高考卷 19 高考数学(文)一轮复习精品资料(解析版) (2)

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1.已知cosα=k,k∈R,α∈π2,π,则sin(π+α)=()A.-1-k2B.1-k2[来源:学科网ZXXK]C.±1-k2D.-k【答案】A【解析】由cosα=k,α∈π2,π得sinα=1-k2,∴sin(π+α)=-sinα=-1-k2.2.已知α为锐角,且tan(π-α)+3=0,则sinα的值是()A.13B.31010C.377D.353[来源:学+科+网Z+X+X+K]【答案】B3.已知tanα=-35,则sin2α=()A.1517B.-1517C.-817D.817【答案】B【解析】sin2α=2sinαcosαsin2α+cos2α=2tanαtan2α+1=2×(-35)(-35)2+1=-1517.[来源:学科网ZXXK]4.已知α和β的终边关于直线y=x对称,且β=-π3,则sinα等于()A.-32B.32C.-12D.12【答案】D[来源:Zxxk.Com]【解析】因为α和β的终边关于直线y=x对称,所以α+β=2kπ+π2(k∈Z).又β=-π3,所以α=2kπ+5π6(k∈Z),即得sinα=12.学#科网[来源:学科网]5.已知sin(π-α)=log412,且α∈-π2,0,则tan(2π-α)的值为()[来源:学科网]A.-33B.33C.±33D.3【答案】B6.若θ∈π4,π2,sin2θ=34,则sinθ的值是()A.7-14B.24C.7+14D.74【答案】C【解析】由θ∈π4,π2,知sinθ+cosθ>0,sinθ-cosθ>0.又(sinθ+cosθ)2=1+2sinθcosθ=74,(sinθ-cosθ)2=1-2sinθcosθ=14,∴sinθ+cosθ=72,且sinθ-cosθ=12,从而sinθ=7+14.7.下列各数中与sin2019°的值最接近的是()A.12B.32C.-12D.-32【答案】C【解析】2019°=5×360°+180°+39°,∴sin2019°=-sin39°和-sin30°接近.选C.8.已知sin(π+θ)=-3cos(2π-θ),|θ|<π2,则θ等于()A.-π6B.-π3C.π6D.π3【答案】D【解析】∵sin(π+θ)=-3cos(2π-θ),∴-sinθ=-3cosθ,∴tanθ=3.∵|θ|<π2,∴θ=π3.9.已知tan(α-π)=34,且α∈π2,3π2,则sinα+π2=()A.45B.-45C.35D.-35【答案】B10.已知f(α)=sinπ-αcos2π-αcos-π-αtanα,则f-31π3的值为()A.12B.-13C.-12D.13【答案】C11.已知sinα+π12=13,则cosα+7π12的值为()A.13B.-13C.-223D.223【答案】B【解析】cosα+7π12=cosπ2+α+π12=-sinα+π12=-13.选B.学#科网12.已知tanx=2,则sin2x+1的值为()A.0B.95C.43D.53【答案】B【解析】sin2x+1=2sin2x+cos2xsin2x+cos2x=2tan2x+1tan2x+1=95.故选B.[来源:Z。xx。k.Com]13.已知1+sinαcosα=-12,则cosαsinα-1的值是()[来源:Zxxk.Com]A.12B.-12C.2D.-2【答案】A【解析】因为1-sin2α=cos2α,cosα≠0,1-sinα≠0,所以(1+sinα)(1-sinα)=cosαcosα,所以1+sinαcosα=cosα1-sinα,所以cosα1-sinα=-12,即cosαsinα-1=12.故选A.14.若A,B是锐角△ABC的两个内角,则点P(cosB-sinA,sinB-cosA)在()A.第一象限B.第二象限C.第三象限D.第四象限【答案】B15.若θ∈π4,π2,sinθcosθ=3716,则sinθ=()A.35B.45C.74D.34【答案】D【解析】∵sinθcosθ=3716,∴(sinθ+cosθ)2=1+2sinθcosθ=8+378,(sinθ-cosθ)2=1-2sinθcosθ=8-378,∵θ∈π4,π2,∴sinθ+cosθ=3+74①,sinθ-cosθ=3-74②,联立①②得,sinθ=34.16.已知cos(75°+α)=513,α是第三象限角,则sin(195°-α)+cos(α-15°)的值为________.【答案】-1713【解析】因为cos(75°+α)=5130,α是第三象限角,所以75°+α是第四象限角,sin(75°+α)=-1-cos275°+α=-1213.所以sin(195°-α)+cos(α-15°)=sin[180°+(15°-α)]+cos(15°-α)=-sin(15°-α)+cos(15°-α)=-sin[90°-(75°+α)]+cos[90°-(75°+α)]=-cos(75°+α)+sin(75°+α)=-513-1213=-1713.17.已知角α的终边上一点P(3a,4a)(a0),则cos()540°-α的值是________.【答案】35【解析】cos(540°-α)=cos(180°-α)=-cosα.因为a0,所以r=-5a,所以cosα=-35,所以cos(540°-α)=-cosα=35.[来源:Zxxk.Com]18.已知sinθ+cosθ=713,θ∈(0,π),则tanθ=________.【答案】-12519.sin4π3·cos5π6·tan-4π3的值是________.【答案】-334【解析】原式=sinπ+π3·cosπ-π6·tan(-π-π3)=-sinπ3·-cosπ6·-tanπ3=-32×-32×(-3)=-334.20.直线2x-y+1=0的倾斜角为θ,则1sin2θ-cos2θ的值为________.【答案】53学#科网【解析】由题意可知,tanθ=2,则1sin2θ-cos2θ=sin2θ+cos2θsin2-cos2θ=tan2θ+1tan2θ-1=53.21.已知θ为锐角,且sin(θ-π4)=210,则tan2θ=________.【答案】-247【解析】由已知sinθ-π4=210得sinθ-cosθ=15,再由θ为锐角且sin2θ+cos2θ=1,得sinθ=45,cosθ=35,所以tanθ=43,tan2θ=2tanθ1-tan2θ=2×431-169=-247.22.已知sin(3π+α)=2sin3π2+α,求下列各式的值:(1)sinα-4cosα5sinα+2cosα;(2)sin2α+sin2α.[来源:学§科§网]23.已知f(α)=sinπ-αcos2π-αcos-α+3π2cosπ2-αsin-π-α.(1)化简f(α);(2)若α是第三象限角,且cosα-3π2=15,求f(α)的值.解(1)f(α)=sinπ-αcos2π-αcos-α+3π2cosπ2-αsin-π-α=sinαcosα-sinαsinαsinα=-cosα.(2)因为α是第三象限角,且cosα-3π2=-sinα=15,sinα=-15.所以cosα=-1-sin2α=-1--152=-265.所以f(α)=-cosα=265.

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