2020届高考数学江苏版二轮习题：基础滚动小练第22讲　三角函数应用题 WORD版含解析.docx

1、第22讲三角函数应用题1.在ABC中,a=8,B=60,C=75,则b=.2.(2019南通期末三县联考,8)已知函数f(x)的周期为4,且当x(0,4时, f(x)=cos x2,0x2,log2x-32,2b0)的离心率为12,右准线方程为x=4,A,B分别是椭圆C的左,右顶点,过右焦点F且斜率为k(k0)的直线l与椭圆C相交于M,N两点.(1)求椭圆C的标准方程;(2)记AFM,BFN的面积分别为S1,S2,若S1S2=32,求k的值;(3)设线段MN的中点为D,直线OD与右准线相交于点E,记直线AM,BN,FE的斜率分别为k1,k2,k3,求k2(k1-k3)的值.答案精解精析1.答案

2、46解析A=180-60-75=45,由正弦定理得bsinB=asinA,b=8sin60sin45=46.2.答案0解析由题意得f-12=f4-12=f72=log272-32=log2 2=1,ff-12=f(1)=cos2=0.3.答案6解析双曲线x24-y22=1的渐近线为y=22x,由对称性,不妨设Px0,22x0(x00),则PO=62x0,62x0=3,x0=6,P到y轴的距离为6.4.答案2或-23解析由题意可得|4+2a|1+a2=|4a|1+a2,则4+2a=4a或4+2a=-4a,解得a=2或a=-23.5.答案(1,2解析y=sin x+3cos x=2sinx+3,x

3、0,2,x+33,56,10).依题意,得ca=12,且a2c=4,解得a=2,c=1.所以b2=a2-c2=3.所以椭圆C的标准方程为x24+y23=1.(2)设点M(x1,y1)(x10,y10),N(x2,y2).由S1S2=32,得12|AF|y1|12|BF|y2|=32,整理可得|y1|y2|=12,所以NF=2FM.所以1-x2=2(x1-1),-y2=2y1,即x2=3-2x1,y2=-2y1.又点M,N在椭圆C上,所以x124+y123=1,(3-2x1)24+(-2y1)23=1,解得x1=74,y1=358.所以直线l的斜率k=35874-1=52.(3)依题意,直线l的

4、方程为y=k(x-1).联立y=k(x-1),x24+y23=1,整理得(4k2+3)x2-8k2x+4k2-12=0,所以x1+x2=8k24k2+3,x1x2=4k2-124k2+3.故xD=x1+x22=4k24k2+3,yD=k(xD-1)=-3k4k2+3,所以直线OD的方程为y=-34kx,令x=4,得yE=-3k,即E4,-3k.所以k3=-3k4-1=-1k.所以k2(k1-k3)=k2k1+1k=y2x2-2y1x1+2+1k=k(x2-1)x2-2k(x1-1)x1+2+1k=k2(x1-1)(x2-1)+(x2-1)(x1+2)(x1+2)(x2-2)=k2x1x2-(x1+x2)+1+x1x2-x1+2x2-2x1x2-2x1+2x2-4=k2x1x2-(x1+x2)+1+x1x2-(x1+x2)-2+3x2x1x2-2(x1+x2)-4+4x2=k24k2-124k2+3-8k24k2+3+1+4k2-124k2+3-8k24k2+3-2+3x24k2-124k2+3-28k24k2+3-4+4x2=3x2-21k2+184k2+34x2-28k2+244k2+3=3x2-7k2+64k2+34x2-7k2+64k2+3=34.