2022年新教材高考數(shù)學(xué)一輪復(fù)習(xí)考點(diǎn)規(guī)范練21三角恒等變換含解析新人教版

1、PAGE PAGE 6考點(diǎn)規(guī)范練21三角恒等變換一、基礎(chǔ)鞏固1.已知,32,且cos =-45,則tan4-等于()A.7B.17C.-17D.-72.2sin47-3sin17cos17等于()A.-3B.-1C.3D.13.(多選)下列選項(xiàng)中,與sin-116的值相等的是()A.2cos215-1B.cos 18cos 42-sin 18sin 42C.2sin 15sin 75D.tan30+tan151-tan30tan154.已知cos-6+sin =435,則sin+76的值為()A.12B.32C.-45D.-125.已知2sin 2=1+cos 2,則tan 2等于()A.43

2、B.-43C.43或0D.-43或06.已知5sin 2=6cos ,0,2,則tan2等于()A.-23B.13C.35D.237.若0yxbcB.bacC.cabD.acb13.(多選)化簡下列各式,與tan 相等的是()A.1-cos21+cos2B.1+cos(+2)21cos,(0,)C.1-cos2sin2D.sin21-cos214.已知sin 10+mcos 10=2cos 140,則m=.15.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.三、探究創(chuàng)新16.已知函數(shù)f(x)=cos x(

3、sin x+3cosx)(0),若存在實(shí)數(shù)x0,使得對任意的實(shí)數(shù)x,都有f(x0)f(x)f(x0+2 020)成立,則的最小值為()A.12020B.14032C.12020D.1403217.已知函數(shù)f(x)=1+1tanxsin2x-2sinx+4sinx-4.(1)若tan =2,求f()的值;(2)若x12,2,求f(x)的取值范圍.考點(diǎn)規(guī)范練21三角恒等變換1.B因?yàn)?32,且cos=-45,所以sin=-35,所以tan=34.所以tan4-=1-tan1+tan=1-341+34=17.2.D原式=2sin47-sin17cos30cos17=2sin(17+30)-sin17

4、cos30cos17=2sin30=1.故選D.3.BCsin-116=sin6=12.A選項(xiàng)中,2cos215-1=cos30=32,不相等;B選項(xiàng)中,cos18cos42-sin18sin42=cos(18+42)=cos60=12,相等;C選項(xiàng)中,2sin15sin75=2sin15cos15=sin30=12,相等;D選項(xiàng)中,tan30+tan151-tan30tan15=tan45=1,不相等.4.Ccos-6+sin=32cos+32sin=435,12cos+32sin=45.則sin+76=-sin+6=-(32sin+12cos)=-45.5.C因?yàn)?sin2=1+cos2,

5、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-tan2=43.綜上所述,故選C.6.B由題意,知10sincos=6cos,0,2,sin=35,cos=45,tan2=sin2cos2=2sin222sin2cos2=1-cossin=1-4535=13.7.B0yxsin12sin11,acb.故選D.13.BC對于A,1-cos21+cos2=1-(1-2sin2)1+2cos2-1=sin2cos2=tan2=|tan|,由1

6、-cos21+cos20,解得-1cos21,即2+2k(kZ),解得2+k(kZ),故A不符合題意;對于B,因?yàn)?0,),所以1+cos(+2)21cos=1-cos221cos=sin21cos=|sin|cos=sincos=tan,故B符合題意;對于C,1-cos2sin2=2sin22sincos=sincos=tan,故C符合題意;對于D,sin21-cos2=2sincos2sin2=cossintan,故D不符合題意.14.-3由sin10+mcos10=2cos140,可得m=2cos140-sin10cos10=-2cos40-sin10cos10=-2cos(30+10)

7、-sin10cos10=-3cos10cos10=-3.15.k-3,k+6(kZ)f(x)=2sin(x+524)cos(x+524)-2cos2x+524+1=sin2x+512-cos(2x+512)=2sin2x+512cos4-cos2x+512sin4=2sin2x+512-4=2sin2x+6,f(x)的最小正周期T=22=.由f(x)=2sin2x+6,得當(dāng)2k-22x+62k+2(kZ),即k-3xk+6(kZ)時,f(x)單調(diào)遞增,即函數(shù)f(x)的單調(diào)遞增區(qū)間是k-3,k+6(kZ).16.C由題意可得,f(x0)是函數(shù)f(x)的最小值,f(x0+2020)是函數(shù)f(x)的

8、最大值.因?yàn)閒(x)=cosx(sinx+3cosx)=12sin2x+31+cos2x2=sin2x+3+32,所以要使取最小值,只需保證區(qū)間x0,x0+2020為一個完整的單調(diào)遞增區(qū)間即可.故2020=122min,求得min=12020,故的最小值為12020,故選C.17.解(1)f(x)=sin2x+sinxcosx+2sinx+4cosx+4=1-cos2x2+12sin2x+sin2x+2=12+12(sin2x-cos2x)+cos2x=12(sin2x+cos2x)+12.由tan=2,得sin2=2sincossin2+cos2=2tantan2+1=45,cos2=cos2-sin2sin2+cos2=1-tan21+tan2=-35.故f(