第1课时诱导公式二、三、四课后篇巩固提升基础巩固1.已知sin(π+θ)=45,则角θ的终边在()A.第一或第二象限B.第二或第三象限C.第一或第四象限D.第三或第四象限解析由已知得-sinθ=45,所以sinθ=-45,故角θ的终边在第三或第四象限.答案D2.若cos(π-α)=-12,则cos(-2π-α)的值为()A.12B.±❑√32C.-12D.±12解析∵cos(π-α)=-cosα=-12,∴cosα=12.∴cos(-2π-α)=cos(-α)=cosα=12.答案A3.sin(-13π6)-cos(-10π3)-tan(15π4)的值为()A.-2B.0C.12D.1解析原式=-sin(2π+π6)-cos(2π+4π3)-tan(2π+7π4)=-sinπ6-cos(π+π3)-tan(2π-π4)=-12+cosπ3+tanπ4=-12+12+1=1.答案D4.已知tan(π-α)=12,则sinα+cosα2sinα-cosα=()A.14B.-14C.12D.-12解析由已知得-tanα=12,所以tanα=-12.于是sinα+cosα2sinα-cosα=tanα+12tanα-1=-12+12×(-12)-1=-14.答案B5.若角7π-α的终边与单位圆的交点坐标是(x,35),则cos(α-2018π)=()A.±45B.±35C.45D.-35解析依题意,sin(7π-α)=35,即sinα=35,于是cosα=±45,故cos(α-2018π)=cosα=±45.答案A6.记cos(-80°)=k,则tan100°等于()A.❑√1-k2kB.-❑√1-k2kC.k❑√1-k2D.-k❑√1-k2解析∵cos(-80°)=cos80°=k,sin80°=❑√1-cos280°=❑√1-k2,∴tan100°=-tan80°=-❑√1-k2k.故选B.答案B7.已知sin(45°+α)=513,则sin(135°-α)=.解析sin(135°-α)=sin[180°-(45°+α)]=sin(45°+α)=513.答案5138.已知tan(π7+α)=5,则tan(6π7-α)=.解析tan(6π7-α)=tan[π-(π7+α)]=-tan(π7+α)=-5.答案-59.设tan(5π+α)=m,则sin(α-3π)+cos(π-α)sin(-α)-cos(π+α)=.解析∵tan(5π+α)=tanα=m,∴原式=-sinα-cosα-sinα+cosα=-tanα-1-tanα+1=-m-1-m+1=m+1m-1.答案m+1m-110.已知sin(3π+α)=13,求:sin(180°+α)cos(720°+α)tan(540°+α)sin(-180°+α)tan(900°+α)sin(-180°-α)cos(-180°-α)的值.解∵sin(3π+α)=13,∴sinα=-13.原式=(-sinα)·cosα·tanα·(-sinα)tanα·sinα·(-cosα)=-sinα=13.能力提升1.❑√1-2sin(π+2)cos(π-2)等于()A.sin2-cos2B.sin2+cos2C.±(sin2-cos2)D.cos2-sin2解析❑√1-2sin(π+2)cos(π-2)=❑√1-2sin2cos2=❑√(sin2-cos2)2=|sin2-cos2|=sin2-cos2.答案A2.(多选题)已知A=sin(kπ+α)sinα+cos(kπ+α)cosα(k∈Z),则A的值是()A.-1B.-2C.1D.2解析当k为偶数时,A=sinαsinα+cosαcosα=2;当k为奇数时,A=-sinαsinα−cosαcosα=-2.故选BD.答案BD3.(一题多空题)已知f(n)=sinnπ4(n∈Z),则f(1)=,f(7)=,f(1)+f(2)+…+f(8)=,f(1)+f(2)+…+f(100)=.解析∵f(n)=sinnπ4(n∈Z),∴f(1)=❑√22,f(2)=1,f(3)=❑√22,f(4)=0,f(5)=-❑√22,f(6)=-1,f(7)=-❑√22,f(8)=0.即sinπ4+sin2π4+sin3π4+…+sin8π4=0,且以8为循环周期.则f(1)+f(2)+…+f(100)=sinπ4+sin2π4+sin3π4+…+sin100π4=sinπ4+sin2π4+sin3π4+sin4π4=1+❑√2.答案❑√22-❑√2201+❑√24.(1)已知sinα是方程5x2-7x-6=0的根,求cos(α+2π)cos(4π+α)tan2(2π+α)tan(6π+α)sin(2π+α)sin(8π+α)的值;(2)已知sin(4π+α)=❑√2sinβ,❑√3cos(6π+α)=❑√2cos(2π+β),且0<α<π,0<β<π,求α和β的值.解(1)因为方程5x2-7x-6=0的两根为2和-35,所以sinα=-35.由sin2α+cos2α=1,得cosα=±❑√1-sin2α=±45.当cosα=45时,tanα=-34;当cosα=-45时,tanα=34.所以原式=cosα·cosα·tan2α·tanαsinα·sinα=tanα=±34.(2)因为sin(4π+α)=❑√2sinβ,所以sinα=❑√2sinβ.①因为❑√3cos(6π+α)=❑√2cos(2π+β),所以❑√3cosα=❑√2cosβ.②①2+②2,得sin2α+3cos2α=2(sin2β+cos2β)=2,所以cos2α=12,即cosα=±❑√22.又0<α<π,所以α=π4或α=3π4.又0<β<π,当α=π4时,由②得β=π6;当α=3π4时,由②得β=5π6.所以α=π4,β=π6或α=3π4,β=5π6.
