高考解答題專(zhuān)項(xiàng)二　三角函數(shù)中的綜合問(wèn)題

1、 高考解答題專(zhuān)項(xiàng)二三角函數(shù)中的綜合問(wèn)題1.已知函數(shù)f(x)=2sin xcosx-6-12(00,0B,fA-B2-12=35,求cosA-B2,并證明sin A255.6.(2021河南鄭州高三三模)在ABC中,AB=2AC,點(diǎn)D在BC邊上,AD平分BAC.(1)若sinABC=55,求cosBAC;(2)若AD=AC,且ABC的面積為7,求BC.高考解答題專(zhuān)項(xiàng)二三角函數(shù)中的綜合問(wèn)題1.解 f(x)=2sin xcosx-6-12=2sin xcos xcos6+sin xsin6-12=3cos xsin x+sin2x-12=32sin 2x-12cos 2x=sin2x-6.若f(x)

2、在a,b上單調(diào)遞增,且b-a的最大值為2,則T=22,故=1,所以f(x)=sin2x-6.由2+2k2x-632+2k(kZ),得3+kx56+k(kZ),令k=0,得3x56;令k=-1,得-23k-6.又-2x2,所以f(x)在-2,2上單調(diào)遞減區(qū)間為-2,-6,3,2.2.解 (1)由2tanBtanA+tanB=bc及正弦定理可知,2sinBcosBsinAcosA+sinBcosB=sinBsinC,所以2sinBcosBcosAcosBsin(A+B)=sinBsinC,因此2cos A=1.又A(0,),所以A=3.(2)由余弦定理a2=b2+c2-2bccos A,得13=9

3、+c2-3c,所以c2-3c-4=0,即(c-4)(c+1)=0,解得c=4.從而SABC=12bcsin A=123432=33.3.解 (1)由正弦定理得sin Bcos C+3sin Bsin C=sin A+sin C=sin(B+C)+sin C=sin Bcos C+cos Bsin C+sin C.因?yàn)镃為三角形內(nèi)角,sin C0,所以3sin B-cos B=1,sinB-6=12.因?yàn)?6B-63,則B-6=6,即B=3.(2)由已知S=12acsin B=34ac=3,得ac=4.又a2+c2-b2=2accos B,即a2+c2-4=ac,解得a=c=2.4.解 (1)連

4、接BD,在RtBAD中,由AB=4,AD=3,BAD=90,得BD=5,sinABD=35,cosABD=45.ABC=45,DBC=45-ABD,sinDBC=sin 45cosABD-cos 45sinABD=2245-2235=210.在RtBCD中,由BCD=90,知CD=BDsinDBC=5210=22.(2)連接AC,由(1)知BD=5,在RtABD中易知sinABD=35,cosABD=45.在RtBCD中,由BC=25,BD=5,得CD=5.易知sinCBD=55,cosCBD=255.cosABC=cos(ABD+CBD)=cosABDcosCBD-sinABDsinCBD=

5、45255-3555=55.在ABC中,由余弦定理得AC2=AB2+BC2-2ABBCcosABC=42+(25)2-242555=20,AC=25.5.解 (1)由f(0)=12,得sin =12.又00,結(jié)合函數(shù)圖象可知122512,所以0125.又kZ,所以k=0,從而=2,因此f(x)=sin2x+6.(2)由fA-B2-12=sin(A-B)=35,因?yàn)?BA2,所以0A-B2,故A=A+B2+A-B24+A-B2.又y=sin x在0,2上單調(diào)遞增,且A0,2,4+A-B20,2,所以sin Asin4+A-B2=sin4cosA-B2+cos4sinA-B2=2231010+10

6、10=255.6.解 (1)令A(yù)BC的邊AC,AB,BC為b,c,a,由題意可得c=2b,ABAC,ABCACB,ABC為銳角,即cosABC=1-15=255.ACsinABC=ABsinACB,sinACB=255.ACB(0,),cosACB=55.cosBAC=-cos(ABC+ACB)=sinABCsinACB-cosABCcosACB.當(dāng)cosACB=55時(shí),cosBAC=55255-25555=0.當(dāng)cosACB=-55時(shí),cosBAC=55255+25555=45.所以cosBAC=0或45.(2)設(shè)CAD=DAB=,由于SABC=SACD+SADB,所以12ACADsin +12ABADsin =12ABACsin 2,由AD=AC,AB=2AC可得3sin =4sin cos .因?yàn)閟in 0,則cos =34,sin =1-cos