7.2.4诱导公式课后篇巩固提升基础巩固1.sin(-π3)+2sin4π3+3sin2π3等于()A.1B.12C.0D.-1答案C2.已知角α的顶点与原点O重合,始边与x轴非负半轴重合,它的终边过点P-35,-45,则sin(π-α)=()A.-45B.45C.-35D.35解析由三角函数的定义可得sinα=-45,则sin(π-α)=sinα=-45.答案A3.已知a=tan(-7π6),b=cos23π4,c=sin(-33π4),则a,b,c的大小关系是()A.b>a>cB.a>b>cC.b>c>aD.a>c>b解析因为a=-❑√33,b=❑√22,c=-❑√22,所以b>a>c.答案A4.(多选)化简❑√1+2sin(π-2)·cos(π-2)的结果是()A.sin2-cos2B.|cos2-sin2|C.±(cos2-sin2)D.无法确定解析原式=|sin(π-2)+cos(π-2)|=|sin2-cos2|=sin2-cos2.答案AB5.已知tanα-π6=12,且α∈0,π2,则cos2π3-α=()A.-❑√55B.❑√55C.-2❑√55D.2❑√55解析由tanα-π6=12>0,且α∈0,π2可得0<α-π6<π3,则sinα-π6=❑√55,故cos2π3-α=cosπ2-α-π6=sinα-π6=❑√55.答案B6.设tan(5π+α)=m,则sin(α-3π)+cos(π-α)sin(-α)-cos(π+α)的值为.解析由题意知tanα=m,原式=-sinα-cosα-sinα+cosα=-tanα-1-tanα+1=m+1m-1.答案m+1m-17.已知sin(π3-α)=12,则cos(π6+α)=.解析∵(π3-α)+(π6+α)=π2,∴cos(π6+α)=sin(π3-α)=12.答案128.化简:(1)1+cos(π2+α)sin(π2-α)tan(π+α);(2)sin(2π-α)cos(π+α)cos(π2+α)cos(11π2-α)cos(π-α)sin(3π-α)sin(-π-α)sin(9π2+α).解(1)原式=1+(-sinα)cosαtanα=1-sin2α=cos2α.(2)原式=(-sinα)(-cosα)(-sinα)cos[5π+(π2-α)](-cosα)sin(π-α)[-sin(π+α)]sin[4π+(π2+α)]=-sin2αcosα[-cos(π2-α)](-cosα)sinα[-(-sinα)]sin(π2+α)=sin2αcosαsinα-cosαsin2αcosα=-sinαcosα=-tanα.9.已知sin(5π+α)=lg13√10,求cos(2π+α)的值.解∵sin(5π+α)=sin(π+α)=-sinα,lg13√10=lg10-13=-13,∴sinα=13,∴cos(2π+α)=cosα=±❑√1-sin2α=±❑√1-(13)2=±2❑√23.能力提升1.已知函数f(x)=cosx2,则下列等式成立的是()A.f(2π-x)=f(x)B.f(2π+x)=f(x)C.f(-x)=f(x)D.f(-x)=-f(x)解析f(-x)=cos(-x2)=cosx2=f(x).答案C2.已知f(cosx)=sinx,设x是第一象限角,则f(sinx)为()A.secxB.cosxC.sinxD.1-sinx解析f(sinx)=f(cos(π2-x))=sin(π2-x)=cosx.答案B3.已知α为锐角,2tan(π-α)-3cos(π2+β)+5=0,tan(π+α)+6sin(π+β)-1=0,则sinα的值是()A.3❑√55B.3❑√77C.3❑√1010D.13解析由已知可知-2tanα+3sinβ+5=0,tanα-6sinβ-1=0,所以tanα=3.又tanα=sinαcosα,所以9=sin2αcos2α=sin2α1-sin2α.所以sin2α=910.因为α为锐角,所以sinα=3❑√1010.答案C4.已知sin(2π+θ)tan(π+θ)tan(3π-θ)cos(π2-θ)tan(-π-θ)=1,则3sin2θ+3sinθcosθ+2cos2θ的值是()A.1B.2C.3D.6解析因为原式=sinθ·tanθ·tan(-θ)-sinθ·tanθ=tanθ=1,所以3sin2θ+3cos2θsin2θ+3sinθ·cosθ+2cos2θ=3tan2θ+3tan2θ+3tanθ+2=1.答案A5.已知tan(π-θ)=3,则sin(π2+θ)-cos(π-θ)sin(π2-θ)-sin(π-θ)=()A.-1B.-12C.1D.12解析由tan(π-θ)=3,得-tanθ=3,即tanθ=-3,则sin(π2+θ)-cos(π-θ)sin(π2-θ)-sin(π-θ)=cosθ+cosθcosθ-sinθ=2cosθcosθ-sinθ=21-tanθ=12.答案D6.tan1°·tan2°·tan3°·…·tan89°=.解析因为tan1°·tan89°=1,所以tan1°·tan2°·…·tan89°=1×1×…×1⏟44个×tan45°=1.答案17.(双空)已知函数f(x)={cosxπ6,x≤0,log12(x+2),x>0,则f(2)=,f[f(2)]=.解析由题意可得f(2)=log12(2+2)=log124=-2,则f[f(2)]=cos-π3=cosπ3=12.答案-2128.已知α是第三象限的角,且f(α)=sin(π-α)cos(2π-α)tan(-α+3π2)cot(-α-π)sin(-α-π).(1)化简f(α);(2)若cos(α-3π2)=15,求f(α)的值.解(1)f(α)=sin(π-α)cos(2π-α)tan(3π2-α)cot(-α-π)sin(-α-π)=sinα·cosα·cotα-cotα·sinα=-cosα.(2)∵cos(α-3π2)=-sinα,∴sinα=-15.又α是第三象限的角,∴cosα=-❑√1-sin2α=-25❑√6,∴f(α)=25❑√6.9.已知tanα,1tanα是关于x的方程3x2-3kx+3k2-13=0的两实根,且3π<α<7π2,求cos(3π+α)+sin(π+α)的值.解tanα,1tanα是关于x的方程3x2-3kx+3k2-13=0的两实根,1=tanα·1tanα=13(3k2-13),所以k2=163(当k2=163时,Δ=9k2-4×3(3k2-13)>0).因为3π<α<7π2,所以tanα>0,sinα<0,cosα<0.又tanα+1tanα=--3k3=k,所以k>0,故取k=4❑√33.于是tanα+1tanα=sinαcosα+cosαsinα=1sinαcosα=4❑√33,即sinαcosα=❑√34.所以(sinα+cosα)2=1+2sinαcosα=2+❑√32.因为sinα+cosα<0,所以sinα+cosα=-❑√3+12.于是cos(3π+α)+sin(π+α)=cos(π+α)+sin(π+α)=-(cosα+sinα)=❑√3+12.
