.【選題明細(xì)表】..簡(jiǎn)、求值的圖象與性質(zhì)的線性運(yùn)算4向量基本定理及應(yīng)用9量的數(shù)量積運(yùn)算1.sin15°cos15°等于(A)(A)(B)(C)(D)π)上為減函數(shù)的是(B)(A)y=cos2x(B)y=2|sinx|ycosxDytanxword輯....(A)4(B)-4(C)(D)-b,CAP=AB,BQ=BC,若=a,=b,則等于(A)(A)a+b(B)-a+bab解析:=-=b-a.所以==a,==b-a.word輯....于(B)(A)1(B)2(C)-1(D)3=1+c2-2c×1×cos=1+c2-c,則實(shí)數(shù)λ的值為(A)abab0,cos(+2α)等于(D)word輯....(A)(B)(C)-(D)-則cos(+2α)=cos(+2α)=-sin2α=-2sinαcosα(A)[-,0](B)[0,]C[,](D)[,]=2,令-+2kπ≤2x+≤+2kπ,k∈Z,可得-+kπ≤x≤+kπ,k∈Z.word輯....且滿足=+,則等于(A)(A)(B)(C)(D)word輯....則·等于(A)-=-1?=5?||=,·=(-)·=2××-×5.....(A)①f(x)的圖象關(guān)于直線x=對(duì)稱x③f(x)的圖象向左平移個(gè)單位,得到一個(gè)偶函數(shù)的圖象fx[0,]上為增函數(shù)(A)③(B)①③(C)②④(D)①③④x上移動(dòng),則·的最大值是(A)word輯....(A)2(B)1+(C)π(D)4故OA=cosθ,OD=sinθ,=cosθ+sinθ,y=sin(-θ)=cosθ,故=(cosθ+sinθ,cosθ),同理可求得C(sinθ,cosθ+sinθ),即=(sinθ,cosθ+sinθ),所以·=(cosθ+sinθ,cosθ)·(sinθ,cosθ+sinθ)word輯....a·c=|a||c|cos=2,word輯....T=-=π,所以=,答案:221cabc值范圍是.word輯....所以|c|的取值范圍是[2-1,2+1].]所以sinα=,cosα=,sinsincoscossin(3,-4),word輯....所以·=3×3+4×(-4)=-7.anA(2)因?yàn)镾=bcsinA=bc=,△ABCword輯....(2)若sin(θ-)=,0<<,求cos的值.QQQ所以sinθ=2cosθ,(2)因?yàn)閟in(θ-)=,0<<,QQθ∈(0,),則-<θ-<,Q則cos(θ-Q)===,=cosθcos(θ-)+sinθsin(θ-)QQ=×+×=.word輯....(2016·江西南昌一模)已知函數(shù)f(x)=(sinωx+cosωx)解:(1)f(x)=sinωxcosωx+cos2ωx-=sin(2ωx+),sBbcosCsinAcosBsinBCsinA,word輯....nBsinCsinBsinAB=sinB+sinAcosB+cosAsinBsinBcosBword輯.