1第5章三角函数5.2任意角的三角函数5.2.3诱导公式第2课时诱导公式五~六课后篇巩固提升必备知识基础练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.若α是第二象限角,且sinπ2-α=-35,则1tan(π2-α)=()A.43B.34C.-43D.-34答案C解析由sinπ2-α=-35可知cosα=-35,结合α是第二象限角可知sinα=√1-cos2α=45,tanα=sinαcosα=-43,1tan(π2-α)=tanα=-43.故选C.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B2解析由sin(180°+α)+cos(90°+α)=-14,得sinα=18,cos(270°-α)+2sin(360°-α)=-sinα-2sinα=-3sinα=-38.4.(2020甘肃白银高一期末)已知cosπ6-α=13,则sin2π3-α=()A.49B.13C.59D.-59答案B解析令θ=π6-α,则α=π6-θ,∴sin2π3-α=sinπ2+θ=cosθ=13.故选B.5.若sinπ2+θ=37,则cos2π2-θ=.答案4049解析sin(π2+θ)=cosθ=37,则cos2(π2-θ)=sin2θ=1-cos2θ=1-949=4049.6.若cosα=13,且α是第四象限角,则1tan(3π2-α)=,cosα+3π2=.答案-2√2-2√23解析因为α是第四象限的角,所以sinα=-√1-cos2α=-2√23.于是cosα+3π2=-cosα+π2=sinα=-2√23.1tan(3π2-α)=1tan(π+π2-α)=1tan(π2-α)=tanα=sinαcosα=-2√2.7.已知cosπ2+α=2sinα-π2,求:(1)tan5π2-α;3(2)sin(π2-α)+cos(3π2-α)5cos(5π2-α)+3sin(7π2-α).解 cosπ2+α=2sinα-π2,∴-sinα=-2cosα,∴tanα=2.(1)tan5π2-α=tan2π+π2-α=tanπ2-α=1tanα=12;(2)原式=cosα-sinα5sinα-3cosα=1-tanα5tanα-3=1-25×2-3=-17.关键能力提升练8.已知π<α<2π,cos(α-9π)=-35,则tan3π2+α=()A.43B.-43C.-34D.34答案D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,tanα=-43,因此tan3π2+α=tanπ2+α=-1tanα=34,故选D.9.sin(α-π2)cos(3π2+α)tan\(π-α\)tan\(-α-π\)sin\(-α-π\)=()A.cosαB.-cosαC.sinαD.-sinα4答案B解析sin(α-π2)cos(3π2+α)tan\(π-α\)tan\(-α-π\)sin\(-α-π\)=sin[-(π2-α)]cos(3π2+α)tan\(π-α\)tan[-\(π+α\)]sin[-\(π+α\)]=-cosαsinα\(-tanα\)\(-tanα\)sinα=-cosα,故选B.10.(多选题)(2021山东青岛二中高一期末)在△ABC中,下列关系恒成立的是()A.tan(A+B)=tanCB.cos(2A+2B)=...
