2020版高考數(shù)學(xué)復(fù)習(xí)第三單元第21講簡單的三角恒等變換練習(xí)文（含解析）新人教A版.docx

第21講簡單的三角恒等變換 1.若cos2-=23,則cos(-2)=()A.29B.59C.-29D.-592.已知sin-4=33,則sin2=()A.13B.-23C.255D.-2333.設(shè)函數(shù)f(x)=sinx+4+cosx-4,則()A.f(x)=-fx+2B.f(x)=f-x+2C.f(x)fx+2=1D.f(x)=-f-x+24.2018宿州一模 若sin6-=14,則cos2-3的值為.5.(1+3tan10)cos40=. 6.已知0,22,且sin,sin2,sin4成等比數(shù)列,則的值為()A.6B.3C.23D.347.2018貴州聯(lián)考 已知sin-2cos=102,則tan2=()A.43B.-34C.34D.-438.2018唐山期末 已知cos36cos72=14,由此可算得cos36=()A.5+14B.5-12C.3+14D.3+249.設(shè)a=cos50cos127+cos40cos37,b=22(sin56-cos56),c=1-tan2391+tan239,則a,b,c的大小關(guān)系是()A.abcB.bacC.cabD.acb10.定義運算abcd=ad-bc.若cos=17,sinsincoscos=3314,02,則等于()A.12B.6C.4D.311. 已知sin=35,2,則cos22sin+4=.12.函數(shù)f(x)=3sin23x-2sin213x2x34的最小值是. 13.2018四川宜賓期中 已知函數(shù)f(x)=cosx-3-sin2-x.(1)求函數(shù)f(x)的最小正周期;(2)若0,2,且f+6=35,求f(2)的值.14.2018湖南衡陽聯(lián)考 已知函數(shù)f(x)=sin54-x-cos4+x.(1)求函數(shù)f(x)的單調(diào)遞增區(qū)間;(2)已知cos(-)=35,cos(+)=-35,00,所以x=5+14,故選A.9.D解析 由三角恒等變換公式,可得a=cos50cos127+cos40cos37=cos50cos127+sin50sin127=cos(50-127)=cos(-77)=cos77=sin13,b=22(sin56-cos56)=22sin56-22cos56=sin(56-45)=sin11,c=1-tan2391+tan239=1-sin239cos2391+sin239cos239=cos239-sin239=cos78=sin12.因為函數(shù)y=sinx,x0,90為增函數(shù),所以sin13sin12sin11,所以acb,故選D.10.D解析 由題設(shè)得sincos-cossin=sin(-)=3314.02,cos(-)=1314.又cos=17,sin=437.故sin=sin-(-)=sincos(-)-cossin(-)=4371314-173314=32,=3.11.-75解析cos22sin(+4)=cos2-sin22(22sin+22cos)=cos-sin.sin=35,2,cos=-45,原式=-75.12.3-1解析f(x)=3sin23x-1-cos23x=2sin23x+6-1,2x34,223x+623,f(x)min=f34=2sin23-1=3-1.13.解:(1)f(x)=12cosx+32sinx-cosx=32sinx-12cosx=sinx-6,函數(shù)f(x)的最小正周期為2.(2)由(1)知f(x)=sinx-6,f+6=sin+6-6=sin=35.0,2,cos=1-sin2=1-(35)2=45,sin2=2sincos=23545=2425,cos2=2cos2-1=2452-1=725,f(2)=sin2-6=32sin2-12cos2=322425-12725=243-750.14.解:(1)f(x)=sin54-x-cos4+x=sinx-4-sin2-4+x=2sinx-4,由-2+2kx-42+2k,kZ,得-4+2kx34+2k,kZ,故函數(shù)f(x)的單調(diào)遞增區(qū)間為-4+2k,34+2k(kZ).(2)方法一:cos(-)=35,cos(+)=-35,且02,sin(-)=-45,sin(+)=45.從而cos2=cos(+)-(-)=cos(+)cos(-)+sin(+)sin(-)=-925-1625=-1,故cos=0,02,=2,f()=2sin4=2.方法二:cos(-)=35,cos(+)=-35,coscos+sinsin=35,coscos-sinsin=-35.由+可得coscos=0,又02,cos=0,=2,f()=f2=2sin2-4=2.15.3解析cos(-310)sin(-5)=sin(-310+2)sin(-5)=sin(+5)sin(-5)=sincos5+cossin5sincos5-cossin5=sincoscos5+sin5sincoscos5-sin5=2tan5cos5+sin52tan5cos5-sin5=3sin5sin5=3.16.x=-6解析y=sin3x+3cosx-6-cos3x