2020版高考數(shù)學(xué)一輪復(fù)習(xí) 課時(shí)規(guī)范練22 三角恒等變換 理 北師大版.doc

課時(shí)規(guī)范練22三角恒等變換基礎(chǔ)鞏固組1.函數(shù)f(x)=(3sin x+cos x)(3cos x-sin x)的最小正周期是()A.2B.C.32D.22.已知sin+5=33,則cos2+25=()A.B.33C.D.323.(2018云南民族中學(xué)一模)已知tan =2,則2sin2+1cos2-4的值是()A.B.-134C.135D.1344.(2018四川成都七中模擬)已知sin76+=33,則cos23-2=()A.-B.-C.D.5.已知f(x)=sin2x+sin xcos x,則f(x)的最小正周期和一個(gè)遞增區(qū)間分別為()A.,0,B.2,-4,34C.,-8,38D.2,-4,46.(2018黑龍江高考仿真(三)已知sin+3+sin =-435,則cos+23=()A.-B.-C.D.7.(2018全國第一次大聯(lián)考)已知sinx+3=13,則sin53-x-cos2x-3的值為.8.設(shè)f(x)=1+cos2x2sin2-x+sin x+a2sinx+4的最大值為2+3,則實(shí)數(shù)a=.9.設(shè)為銳角,若cos+6=45,則sin2+12的值為.10.(2018湖北百所重點(diǎn)校聯(lián)考)設(shè)0,3,滿足6sin +2cos =3.(1)求cos+6的值;(2)求cos2+12的值.綜合提升組11.已知函數(shù)f(x)=sin(x+)+1>0,0<2的圖像的相鄰兩對稱軸之間的距離為,且在x=6時(shí)取得最大值2,若f()=,且6<<23,則sin2+23的值為()A.1225B.-1225C.2425D.-242512.已知-3,0,cos+6-sin =435,則sin+12的值是()A.-235B.-210C.235D.-13.(2018湖南長郡中學(xué)一模,17改編)已知函數(shù)f(x)=2sin xcos22+cos xsin -sin x(0<<)在x=處取最小值.則的值為.14.(2018安徽合肥二模)已知a=(sin x,3cos x),b=(cos x,-cos x),函數(shù)f(x)=ab+32.(1)求函數(shù)y=f(x)圖像的對稱軸方程;(2)若方程f(x)=在(0,)上的解為x1,x2,求cos(x1-x2)的值.創(chuàng)新應(yīng)用組15.已知m=tan(+)tan(-+),若sin 2(+)=3sin 2,則m=()A.-1B.C.D.216.函數(shù)y=sin +cos -4sin cos +1,且2sin2+sin21+tan=k,4<2,(1)把y表示成k的函數(shù)f(k);(2)求f(k)的最大值.參考答案課時(shí)規(guī)范練22三角恒等變換1.Bf(x)= 2sinx+62cosx+6=2sin2x+3,故最小正周期T=22=,故選B.2.A由題意sin+5=33,cos2+25=cos 2+5=1-2sin2+5=1-2332=13.故選A.3.Dtan =2,2sin2+1cos2-4=2sin2+sin2+cos2cos2-2=3sin2+cos2sin2=3sin2+cos22sincos=3tan2+12tan=322+122=134.4.B由題意sin76+=sin+6+=-sin6+,所以sin6+=-33,由于cos23-2=cos-3+2=-cos3+2=-cos26+=2sin26+-1=2-332-1=-13,故選B.5.C由f(x)=sin2x+sin xcos x=1-cos2x2+sin 2x=12+2222sin2x-22cos2x=12+22sin2x-4,則T=22=.又2k-22x-42k+2(kZ),k-8xk+38(kZ)為函數(shù)的遞增區(qū)間.故選C.6.Dsin3+sin =sincos +cossin +sin =-435,32sin +32cos =-435,即32sin +12cos =-45.sin+6=-45.故cos+23=cos+2+6=-sin+6=45.7.sin53-x-cos2x-3=sin2-x+3-cos 2x+3-2=-sinx+3+cos 2x+3=-sinx+3+1-2sin2x+3=-+1-=.8.3f(x)=1+2cos2x-12cosx+sin x+a2sinx+4=cos x+sin x+a2sinx+4=2sinx+4+a2sinx+4=(2+a2)sinx+4.依題意有2+a2=2+3,則a=3.9.17502為銳角,cos+6=,sin+6=35,sin2+3=2sin+6cos+6=2425,cos2+3=2cos2+6-1=725,sin2+12=sin2+3-4=22sin2+3-cos2+3=17250.10.解 (1)6sin +2cos =3,sin+6=64.0,3,+66,2,cos+6=104.(2)由(1)可得cos2+3=2cos2+6-1=21042-1=14.0,3,2+33,sin2+3=154.cos2+12=cos2+3-4=cos2+3cos4+sin2+3sin4=30+28.11.D由題意,T=2,即T=2=2,即=1.又當(dāng)x=6時(shí),f(x)取得最大值,即6+=2+2k,kZ,即=3+2k,kZ.0<2,=3,f(x)=sinx+3+1.f()=sin+3+1=95,可得sin+3=45.6<<23,可得2<+3<,cos+3=-35.sin2+23=2sin+3cos+3=245-35=-2425.故選D.12.B由cos+6-sin =435,可得32cos -32sin =435,12cos -32sin =45,cos+3=45.-3,0,+30,3,sin+3=35,sin+12=sin+3-4=22sin+3-22cos+3=2235-45=-210,故選B.13.f(x)=2sin x1+cos2+cos xsin -sin x=sin x+sin xcos +cos xsin -sin x=sin xcos +cos xsin =sin(x+).因?yàn)楹瘮?shù)f(x)在x=處取最小值,所以sin(+)=-1,由誘導(dǎo)公式知sin =1,因?yàn)?<<,所以=2.14.解 (1)f(x)=ab+32=(sin x,3cos x)(cos x,-cos x)+32=sin xcos x-3cos2x+32=12sin 2x-32cos 2x=sin2x-3.令2x-3=k+2,得x=512+k2(kZ),即y=f(x)的對稱軸方程為x=512+k2(kZ).(2)由條件知sin2x1-3=sin2x2-3=13>0,且0<x1<512<x2<23,易知(x1,f(x1)與(x2,f(x2)關(guān)于x=512對稱,則x1+x2=56,cos(x1-x2)=cosx1-56-x1=cos2x1-56=cos2x1-3-2=sin2x1-3=13.15.Dsin 2(+)=3sin 2,sin(+)-(-)=3sin(+)-(+-),sin(+)cos(-)-cos(+)sin(-)=3sin(+)cos(+-)-3cos(+)sin(+-),即-2sin(+)cos(+-)=-4cos(+)sin(+-),12tan(+)=tan(+-),故m=tan(+)tan(-+)=2,故選D.16.解 (1)k=2sin2+sin21+tan=2sin2+2sincos1+sincos=2sin(sin+cos)cos+sincos=2sin cos ,(sin +cos )2=1+2sin cos =1+k.4<2,sin +cos >0.sin +cos =1+k.y=1+k-2k+1.由于k=2sin cos =sin 2,4<2,0k<1.f(k)=1+k-2k+1(0k<1).(2)設(shè)1+k=t,則k=t2-1,1t<2.y=t-(2t2-2)+1,即y=-2t2+t+3(1t<2).關(guān)于t的二次函數(shù)在區(qū)間1,2)內(nèi)是減少的,t=1時(shí),y取最大值2.