3.1.2 第1课时.docx
第三章3.13.1.1第1课时A级基础巩固一、选择题1.若AABC中，C=90。，AC=3,BC=4,则Sin(A-8)的值是(D)3n4A.§B.§-24725,253434解析由条件可知cosA=,SinA=Sin8=g,COSB=sin(AB)=SinAcosB-44337cos.sinB=5×55×5=252 .设W(0,芬若Sina=则gCoS(a+；)等于(B)A.IB.解析2cos(a+j)=2(cosa2-sina*9)=55=5,3 .COsJ的值等于（C）B.A.D.小+啦4解析CO招=cos居=_(COSWCOs；_Si吟sin.0v232>62一6X2-2,2厂4'B. 2D. 245co哈si哈的值是(B)A.02-2s1n'3cosj- Sin 强=C.y2解析.12)=2(smjc0sj2-s3snj2=2sm(1一五)=2sin；=啦.5.85(工+2)+2§皿；1+必由可化简为(A)A.cosxB.sinxC.s(x÷y)D.s(-y)解析原式=cos(x+y)+y+2sin(x+y)siny=cos(x+y)coSy-Sin(X+y)siny+2sin(x+y)siny=cos(÷y)sy÷sin(x÷j)siny=cosx.356 .已知COSa=g,cos(ct÷)=-,q、夕都是锐角，则COS夕=(C)解析V,是锐角，0<。+夕<乂又cos(+0=一,<0,<a÷<,sin(12453124+£)=万，Sina=亍又Co4=cos(+6-)=cos(+6)cos+sin(+A)Sina=×÷-j×-_33-65,二、填空题7 .sinl50+sin750的值是_孝_.解析sinl50÷sin750=sin(450-300)+sin(450+300)=2sin450cos300=.8 .已知sin(-)cosacos(-)sina=m,且为第三象限角，则CoSB=一1一加.解析由sin(-)cosacos()sin«=m,得sin(-jff)=m,即SiM=一?，又：为第三象限角,三、解答题9 .化简求值：(1)cos44°sin14osin44ocos14°；(2)sin(54o-x)cos(360÷x)÷cos(540-x)sin(36o÷x)解析(1)原式=Sin(14。-44。)=sin(-30°)=(2)原式=sin(54o-)+(36o+x)=sin90o=1.10 .已知cos。=H，g(,半)，求CoS伍+/的值.解析cos=H，e(,竽)，/.Sino=-K.*.COS(e+§=cos0cos：sinsin;=-122rA2=-72一132I132-26,B级素养提升一、选择题1.在AABC中，已知Sin(A-B)cosB+cos(A-5)sin821,则AABC是(C)A.锐角三角形B.钝角三角形C.直角三角形D.等腰非直角三角形解析由题设知sin(AB)+B21,sinA21而SirL4W1,sinA=l,A=菱,ZXABC是直角三角形.2、行32.若a、均为锐角，Sina=?-,sin(+份=则COSS等于(B)A.C.255童维5 -X 25B.D.252525 一 25解析与夕均为锐角，且Sina=2Z>sin(+6)=,.+/?为钝角,34又由sin(+份=§得，cos(+A)=-g,由Sina=邛得，COSQ=培，X手=需故4/.CoS"=CoS(+£)=cos(+0)CoSa+sin(+为Sina=-§X选B.443.已知COS(+0)=g,cos(a-fi)=-jf则COSaCOS夕的值为(A)4A.0B.gC.0或3D.0或与4解析由条件得，cosacosQ-sinsi印=子4cosacos÷sinasin/?=一亍左右两边分别相加可得coscosy7=0.sin470-SinI7°cos3004cos17°(C)A.乎B.VC.ID.当解析Sin470-sinl708s30°CoSl70sin(300+17°)-sinl70cos300cos17°sin300cos170+cos300sin17。一Sin17。COS30。cos17°皿端j30。=；.二、填空题45.若cos(a÷)cosa÷sin(a÷)sina=,且4500<少<540°,3+4310一4解析由已知得cos(a+0)a=COS尸=一§,3V450o<<540o,.siM=g,.sin(60f=坐X(-力宝=-若黑316.若cos(a+0=g,cos(a-则tanatanS=_/_.sin(60o-)=-解析由已知，得R1COS(Q+6)=§,3cos(a一4)=亍CoSaCOs£-sinasiM=g,即彳3CoSaCOSA+sinasiM=.(COS«COS=T,；sinasin/?=.CSinaSinS1所以tantaM=CoSacosA=三、解答题7 .已知向量=(cos,sina),力=(CoS夕，Sins),|。一例二=求cos(一£)的值;(2)若一5<kO<<g,且Si邛=一=求Sina的值.25解析"Tl=等，-4.*.a12ab+Z>2=予又=(cos,sina),O=(COS6，sin/?),.*.a1=b2=1,ab=cosacos÷sinasin=cos(-),3：,cos(a-)=.(2),5<<0<tt<,，0<a0<,34由(1)得cos()=§,sin(-5)=j,512又SiM=一舌，:.cos"=百，.Sina=Sin(a0)+0=sin(a)cos÷cos(a一mSin=5×T3+5×(-)=8 .已知CoSa=乎，sin(a且a、SS(O,外求:(I)COS(2a彼)的值；Q)P的值.解析因为a、(0,分，所以a-Q(苫，芬又Sinm一份=>0所以sina=Icos2a=,cos(a)=1-sin2(a-?)=cos(2-)cos÷()=coscos(-)SinaSin(a一4)_a/53K)25VT0V2-510510-10,(2)cos=cosa(a)=cosacos(a-)÷SinaSin(a一尸)5v310,25vIO2-510510-2,又因为夕(0,2),所以C级实力拔高已知cos(a一份=-H，cos(a+S)=I且a),a+Q管，2兀).求cos2a,cos2£及角尸的值.思路分析探讨角的范围时，a一般看作。+(一为，先求出一夕的范围，再求a÷(-.)的范围.解析由a6£像),且cos(a-4)=-H，得sin(a-£)=.由。+尸£(即，2),且cos(a+)=.得sin(a+用=一卷.*.cos2a=cos(a+0)+(X)=cos(a÷)cos(a-)sin(a+为Sin(Q-)=_1212_CJAA=_1191313I1313-169,cos2=cos(a+)-(a-)=cos(a÷夕)cos(a-)÷sin(a÷6)sin(a-)=13×13+()×=1
