2020版高考数学总复习第三篇三角函数、解三角形第3节三角恒等变换应用能力提升.docx

第3节三角恒等变换【选题明细表】知识点、方法题号三角函数的化简求值5,6,7,13给值求值1,2,4,8给值求角3,14综合应用9,10,11,12,15基础巩固(建议用时:25分钟)1.若cos=,则sin 2等于(D)(A)(B)(C)-(D)-解析:cos=(cos +sin )=cos +sin =1+sin 2=,所以sin 2=-.故选D.2.若tan =,则cos2+2sin 2等于(A)(A)(B)(C)1(D)解析:当tan =时,原式=cos2+4sin cos =,故选A.3.已知在ABC中,3sin A+4cos B=6,3cos A+4sin B=1,则角C的大小为(A)(A)(B)(C)或(D)或解析:已知式平方和得9+16+24sin(A+B)=37,因而sin(A+B)=.在ABC中,sin C=sin-(A+B)=sin(A+B)=,因而C=或,又3cos A+4sin B=1化为4sin B=1-3cos A0,所以cos A,故C=,故选A.4.(2018安阳二模)已知为第二象限角,且sin 2=-,则cos -sin 的值为(B)(A)(B)-(C)(D)-解析:因为sin 2=2sin cos =-,即1-2sin cos =,所以(sin -cos )2=,又为第二象限角,所以cos sin ,则cos -sin =-.故选B.5.(2018湖北黄冈二模)若cos(-)=-,则cos(-)+cos 等于(C)(A)- (B) (C)-1 (D)1解析:由cos(-)+cos =cos +sin +cos =cos(-)= -1,故选C.6.已知向量m=(sin ,1),n=(cos ,cos2),f(x)=mn,若f(x)=1,则cos(x+)的值为(A)(A) (B) (C)- (D)-解析:因为f(x)=mn=sin cos +cos2=sin +cos +=sin(+)+,而f(x)=1,所以sin(+)=,所以cos(x+)=cos 2(+)=1-2sin2(+)=.故选A.7.已知,为锐角,且cos(+)=,sin =,则cos 的值为(A)(A)(B)(C)(D)解析:根据题意,为锐角,且sin =,则cos =,若cos(+)=,则+也为锐角,则sin(+)=,则cos =cos(+)-=cos(+)cos +sin(+)sin =+=,故选A.8.(2018昆明质检)已知0,-0,cos(+)=,cos(-)=,则cos(+)=.解析:因为0,-0,所以+,-,又cos(+)=, cos(-)=,所以sin(+)=,sin(-)=,则cos(+)=cos(+ )-(-)=.答案:9. 如图,O与x轴的正半轴的交点为A,点B,C在O上,且B(,-),点C在第一象限,AOC=,BC=1,则cos(-)=.解析:由B(,-),得OB=OC=1,又BC=1,所以BOC=,由三角函数的定义,得sinAOB=,cosAOB=,所以sin =sin(-AOB)=sin cosAOB-cos sinAOB=-=,同理cos =,所以cos(-)=cos cos +sin sin =-+=-.答案:-能力提升(建议用时:25分钟)10.已知在ABC中,sin Asin B=cos2,则下列结论一定成立的是(A)(A)A=B(B)A=C(C)B=C(D)A=B=C解析:因为sin Asin B=cos2=,所以2sin Asin B=1-cos Acos B+sin Asin B,所以cos(A-B)=1,又0A,0B,所以- A-Bbcd(B)badc(C)dabc(D)cadb解析:a=(sin 56-cos 56)=sin(56-45)=sin 11,b=cos 50cos 128+cos 40cos 38=sin 50cos 38-cos 50sin 38=sin 12,c=cos 81=sin 9,d=(cos 80-2cos250+1)=(cos 80-cos 100)=sin 10.得badc.故选B.12.(2018岳阳质检)若tancos=sin-msin,则实数m的值为(A)(A)2 (B) (C)2 (D)3解析:由tancos=sin-msin,可得sincos=cossin-msincossin(-)=-msincos2sin=msinm=2.故选A.13.(2018洛阳二模)已知tan(+)=,且为第二象限角,若=,则sin(-2)cos 2-cos(-2)sin 2等于(D)(A)-(B)(C)-(D)解析:tan(+)=,所以tan =-,又为第二象限角,所以cos =-,sin(-2)cos 2-cos(-2)sin 2=sin(-4)= sin(-)=-cos =,故选D.14.已知sin =,cos(+)=-,若,是锐角,则=.解析:sin =,cos(+)=-,是锐角,则cos =,sin(+ )=,所以cos =cos(+)-=cos(+)cos +sin(+)sin =,所以=.答案:15.(2018长春二模)已知关于x的方程2x2-(+1)x+m=0的两个根为sin 和cos ,(0,2),求:(1)+的值;(2)m的值;(3)方程的两根及的值.解: