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1、考點(diǎn)規(guī)范練22三角恒等變換一、基礎(chǔ)鞏固1.2sin47-3sin17cos17=()A.-3B.-1C.3D.1答案D解析原式=2sin47-sin17cos30cos17=2sin(17+30)-sin17cos30cos17=2sin30=1.故選D.2.已知2sin 2=1+cos 2,則tan 2=()A.43B.-43C.43或0D.-43或0答案C解析因?yàn)?sin2=1+cos2,所以2sin2=2cos2.所以2cos(2sin-cos)=0,解得cos=0或tan=12.若cos=0,則=k+2,kZ,2=2k+,kZ,所以tan2=0.若tan=12,則tan2=2tan1-

2、tan2=43.綜上所述,故選C.3.已知函數(shù)f(x)=3sin xcos x+3cos2x(0)的最小正周期為2,將函數(shù)f(x)的圖象向左平移(0)個(gè)單位后,得到的函數(shù)圖象的一條對(duì)稱軸為x=8,則的值不可能為()A.524B.1324C.1724D.2324答案B解析f(x)=3sinxcosx+3cos2x=32sin2x+31+cos2x2=3sin2x+6+32,22=2,即=2,f(x)=3sin4x+6+32.平移后的函數(shù)為g(x)=3sin4(x+)+6+32=3sin4x+4+6+32.由題意,得48+4+6=k+2,kZ,解得=k4-24,kZ,故選B.4.已知f(x)=si

3、n2x+sin xcos x,則f(x)的最小正周期和一個(gè)單調(diào)遞增區(qū)間分別為()A.,0,B.2,-4,34C.,-8,38D.2,-4,4答案C解析由f(x)=sin2x+sinxcosx=1-cos2x2+12sin2x=12+2222sin2x-22cos2x=12+22sin2x-4,則T=22=.又2k-22x-42k+2(kZ),k-8xk+38(kZ)為函數(shù)的單調(diào)遞增區(qū)間.故選C.5.已知5sin 2=6cos ,0,2,則tan 2=()A.-23B.13C.35D.23答案B解析由題意,知10sincos=6cos,又0,2,sin=35,cos=45,tan2=sin2co

4、s2=2sin222sin2cos2=1-cossin=1-4535=13.6.已知tan+4=-12,且2,則sin2-2cos2sin-4等于()A.255B.-3510C.-255D.-31010答案C解析sin2-2cos2sin-4=2sincos-2cos222(sin-cos)=22cos,由tan+4=-12,得tan+11-tan=-12,解得tan=-3.因?yàn)?0,02的圖象的相鄰兩對(duì)稱軸之間的距離為,且在x=6時(shí)取得最大值2,若f()=95,且623,則sin2+23的值為()A.1225B.-1225C.2425D.-2425答案D解析由題意知,T=2,即T=2=2,即

5、=1.又當(dāng)x=6時(shí),f(x)取得最大值,即6+=2+2k,kZ,即=3+2k,kZ.02,=3,f(x)=sinx+3+1.f()=sin+3+1=95,可得sin+3=45.623,可得2+30),則A=,b=.答案21解析因?yàn)?cos2x+sin2x=1+cos2x+sin2x=2sin2x+4+1,所以A=2,b=1.9.(2018江蘇,16)已知,均為銳角,且tan =43,cos(+)=-55.(1)求cos 2的值;(2)求tan(-)的值.解(1)因?yàn)閠an=43,tan=sincos,所以sin=43cos.因?yàn)閟in2+cos2=1,所以cos2=925,因此cos2=2co

6、s2-1=-725.(2)因?yàn)?為銳角,所以+(0,).又因?yàn)閏os+=-55,所以sin(+)=1-cos2(+)=255,因此tan(+)=-2.因?yàn)閠an=43,所以tan2=2tan1-tan2=-247.因此,tan(-)=tan2-(+)=tan2-tan(+)1+tan2tan(+)=-211.10.已知函數(shù)f(x)=sinx-6+cosx-3-2sin2x2(0)的周期為.(1)求的值;(2)若x0,2,求f(x)的最大值與最小值.解(1)函數(shù)f(x)=sinx-6+cosx-3-2sin2x2=sinxcos6-cosxsin6+cosxcos3+sinxsin3-21-co

7、sx2=3sinx+cosx-1=2sinx+6-1(0),f(x)的周期為2=,=2.(2)x0,2,2x+66,76.sin2x+6-12,1.f(x)的最大值為1,最小值為-2.11.已知點(diǎn)4,1在函數(shù)f(x)=2asin xcos x+cos 2x的圖象上.(1)求a的值和f(x)的最小正周期;(2)求函數(shù)f(x)在(0,)內(nèi)的單調(diào)遞減區(qū)間.解(1)函數(shù)f(x)=2asinxcosx+cos2x=asin2x+cos2x.f(x)的圖象過(guò)點(diǎn)4,1,即1=asin2+cos2,可得a=1.f(x)=sin2x+cos2x=2sin2x+4.函數(shù)f(x)的最小正周期T=22=.(2)由2k

8、+22x+432+2k,kZ,可得k+8x58+k,kZ.函數(shù)f(x)的單調(diào)遞減區(qū)間為k+8,58+k,kZ.x(0,),當(dāng)k=0時(shí),可得單調(diào)遞減區(qū)間為8,58.二、能力提升12.已知m=tan(+)tan(-+),若sin2(+)=3sin 2,則m=()A.-1B.34C.32D.2答案D解析sin2(+)=3sin2,sin(+)-(-)=3sin(+)-(+-),sin(+)cos(-)-cos(+)sin(-)=3sin(+)cos(+-)-3cos(+)sin(+-),即-2sin(+)cos(+-)=-4cos(+)sin(+-),tan(+)=2tan(+-),故m=tan(+

9、)tan(-+)=2,故選D.13.已知cos =13,cos(+)=-13,且,0,2,則cos(-)的值等于()A.-12B.12C.-13D.2327答案D解析0,2,2(0,).cos=13,cos2=2cos2-1=-79,sin2=1-cos22=429,又,0,2,+(0,),sin(+)=1-cos2(+)=223,cos(-)=cos2-(+)=cos2cos(+)+sin2sin(+)=-79-13+429223=2327.14.已知函數(shù)f(x)=2sinx+524cosx+524-2cos2x+524+1,則f(x)的最小正周期為;函數(shù)f(x)的單調(diào)遞增區(qū)間為.答案k-3

10、,k+6(kZ)解析f(x)=2sinx+524cosx+524-2cos2x+524+1=sin2x+512-cos2x+512=2sin2x+512cos4-cos2x+512sin4=2sin2x+512-4=2sin2x+6.f(x)的最小正周期T=22=.因此f(x)=2sin2x+6.當(dāng)2k-22x+62k+2(kZ),即k-3xk+6(kZ)時(shí),函數(shù)f(x)的單調(diào)遞增區(qū)間是k-3,k+6(kZ).15.已知函數(shù)f(x)=23sinx+6cos x(02),且f(x)的圖象過(guò)點(diǎn)512,32.(1)求的值及函數(shù)f(x)的最小正周期;(2)將y=f(x)的圖象向右平移6個(gè)單位長(zhǎng)度,得到

11、函數(shù)y=g(x)的圖象,已知g2=536,求cos2-3的值.解(1)函數(shù)f(x)=23sinx+6cosx=23sinx32+23cosx12cosx=3sin2x+6+32.f(x)的圖象過(guò)點(diǎn)512,32,3sin2512+6+32=32,2512+6=k,kZ,即=6k-15.再結(jié)合00,所以sinA=32,所以A=3.(2)因?yàn)锳+B+C=,所以sinB=sin-(A+C)=sin(A+C),所以sinB+sinC=sin3+C+sinC=32cosC+32sinC=3sinC+6.因?yàn)樵阡J角三角形ABC中,A=3,所以B+C=23,B=23-C,所以023-C2,0C2,故6C2,由正弦函數(shù)的單調(diào)性可知,sinB+sinC的取值范圍為32,3.9