资源描述
课时规范练21两角和与差的正弦、余弦与正切公式基础巩固组1.若cos4-=35,则sin 2=()A.725B.C.-D.-7252.(2018河北衡水中学三调)若2,且3cos 2=sin4-,则sin 2的值为()A.-118B.118C.-1718D.17183.对于锐角,若sin-12=35,则cos2+3=()A.2425B.C.28D.-24254.设sin4+=13,则sin 2=()A.-B.-C.D.5.若tan =2tan5,则cos-310sin-5=()A.1B.2C.3D.46.(2018河北衡水中学16模,5)已知满足sin =,则cos4+cos 4-=()A.718B.2518C.-718D.-25187.(2018河北衡水中学17模,6)已知sin =1010,0,2,则cos2+6的值为()A.43-310B.43+310C.4-3310D.33-4108.设sin 2=-sin ,2,则tan 2的值是.9.已知0,2,tan =2,则cos-4=.10.若sin 2=55,sin(-)=1010,且4,32,则+=.综合提升组11.(2018宁夏石嘴山一模)若tan+4=-3,则cos 2+2sin 2=()A.B.1C.-D.-12.(2018福建百校临考冲刺)若(0,),且3sin +2cos =2,则tan2=()A.32B.34C.233D.43313.(2018北京怀柔区模拟)已知函数f(x)=(sin x+cos x)2+cos 2x-1.(1)求函数f(x)的最小正周期;(2)求函数f(x)在区间-4,4上的最大值和最小值.创新应用组14.(2018重庆巴蜀中学月考)已知sin+12=24,则sin3-2=()A.24B.C.74D.-15.(2018河北衡水中学押题二,10)已知函数f(x)=3sin xcos x-4cos2x(0)的最小正周期为,且f()=,则f+2=()A.-B.-C.-112D.-13216.已知sin+4=14,-32,-,则cos+712的值为.参考答案课时规范练21两角和与差的正弦、余弦与正切公式1.D(法一)cos24-=2cos24-1=2352-1=-725,且cos24-=cos2-2=sin 2,故选D.(法二)由cos4-=35,得22cos +22sin =35,即22(cos +sin )=35,两边平方得12(cos2+sin2+2cos sin )=925,整理得2sin cos =-725,即sin 2=-725,故选D.2.C由3cos 2=sin4-,得3(cos2-sin2)=22(cos -sin ).2,cos -sin 0,cos +sin =26.两边平方,得1+2sin cos =118,sin 2=-1718.故选C.3.D由为锐角,且sin-12=,可得cos-12=,sin2-6=2=2425,cos2+3=cos2+2-6=-sin2-6=-2425,故选D.4.Asin 2=-cos2+2=2sin24+-1=2132-1=-79.5.C因为tan =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,故选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.故选A.8.3sin 2=2sincos =-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.因为0,2,所以cos =55,sin =255.因为cos-4=cos cos4+sin sin4,所以cos-4=5522+25522=31010.10.74因为4,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,化简可得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,函数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.故函数f(x)在区间-4,4上的最大值和最小值分别为2,-1.14.Bsin3-2=sin-2=cos+2=1-2sin2+12=1-2242=1-=.15.B函数f(x)=3sin xcos x-4cos2x=sin 2x-2(1+cos 2x)= sin(2x-)-2,其中tan =,所以f(x)的最小正周期为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,故选B.16.-15+38由-32,-,得+-54,-34,又sin+4=14,所以cos+4=-154.cos+712=cos+4+3=cos+4cos3-sin+4sin3=-15412-1432=-15+38.
展开阅读全文