1第2课时诱导公式五、六课后篇巩固提升合格考达标练1.若α∈(π,3π2),则√1-sin2(3π2-α)=()A.sinαB.-sinαC.cosαD.-cosα答案B解析 α∈(π,3π2),∴sinα<0.∴√1-sin2(3π2-α)=√1-cos2α=√sin2α=-sinα.2.已知sin(π-α)=-2sin(π2+α),则sinαcosα=()A.25B.-25C.25或-25D.-15答案B解析 sin(π-α)=-2sin(π2+α),∴sinα=-2cosα.再由sin2α+cos2α=1可得sinα=2√55,cosα=-√55,或sinα=-2√55,cosα=√55,∴sinαcosα=-25.故选B.3.若sin(180°+α)+cos(90°+α)=-14,则cos(270°-α)+2sin(360°-α)的值为()A.-16B.-38C.16D.38答案B解析由sin(180°+α)+cos(90°+α)=-14,得sinα=18,cos(270°-α)+2sin(360°-α)=-sinα-2sinα=-3sinα=-38.4.(2021黑龙江哈尔滨南岗高一期末)已知sinπ6+α=-45,则cosπ3-α=()A.45B.35C.-45D.-35答案C2解析 sinπ6+α=-45,∴cosπ3-α=cosπ2-π6+α=sinπ6+α=-45,故选C.5.(多选题)(2021北京海淀高一期末)已知3sinπ2-α-sin(π+α)=-√2,则cosα-sinα的取值可以为()A.8√25B.√2C.-√2D.3√25答案CD解析因为3sinπ2-α-sin(π+α)=3cosα+sinα=-√2,所以{3cosα+sinα=-√2,cos2α+sin2α=1,整理得10cos2α+6√2cosα+1=0,解得cosα=-√210或-√22,当cosα=-√210时,sinα=-7√210,则cosα-sinα=3√25;当cosα=-√22时,sinα=√22,则cosα-sinα=-√2.故选CD.6.若cosα=13,且α是第四象限的角,则sinα=,cos(α+3π2)=.答案-2√23-2√23解析因为α是第四象限的角,所以sinα=-√1-cos2α=-2√23.于是cos(α+3π2)=-cos(α+π2)=sinα=-2√23.7.若sin(π2+θ)=37,则cos2(π2-θ)=.答案4049解析sin(π2+θ)=cosθ=37,则cos2(π2-θ)=sin2θ=1-cos2θ=1-949=4049.38.(2021天津东丽高一期末)已知sin(π+α)=-45,α∈π2,π.求值:sin(π2+α)+2cos\(π-α\)sin(π2-α)+sin\(-α\).解 sin(π+α)=-sinα=-45,α∈π2,π,则sinα=45.∴cosα=-√1-sin2α=-35,故tanα=-43.sin(π2+α)+2cos\(π-α\)sin(π2-α)+sin\(-α\)=cosα-2cosαcosα-sinα=-cosαcosα-sinα=-37.等级考提升练9.(2021吉林公主岭高一期末)已知角θ终边经过点(3,-4),则sin(3π2-θ)·cos\(π+θ\)sin(π2+θ)·cos(5π2+θ)=()A.34B.43C.-43D.-34答案A解析 角θ终边经过点(3,-4),∴tanθ=-43,则sin(3π2-θ)·cos\(π+θ\)sin(π2+θ)·cos(5π2+θ)=-cosθ·\(-cosθ\)cosθ·\(-sinθ\)=-1tanθ=34,故选A.10.已知π<...
