2020版高考數(shù)學(xué)一輪復(fù)習(xí) 第四章 三角函數(shù)、解三角形 課時(shí)規(guī)范練20 兩角和與差的正弦、余弦與正切公式 文 北師大版.doc
課時(shí)規(guī)范練20兩角和與差的正弦、余弦與正切公式基礎(chǔ)鞏固組1.若cos4-=35,則sin 2=()A.725B.C.-D.-7252.(2018河北衡水中學(xué)三調(diào))若2,且3cos 2=sin4-,則sin 2的值為()A.-118B.118C.-1718D.17183.對(duì)于銳角,若sin-12=35,則cos2+3=()A.2425B.C.28D.-24254.設(shè)sin4+=13,則sin 2=()A.-B.-C.D.5.若tan =2tan5,則cos-310sin-5=()A.1B.2C.3D.46.(2018河北衡水中學(xué)16模,5)已知滿足sin =,則cos4+cos 4-=()A.718B.2518C.-718D.-25187.(2018河北衡水中學(xué)17模,6)已知sin =1010,0,2,則cos2+6的值為()A.43-310B.43+310C.4-3310D.33-4108.設(shè)sin 2=-sin ,2,則tan 2的值是.9.已知0,2,tan =2,則cos-4=.10.若sin 2=55,sin(-)=1010,且4,32,則+=.綜合提升組11.(2018寧夏石嘴山一模)若tan+4=-3,則cos 2+2sin 2=()A.B.1C.-D.-12.(2018福建百校臨考沖刺)若(0,),且3sin +2cos =2,則tan2=()A.32B.34C.233D.43313.(2018北京懷柔區(qū)模擬)已知函數(shù)f(x)=(sin x+cos x)2+cos 2x-1.(1)求函數(shù)f(x)的最小正周期;(2)求函數(shù)f(x)在區(qū)間-4,4上的最大值和最小值.創(chuàng)新應(yīng)用組14.(2018重慶巴蜀中學(xué)月考)已知sin+12=24,則sin3-2=()A.24B.C.74D.-15.(2018河北衡水中學(xué)押題二,10)已知函數(shù)f(x)=3sin xcos x-4cos2x(>0)的最小正周期為,且f()=,則f+2=()A.-B.-C.-112D.-13216.已知sin+4=14,-32,-,則cos+712的值為.課時(shí)規(guī)范練20兩角和與差的正弦、余弦與正切公式1.D(法一)cos24-=2cos24-1=2352-1=-725,且cos24-=cos2-2=sin 2,故選D.(法二)由cos4-=35,得22cos +22sin =35,即22(cos +sin )=35,兩邊平方得12(cos2+sin2+2cos sin )=925,整理得2sin cos =-725,即sin 2=-725,故選D.2.C由3cos 2=sin4-,得3(cos2-sin2)=22(cos -sin ).2,cos -sin 0,cos +sin =26.兩邊平方,得1+2sin cos =118,sin 2=-1718.故選C.3.D由為銳角,且sin-12=35,可得cos-12=45,sin2-6=23545=2425,cos2+3=cos2+2-6=-sin2-6=-2425,故選D.4.Asin 2=-cos2+2=2sin24+-1=2132-1=-79.5.C因?yàn)閠an =2tan,所以cos-310sin-5=sin-310+2sin-5=sin+5sin-5=sincos5+cossin5sincos5-cossin5=tan+tan5tan-tan5=3tan5tan5=3.6.Acos4+cos4-=cos -cos-=sin-cos-=sin-2=cos 2= (1-2sin2)=121-219=718,故選A.7.Asin =1010,0,2,cos =1-sin2=31010,sin 2=2sin cos =2101031010=35,cos 2=1-2sin2=1-210102=45.cos2+6=32cos 2-12sin 2=3245-1235=43-310.故選A.8.3sin 2=2sin cos =-sin ,cos =-12,又2,sin =32,tan =-3,tan 2=2tan1-tan2=-231-(-3)2=3.9.31010由tan =2,得sin =2cos .又sin2+cos2=1,所以cos2=15.因?yàn)?,2,所以cos =55,sin =255.因?yàn)閏os-4=cos cos4+sin sin4,所以cos-4=5522+25522=31010.10.74因?yàn)?,所以22,2.又sin 2=55,故22,4,2,所以cos 2=-255.又,32,故-2,54,于是cos(-)=-31010,所以cos(+)=cos2+(-)=cos 2cos(-)-sin 2sin(-)=-255-31010-551010=22,且+54,2,故+=74.11.Btan+4=tan+11-tan=-3,tan =2,cos 2+2sin 2=cos2-sin2cos2+sin2+4sincoscos2+sin2=1-tan21+tan2+4tan1+tan2=-35+85=1.12.A由二倍角公式,得3sin +2cos =23sincos+21-2sin2=2,化簡(jiǎn)可得23sin2cos2=4sin22.(0,),20,2,sin20,3cos2=2sin2,tan2=32.13.解 (1)f(x)=(sin x+cos x)2+cos 2x-1=2sin xcos x+cos 2x=sin 2x+cos 2x=2sin2x+4,函數(shù)f(x)的最小正周期T=22=.(2)由(1)可知,f(x)=2sin2x+4.x-4,4,2x+4-4,34,sin2x+4-22,1.故函數(shù)f(x)在區(qū)間-4,4上的最大值和最小值分別為2,-1.14.Bsin3-2=sin2-6-2=cos+2=1-2sin2+12=1-2242=1-14=34.15.B函數(shù)f(x)=3sin xcos x-4cos2x=sin 2x-2(1+cos 2x)= sin(2x-)-2,其中tan =,所以f(x)的最小正周期為T=22=,解得=1,所以f(x)=52sin(2x-)-2,又由f()=12,即f()=52sin(2-)-2=12,即sin(2-)=1,所以f+2=52sin2+2-2=-52sin(2-)-2=-521-2=-92,故選B.16.-15+38由-32,-,得+4-54,-34,又sin+4=14,所以cos+4=-154.cos+712=cos+4+3=cos+4cos3-sin+4sin3=-15412-1432=-15+38.