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1、[学业水平训练]1•sin330。等于•解析:sin330°=sin(360°-30°)=-sin30。=—小、sina..tan(27i—a)=—tana=—.答案:
2、3.已知sin(45°+a)=-A',则sin(225°+«)=.解析:sin(225°+a)=sin(180。+45。+a)=—sin(45°+a)=—备.答案:-春4.已知cos(a—兀)=—咅,且a是第四象限角,贝ljsin(—2兀+°)=解析:由cos(«—7t)=—yy,易得cosa=又因为sin(—2兀+a)=sina,所以只需求出sina即可.19•:a
3、是第四象限角,a=—/1—cos2(z=一盲・答案:-1235.在厶ABC中,若cosA=VI则sin(7L/)=解析:VA是中的内角,・:sin(7i—A)=sin/=彳1—cos?/=*,fZcosA=±/1—sin2z4=土专.答案:
4、土爭I兀6.已知sin(7T—a)=Iog叼,且么丘(一刁0),则tan(27i—a)的值为12_2~32^5pl—sin%5cosa解析:Tsin(7t—a)=sina=log8^=—答案:爭6.求下列三角函数式的值:(1)sin(-330o)cos210°;(2)V3sin(-l200°
5、)tan(-30°)-cos585°tan(-l665°).解:(l)sin(-330°)cos210°=sin(30°-360°)cos(180°+30°)=sin30°(—cos30°)=^x(—(2)V3sin(-l200°)tan(-30°)-cos585°tan(-l665°)=一迈sin1200°(-^)-cos(7200-1350)tan(-9xl800-45°)=sin(l080°+120°)-cos135°tan(-45°)爭(爭十呼8.化简下列各式.(1)sin[a+(2n+l)jc]+sin[a—(2幵+1)兀]
6、sin(«+2/77t)・cos(a—2/m)(用Z);cos190°-sin(-210°)(2)cos(-350°)-tan(-585°)*解:⑴原式=sin(兀+a)+sin(a—兀)sinacosa—2sina_2sinacosacosa.cos(180。+10。)[—Sin(180。+30。)](2)原A=cos(360。一10。)[—tan(360。+225。)]—cos10°sin30°=cos10°-[-tan(180°+45°)]1~2_1—tan45°工[高考水平训练]1・已知tan(3jr—a)=2.rltsin(
7、a—3兀)+cos(兀~a),,z.u则sin(-.)-cos(k+.)为解析:Vtan(3n—a)=2t/.tana=—29原式可化为:—sin«—cosa—tana—12—11—sina+cosa—tana+12+13答案:I2.(2014-抚州质检)若函数f(x)=6/sin(7tr+a)+/>cos(7tr+^),其中a,b,a,0都是非零实数,且满足./(2013)=2,则,/(2014)=.解析:・・刃2013)=asin(2013兀+a)+bcos(2013兀+0)=2,/./(2014)=asin(20147T+«)+
8、/?cos(2014兀+0)=asin[兀+(2013;r+a)]+bcos[兀+(2013兀+0)]=-[asin(2013兀+a)+bcos(2013兀+0)]=—2.答案:一2Ql+2sin290°cos430。3,化间:sin250°+cos790°-pl+2sin(360。一70。)cos(360。+70。)解:丿小式二sin(180°+70°)+cos(2x360°+70°)•/l-
9、-2sin(—70°)cos7帀y](sin70°—cos70°)?=-sin70°+cos70°=cos70°-sin70°sin70°
10、—cos70°cos70°—sin70°=一1・4.已知tan(x+#7t)=a.1513sin(亏兀+x)+3cos(x—石ti)丄°圭'T77_a+3巫叱・,20、(丄22、一Sin(_7T—X)—COS(X十〒71)sin(号兀+x)+3cos(x-学tt)证明s2022sin(亍r—x)—cos(x+〒7T)sin[兀+(x8]+3cos(x+尹一3兀)88sin[4n—(x+^jc)]—cos[2兀+(x+yn)]88—sin(x+y7t)+3cos[(x+尹)~n]sin[—(8)]—cos(x+尹)8—sin(x+y7t
11、)—3cosCx—sin(x+号兀)—cos(xtan()+3_°+3tan&+%)我的写字心得体会从小开始练习写字,几年来我认认真真地按老师的要求去练习写字。以前练习写字,大多是在印有田字格或米字格的练习本上进行。教材