tan1,故A正确;sin(-α)tan(360°-sin(π-α)cos(πcos(π-α)tan(-π-15.2-2原式=cos=cos225=-cos45°sin4516.偶22f(x)=|sin(3π+x)|+4cos2x+π2=|sinx|+4cos(2x+π)=|sinx|-4cos2x,则f(-x)=|sin(-x)|-4cos2(-x)=|sinx|-4cos2x=f(x),∴f(x)为偶函数.f5π4=sin5π4-4cos5π2=sinπ4-4cosπ17.解(1)当n为偶数,即n=2k(k∈Z)时,f(x)=coscos2x(-sin当n为奇数,即n=2k+1(k∈Z)时,f(x)=co=cos2(π+综上,f(x)=sin2x.(2)由(1)知f20203π=sin22020π3=sin2672π+4π3=sin24π3=sin2π+π3=sin218.BCsin(A+B)+sinC=2sinC,故A错误;cos(A+B)+cosC=-cosC+cosC=0,故B正确;sin(2A+2B)+sin2C=sin[2(A+B)]+sin2C=sin[2(π-C)]+sin2C=sin(2π-2C)+sin2C=-sin2C+sin2C=0,故C正确;cos(2A+2B)+cos2C=cos[2(A+B)]+cos2C=cos[2(π-C)]+cos2C=cos(2π-2C)+cos2C=cos2C+cos2C=2cos2C,故D错误.19.12f(x)=sinπ6当x=1时,f(1)=sinπ6当x=2时,f(2)=sinπ3当x=3时,f(3)=sinπ2=当x=4时,f(4)=sin2π当x=5时,f(5)=sin5π当x=6时,f(6)=sinπ=0;当x=7时,f(7)=sin7π6=-当x=8