第2课时诱导公式五、六课后篇巩固提升基础巩固1.若α∈(π,3π2),则❑√1-sin2(3π2-α)=()A.sinαB.-sinαC.cosαD.-cosα解析∵α∈(π,3π2),∴sinα<0.∴❑√1-sin2(3π2-α)=❑√1-cos2α=❑√sin2α=-sinα.答案B2.已知P(sin40°,-cos140°)为锐角α终边上的点,则α=()A.40°B.50°C.70°D.80°解析∵P(sin40°,-cos140°)为角α终边上的点,因而tanα=-cos140°sin40°=-cos(90°+50°)sin(90°-50°)=sin50°cos50°=tan50°,又α为锐角,则α=50°,故选B.答案B3.已知sin(π-α)=-2sin(π2+α),则sinαcosα=()A.25B.-25C.25或-25D.-15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.答案B4.在△ABC中,若sinA+B2=45,则cosC2=()A.-35B.-45C.35D.45解析∵A+B+C=π,∴A+B2=π2−C2.∴sinA+B2=sin(π2-C2)=cosC2=45.答案D5.已知cos(60°+α)=13,且-180°<α<-90°,则cos(30°-α)的值为()A.-2❑√23B.2❑√23C.-❑√23D.❑√23解析由-180°<α<-90°,得-120°<60°+α<-30°.又cos(60°+α)=13>0,所以-90°<60°+α<-30°,即-150°<α<-90°,所以120°<30°-α<180°,cos(30°-α)<0,所以cos(30°-α)=sin(60°+α)=-❑√1-cos2(60°+α)=-❑√1-(13)2=-2❑√23.答案A6.(一题多空题)若cosα=13,且α是第四象限的角,则sinα=,cos(α+3π2)=.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.答案-2❑√23-2❑√237.若sin(π2+θ)=37,则cos2(π2-θ)=.解析sin(π2+θ)=cosθ=37,则cos2(π2-θ)=sin2θ=1-cos2θ=1-949=4049.答案40498.求值:sin2(π4-α)+sin2(π4+α)=.解析∵π4-α+π4+α=π2,∴sin2(π4+α)=sin2[π2-(π4-α)]=cos2(π4-α).∴sin2(π4-α)+sin2(π4+α)=sin2(π4-α)+cos2(π4-α)=1.答案19.化简:sin(-α-3π2)·sin(3π2-α)·tan2(2π-α)cos(π2-α)·cos(π2+α)·cos2(π-α).解原式=sin(-α+π2)·[-sin(π2-α)]·tan2(2π-α)cos(π2-α)·cos(π2+α)·cos2(π-α)=cosα·(-cosα)·tan2αsinα·(-sinα)·cos2α=tan2αsin2α=1cos2α.10.已知角α的终边经过点P(45,-35).(1)求sinα的值;(2)求sin(π2-α)tan(α-π)sin(α+π)cos(3π-α)的值.解(1)∵P(45,-35),|OP|=1,∴sinα=-35.(2)sin(π2-α)tan(α-π)sin(α+π)cos(3π-α)=cosαtanα-sinα(-cosα)=1cosα,由三角函数定义知cosα=45,故所求式子的值为54.能力提升1.已知π<α<2π,cos(α-9π)=-35,则cos(α-11π2)的值为()A.35B.-35C.-45D.45解析因为cos(α-9π)=-cosα=-35,所以cosα=35.又因为α∈(π,2π),所以sinα=-❑√1-cos2α=-45,cos(α-11π2)=-sinα=45.答案D2.已知角α的终边上有一点P(1,3),则sin(π-α)-sin(π2+α)cos(3π2-α)+2cos(-π+α)的值为()A.-25B.-45C.-47D.-4解析sin(π-α)-sin(π2+α)cos(3π2-α)+2cos(-π+α)=sinα-cosα-sinα-2cosα=tanα-1-tanα-2.因为角α终边上有一点P(1,3),所以tanα=3,所以原式=3-1-3-2=-25.故选A.答案A3.已知α为第二象限角,则cosα❑√1+tan2α+sinα❑√1+1tan2α=.解析原式=cosα❑√sin2α+cos2αcos2α+sinα❑√sin2α+cos2αsin2α=cosα1|cosα|+sinα1|sinα|.因为α是第二象限角,所以sinα>0,cosα<0,所以cosα1|cosα|+sinα1|sinα|=-1+1=0,即原式等于0.答案04.sin21°+sin22°+sin23°+…+sin289°=.解析sin21°+sin22°+sin23°+…+sin289°=sin21°+sin22°+sin23°+…+sin245°+cos244°+…+cos21°=(sin21°+cos21°)+(sin22°+cos22°)+…+(sin244°+cos244°)+sin245°=44+12=892.答案8925.已知函数f(x)=❑√2cosx-π12,x∈R.若cosθ=35,θ∈3π2,2π,则fθ-5π12=.解析fθ-5π12=❑√2cosθ-5π12−π12=❑√2cosθ-π2=❑√2cosπ2-θ=❑√2sinθ,由已知可得θ为第四象限角,所以sinθ<0,故sinθ=-❑√1-cos2θ=-45,fθ-5π12=❑√2sinθ=❑√2×-45=-4❑√25.答案-4❑√256.是否存在角α,β,α∈(-π2,π2),β∈(0,π),使等式sin(3π-α)=❑√2cos(π2-β),❑√3cos(-α)=-❑√2cos(π+β)同时成立?若存在,求出α,β的值;若不存在,请说明理由.解由条件,得{sinα=❑√2sinβ,❑√3cosα=❑√2cosβ,①②①2+②2得sin2α+3cos2α=2,∴sin2α=12.又α∈(-π2,π2),∴α=π4或α=-π4.将α=π4代入②,得cosβ=❑√32.又β∈(0,π),∴β=π6,代入①可知符合.将α=-π4代入②得cosβ=❑√32,又β∈(0,π),∴β=π6,代入①可知不符合.综上可知,存在α=π4,β=π6满足条件.
