1、2.3两角和与差的正切函数课后篇巩固探究1.已知-2,0,sin =-45,则tan+4=()A.-7B.-17C.17D.7解析-2,0,cos =35,tan =-43.tan+4=tan+11-tan=-17.答案B2.已知tan(+)=25,tan-4=14,那么tan+4=()A.1318B.1322C.322D.-322解析因为+4=(+)-4,所以tan+4=tan(+)-4=tan(+)-tan-41+tan(+)tan-4=25-141+2514=322,故选C.答案C3.若A=15,B=30,则(1+tan A)(1+tan B)的值为()A.1B.2C.-1D.-2解析由
2、结论A+B=45,则(1+tan A)(1+tan B)=2.答案B4.若tan =lg(10a),tan =lg1a,且+=4,则实数a的值为()A.1B.110C.1或110D.1或10解析tan +tan =lg(10a)+lg1a=lg 10=1.+=4,tan4=tan(+)=tan+tan1-tantan=11-tantan=1,tan tan =0,则有tan =lg(10a)=0或tan =lg1a=0,10a=1或1a=1,即a=110或1,故选C.答案C5.若锐角,使+2=23,tan2tan =13同时成立,则+的值为()A.512B.2C.712D.56解析+2=23,
3、2+=3,tan2+=tan2+tan1-tan2tan=tan2+tan1-13=3,即tan2+tan =233,tan2,tan 是x2-233x+13=0的两个根,解得tan2=tan =33.又,均为锐角,2=6,故+=2.答案B6.(2018全国高考)已知tan-54=15,则tan =.解析tan-54=tan-tan541+tantan54=tan-11+tan=15,5tan -5=1+tan .tan =32.答案327.tan 23+tan 37+3tan 23tan 37的值是.解析tan 60=3=tan23+tan371-tan23tan37,tan 23+tan
4、37=3-3tan 23tan 37,tan 23+tan 37+3tan 23tan 37=3.答案38.已知0,2,且tan+4=3,则log5(sin +2cos )+log5(3sin +cos )=.解析利用两角和的正切公式得tan+4=tan+11-tan=3,tan =12.log5(sin +2cos )+log5(3sin +cos )=log53sin2+7sincos+2cos2sin2+cos2=log53tan2+7tan+2tan2+1=log55=1.答案19.导学号93774095已知tan =3.(1)求tan-4的值;(2)求sin+cossin-2cos的
5、值.解(1)tan-4=tan-tan41+tantan4=3-11+31=12.(2)由tan =3,得cos 0,所以sin+cossin-2cos=tan+1tan-2=3+13-2=4.10.导学号93774096已知tan =-13,cos =55,(0,).(1)求tan(+)的值;(2)求函数f(x)=2sin(x-)+cos(x+)的最大值.解(1)cos =55,(0,),sin =255,tan =2,tan(+)=tan+tan1-tantan=-13+21-132=1.(2)tan =-13,(0,),sin =1010,cos =-31010,f(x)=2(sin x
6、cos -cos xsin )+(cos xcos -sin xsin )=-355sin x-55cos x+55cos x-255sin x=-5sin x.又-1sin x1,f(x)的最大值为5.11.在ABC中,求证:tanA2tanB2+tanB2tanC2+tanC2tanA2=1.证明左边=tanA2tanB2+tanC2+tanB2tanC2=tanA2tanB+C21-tanB2tanC2+tanB2tanC2=tanA2tan2-A21-tanB2tanC2+tanB2tanC2=tanA21tanA21-tanB2tanC2+tanB2tanC2=1=右边.故原等式成立.
