1、2.2两角和与差的正弦、正切公式及其应用课后篇巩固提升基础达标练1.(2020北京昌平区新学道临川学校期中)sin 20cos 10-cos 160sin 10=()A.-32B.32C.-12D.12解析sin20cos10-cos160sin10=sin20cos10+cos20sin10=sin30=12.故选D.答案D2.若tan =12,tan =13,且,32,0,2,则+的大小等于()A.4B.54C.74D.94解析由已知得tan(+)=tan+tan1-tantan=12+131-1213=1.又因为,32,0,2,所以+(,2),于是+=54.答案B3.若tan(+)=25
2、,tan(-)=14,则tan 2=()A.16B.2213C.322D.1318解析tan2=tan(+)+(-)=tan(+)+tan(-)1-tan(+)tan(-)=25+141-2514=1318.答案D4.sin(+75)+cos(+45)-3cos(+15)的值等于()A.1B.1C.-1D.0解析原式=sin(+45)+30+cos(+45)-3cos(+45)-30=32sin(+45)+12cos(+45)+cos(+45)-332cos(+45)+12sin(+45)=32sin(+45)+32cos(+45)-32cos(+45)-32sin(+45)=0.答案D5.设
3、0,2,0,2,且tan =1+sincos,则()A.3-=2B.3+=2C.2-=2D.2+=2解析由tan=1+sincos,得sincos=1+sincos,得sincos-cossin=cos,sin(-)=sin2-.又0,2,0,2,故-=2-,即2-=2.答案C6.已知tan =12,则tan(4+)-11+tan(4+)的值是()A.2B.12C.-1D.-3解析tan(4+)-11+tan(4+)=tan(4+)-tan41+tan(4+)tan4=tan4+-4=tan=12.答案B7.tan 23+tan 37+3tan 23tan 37的值是.解析因为tan60=3=
4、tan23+tan371-tan23tan37,所以tan23+tan37=3-3tan23tan37,所以tan23+tan37+3tan23tan37=3.答案38.已知,都是锐角,sin =45,cos(+)=513,则cos =;sin =.解析因为,都是锐角,所以+(0,),又sin=45,cos(+)=513,所以cos=35,sin(+)=1213,所以sin=sin(+)-=sin(+)cos-cos(+)sin=121335-51345=1665.答案3516659.化简求值:(1)sin(+)cos(-)+cos(+)sin(-);(2)cos(70+)sin(170-)-
5、sin(70+)cos(10+);(3)cos 21cos 24+sin 159sin 204.解(1)原式=sin(+-)=sin2.(2)原式=cos(70+)sin(10+)-sin(70+)cos(10+)=sin(10+)-(70+)=sin(-60)=-32.(3)原式=cos21cos24+sin(180-21)sin(180+24)=cos21cos24-sin21sin24=cos(21+24)=cos45=22.能力提升练1.已知2,tan+4=17,则sin +cos =()A.-17B.-25C.-15D.15解析因为tan+4=1+tan1-tan=17,所以tan=
6、-34,即sin=-34cos,由平方关系得(-34cos)2+cos2=1,解得cos=-45,sin=35,sin+cos=35-45=-15.答案C2.在ABC中,角A,B,C的对边分别是a,b,c,且acos B=(2c-b)cos A,则角A的大小为()A.6B.4C.3D.2解析由正弦定理得sinAcosB=(2sinC-sinB)cosA,即sin(A+B)=2sinCcosA,即sinC=2sinCcosA,即cosA=22,故A=4.答案B3.设,都为锐角,且cos =55,sin(+)=35,则sin 等于()A.2525B.11525C.55D.-55或11525解析因为
7、为锐角,cos=55,所以sin=255.因为,都为锐角,所以0+0,tan0;又,(0,),所以,0,2,即+(0,),因此+=4.答案47.已知cosx-4=210,x2,34.(1)求sin x的值;(2)求sinx+3的值.解(1)sinx=sinx-4+4=sinx-4cos4+cosx-4sin4,=22sinx-4+cosx-4=22sinx-4+210,因为x2,34,所以x-44,2,所以sinx-4=1-2100=7210,所以sinx=228210=45.(2)因为sinx=45,x2,34,故cosx=-35,sinx+3=sinxcos3+cosxsin3=1245+
8、32-35=4-3310.素养培优练1.定义运算abcd=ad-bc.若cos =17,sin sin cos cos =3314,02,则=.解析依题设得,sinsincoscos=sincos-cossin=sin(-)=3314.因为02,所以cos(-)=1314.又因为cos=17,所以sin=437,所以sin=sin-(-)=sincos(-)-cossin(-)=4371314-173314=32,所以=3.答案32.是否存在锐角,使得+2=23,tan2tan =2-3同时成立?若存在,求出锐角,的值;若不存在,请说明理由.解假设存在锐角,使得+2=23,tan2tan=2-3同时成立.由得2+=3,所以tan2+=tan2+tan1-tan2tan=3.又tan2tan=2-3,所以tan2+tan=3-3,因此tan2,tan可以看成是方程x2-(3-3)x+2-3=0的两个根,解得,x1=1,x2=2-3.若tan2=1,则=2,这与为锐角矛盾,所以tan2=2-3,tan=1,所以=6,=4,所以满足条件的,存在,且=6,=4.
