3.1.1两角差的余弦公式主动成长夯基达标1.cos(-15°)的值是()A.B.C.D.解析:cos(-15°)=cos15°=cos(60°-45°)=cos60°cos45°+sin60°sin45°=×+×=.答案:D2.若sinα-sinβ=1-,cosα-cosβ=,则cos(α-β)的值为()A.B.C.D.1解析:由sinα-sinβ=1-,得sin2α-2sinαsinβ+sin2β=.①由cosα-cosβ=,得cos2α-2cosαcosβ+cos2β=.②①+②,得1+1-2(cosαcosβ+sinαsinβ)=2--2cos(α-β)=-.∴cos(α-β)=.答案:B3.下列式子中正确的个数为()①cos(α-β)=cosα-cosβ②cos(α-β)=cosα+cosβ③cos(α-β)=cosαcosβ④cos(α-β)=cosαcosβ-sinαsinβ⑤cos(α-β)=cosαcosβ+sinαsinβA.0B.1C.2D.3解析:利用特殊值确定①②③④是错误的.答案:B4.cos57°cos12°+sin57°sin12°的值是()A.0B.C.D.解析:cos57°cos12°+sin57°sin12°=cos(57°-12°)=cos45°=.答案:D5.cosα=,则cos(α-)的值为()A.B.-C.D.或-解析:∵cosα=,∴α为第一或第四象限角.∴sinα=±.∴cos(α-)=cosαcos+sinαsin=(sinα+cosα)=或.答案:D6.cos(α+β)=,sin(β-)=,α、β∈(0,),那么cos(α+)的值为()A.B.C.D.解析:∵α,β∈(0,),cos(α+β)=>0,sin(β-)=>0,∴sin(α+β)=,cos(β-)=.∴cos(α+)=cos[(α+β)-(β-)]=cos(α+β)cos(β-)+sin(α+β)sin(β-)=×+×=.答案:C7.满足cosα·cosβ=-sinαsinβ的一组α、β的值是()A.α=,β=B.α=,β=C.α=,β=D.α=,β=解析:由已知得cosα·cosβ+sinαsinβ=.∴cos(α-β)=,因此只有A项中两角满足条件.答案:A8.cos(α-35°)cos(25°+α)+sin(α-35°)·sin(25°+α)=____________.解析:原式=cos[(α-35°)-(25°+α)]=cos60°=.答案:9.已知cosα=,cos(α+β)=-,且α、β∈(0,),求cosβ的值.解:由cosα=,α∈(0,),∴sinα=.又cos(α+β)=,0<α+β<π,∴sin(α+β)=.∴cosβ=cos[(α+β)-α]=cos(α+β)cosα+sin(α+β)sinα=()×+=.10.已知cos(α+β)=,cos(α-β)=-,<α+β<2π,<α-β<π,求cos2β.解:因为<α+β<2π,cos(α+β)=,所以sin(α+β)=-.又<α-β<π,cos(α-β)=-,所以sin(α-β)=,cos2β=cos[(α+β)-(α-β)]=cos(α+β)cos(α-β)+sin(α+β)sin(α-β)=-1.11.已知cos(+α)=,sin(-β)=,且0<α<<β<π,求cos(β-α)的值.解:∵0<α<<β<π,∴<+α<<+β<.∵cos(+α)=-<0,∴<+α<.∴sin(+α)=.∵sin(-β)=sin[π-(+β)]=sin(+β)=>0,∴<+β<π.∴cos(+β)=.∴cos(β-α)=cos[(+β)-(+α)]=cos(+β)cos(+α)+sin(+β)sin(+α)=-×()+×()=.走近高考12.(经典回放)已知sin(-α)=(0<α<),求的值.解:注意所求问题中的角与已知角之间的关系,通过变角,拆角等来完成.(+α)+(-α)=,2α=(+α)-(-α),∵cos(+α)=cos[-(-α)]=sin(-α)=,又∵0<α<,∴0<-α<,<+α<.∴cos(-α)=,sin(+α)=.∴cos2α=cos[(+α)-(-α)]=cos(+α)cos(-α)+sin(+α)sin(-α)=×+×=.∴.