第一章1.3第2课时诱导公式五、六A级基础巩固一、选择题1.已知sin(+α)=,那么cosα=(C)A.-B.-C.D.2.已知sinα=,则cos(π+α)等于(A)A.B.C.-D.-[解析]cos(π+α)=sinα=.3.若sin(3π+α)=-,则cos(-α)等于(A)A.-B.C.D.-[解析]由已知,得sinα=,则cos(-α)=-sinα=-.4.已知cos(+α)=-,且α是第四象限角,则cos(-3π+α)(B)A.B.-C.±D.[解析]∵cos(+α)=-,∴sinα=-,∴cos(-3π+α)=-cosα=-=-.5.若sin(180°+α)+cos(90°+α)=-a,则cos(270°-α)+2sin(360°-α)的值是(B)A.-B.-C.D.[解析]由sin(180°+α)+cos(90°+α)=-a,得:-sinα-sinα=-a,即sinα=,cos(270°-α)+2sin(360°-α)=-sinα-2sinα=-3sinα=-a.6.若sin(-α)=,则cos(-α)的值为(B)A.B.-C.D.-[解析]cos(-α)=cos[+(-α)]=-sin(-α)=-.二、填空题7.化简=__-1__.[解析]原式===-=-1.8.(2016·成都高一检测)已知sin(α-)=,那么cos(α+)的值是-.[解析]∵(α+)-(α-)=,∴α+=+(α-),∴cos(α+)=cos[+(α-)]=-sin(α-)=-.三、解答题9.化简:.[解析]原式=====tanα.10.已知sinα是方程5x2-7x-6=0的根,α是第三象限角,求的值.[解析]由已知得sinα=-.∵α是第三象限角,∴cosα=-=-.∴原式===.B级素养提升一、选择题1.若角A、B、C是△ABC的三个内角,则下列等式中一定成立的是(D)A.cos(A+B)=cosCB.sin(A+B)=-sinCC.cos(+C)=sinBD.sin=cos[解析]∵A+B+C=π,∴A+B=π-C,∴cos(A+B)=-cosC,sin(A+B)=sinC.所以A,B都不正确;同理,B+C=π-A,所以sin=sin(-)=cos,因此D是正确的.2.α为锐角,2tan(π-α)-3cos(+β)=-5,tan(π+α)+6sin(π+β)=1,则sinα=(C)A.B.C.D.[解析]由已知可得,-2tanα+3sinβ+5=0,tanα-6sinβ=1解得tanα=3,故sinα=,选C.3.已知sin(-α)=,那么cos(-α)=(D)A.B.-C.D.-[解析]cos(-α)=cos[+(-α)]=-sin(-α)=-.4.若f(sinx)=3-cos2x,则f(cosx)等于(C)A.3-cos2xB.3-sin2xC.3+cos2xD.3+sin2x[解析]f(cosx)=f[sin(-x)]=3-cos2(-x)=3-cos(π-2x)=3+cos2x二、填空题5.已知sin(+α)=,则sin(-α)=.[解析]∵sin(+α)=cosα=,∴sin(-α)=cosα=.6.化简=__-1__.[解析]原式====-1.三、解答题7.若sin(180°+α)=-,0°<α<90°.求的值.[解析]由sin(180°+α)=-,α∈(0°,90°),得sinα=,cosα=,∴原式====2.8.已知cos(-α)=,求cos(π+α)sin(-α)的值.[解析]cos(π+α)·sin(-α)=cos[π-(-α)]·sin[π-(+α)]=-cos(-α)·sin(+α)=-sin[-(-α)]=-cos(-α)=-.C级能力拔高是否存在α∈,β∈(0,π),使等式sin(3π-α)=cos,cos(-α)=-cos(π+β)同时成立?若存在,求出α、β的值;若不存在,说明理由.[思路分析]题中所给条件式比较繁琐,故先化简,然后利用平方关系消去α(或β)解方程可求出角α与β的一个三角函数值和其范围,进一步求出角.[解析]由条件得,①2+②2得,sin2α+3cos2α=2③又∵sin2α+cos2α=1④由③,④得cos2α=即cosα=±,∵α∈,∴α=或α=-.当α=时,代入②得cosβ=,又β∈(0,π),∴β=,代入①可知符合.当α=-时,代入②得cosβ=,又β∈(0,π),∴β=,代入①可知不符合.综上所述,存在α=,β=满足条件.