2020-2021学年高中数学北师大版（2019）必修第二册课后习题：第四章　2-4　积化和差与和差化积公式 WORD版含解析.docx

1、2.4积化和差与和差化积公式课后篇巩固提升基础达标练1.已知cos -cos =12,sin -sin =-13,则tan+2=.解析因为cos-cos=12,所以-2sin+2sin-2=12,因为sin-sin=-13,所以2cos+2sin-2=-13,因为sin-20,cos+20,所以-tan+2=-32,即tan+2=32.答案322.把tan x-tan y化为积的形式为()A.cos(x-y)cosxcosyB.sin(x+y)sinxcosyC.sin(x-y)cosxcosyD.cos(x+y)cosxsiny解析tanx-tany=sinxcosx-sinycosy=si

2、nxcosy-sinycosxcosxcosy=12sin(x+y)+sin(x-y)-12sin(y+x)+sin(y-x)cosxcosy=sin(x-y)cosxcosy.答案C3.cos 20-cos 50=()A.cos 35cos 15B.sin 35sin 15C.2sin 15sin 35D.2sin 15cos 35解析cos20-cos50=-2sin20+502sin20-502=-2sin35sin(-15)=2sin15sin35.答案C4.sin220+cos250+sin 20cos 50=()A.-1B.2C.43D.34解析原式=-12cos(20+20)-c

3、os(20-20)+12cos(50+50)+cos(50-50)+12(sin70-sin30)=12(1-cos40)+12(1+cos100)+12(sin70-sin30)=1-12cos40+12cos100+12sin70-12sin30=34+12sin70+12(cos100-cos40)=34+12sin70-sin100+402sin100-402=34+12sin70-sin30sin70=34.答案D5.cos(x+2 020)-cos(x-2 020)=.解析原式=-2sinx+2020+x-20202sinx+2020-(x-2020)2=-2sinxsin2020

4、.答案-2sin xsin 2 0206.cos 37.5cos 22.5=.解析cos37.5cos22.5=12(cos60+cos15)=14+12cos15=2+6+28.答案2+6+287.cos 15cos 60cos 75=.解析原式=12cos15cos75=14cos90+cos(-60)=18.答案188.求值:sin 42-cos 12+sin 54.解原式=sin42-sin78+sin54=-2cos60sin18+sin54=sin54-sin18=2cos36sin18=22cos36sin18cos182cos18=2cos36(sin18cos18+sin18

5、cos18)2cos18=2sin36cos362cos18=sin36cos36+sin36cos362cos18=sin722cos18=12.9.求证:2cos20+2sin20-12cos20-2sin20-1tan 25=cos15sin15.证明左边=2cos20sin25+2sin20sin25-sin252cos20cos25-2sin20cos25-cos25=sin45-sin(-5)-cos45+cos(-5)-sin25cos45+cos(-5)-sin45-sin(-5)-cos25=sin5+cos5-sin25sin5+cos5-cos25=sin5+sin85-

6、sin25cos85+cos5-cos25=sin5+2cos55sin30-2sin55sin30+cos5=sin5+cos55cos5-sin55=sin5+sin35cos5-cos35=sin20cos(-15)-sin20sin(-15)=cos15sin15=右边.所以原等式成立.能力提升练1.已知sin(+)sin(-)=m,则cos2-cos22等于()A.-mB.mC.-4mD.4m解析sin(+)=sin(-)=cos2-cos22=m.答案B2.在ABC中,若sin Asin B=12(1+cos C),则ABC是()A.等边三角形B.等腰三角形C.不等边三角形D.直角

7、三角形解析由已知得sinAsinB=-12cos(A+B)-cos(A-B)=12(1+cosC).又A+B=-C,所以cos(A-B)-cos(-C)=1+cosC.所以cos(A-B)=1.又-A-B,所以A-B=0.所以A=B,故ABC为等腰三角形.答案B3.cos(x+3)-cos(x-3)+sin(x+3)-sin(x-3)=()A.2cos 3cosx-4B.22sin 3cosx-4C.-22sin 3sinx+4D.-22sin 3sinx-4解析原式=-2sinx+3+x-32sinx+3-(x-3)2+2cosx+3+x-32sinx+3-(x-3)2=-2sinxsin3

8、+2cosxsin3=-2sin3(sinx-cosx)=-22sin3sinx-4.答案D4.若sin +sin =33(cos -cos ),且(0,),(0,),则tan-2=;-=.解析由已知得2sin+2cos-2=33-2sin+2sin-2,因为0+2,-2-20.所以tan-2=3.所以-2=3.所以-=23.答案3235.cos80+2cos40sin80=.解析cos80+2cos40sin80=cos80+cos40+cos40sin80=2cos60cos20+cos40sin80=cos20+cos40sin80=2cos30cos10sin80=3sin80sin8

9、0=3.答案36.计算:sin 20cos 70+sin 10sin 50.解sin20cos70+sin10sin50=12sin(20+70)+sin(20-70)-12cos(10+50)-cos(10-50)=12(sin90-sin50)-12(cos60-cos40)=12-12sin50-14+12cos40=12-12sin50-14+12sin50=14.7.已知向量a=(sin B,1-cos B)与向量b=(2,0)的夹角为3,其中A,B,C是ABC的内角.(1)求B的大小;(2)求sin A+sin C的取值范围.解(1)由题意,得|a|=sin2B+(1-cosB)2

10、=2-2cosB,|b|=2,ab=2sinB.由夹角公式,得cos3=2sinB22-2cosB,整理得2sin2B+cosB-1=0,即2cos2B-cosB-1=0.所以cosB=1(舍去)或cosB=-12.又因为0B,所以B=23.(2)因为A+B+C=,B=23,所以A+C=3.所以-3A-C3.所以-6A-C26.所以sinA+sinC=2sinA+C2cosA-C2=2sin6cosA-C2=cosA-C2.所以sinA+sinC的取值范围是32,1.素养培优练1.2cos2x+3sin2x-3=()A.12+cos 4xB.12-sin 4xC.32+cos 4xD.32+s

11、in 4x解析2cos2x+3sin2x-3=sin2x+3+2x-3-sin2x+3-2x-3=sin4x+sin23=32+sin4x.答案D2.已知ABC的三个内角A,B,C满足A+C=2B,1cosA+1cosC=-2cosB,求cosA-C2的值.解由题设条件知B=60,A+C=120,所以1cosA+1cosC=-2cos60=-22,即cosA+cosC=-22cosAcosC,则2cosA+C2cosA-C2=-2cos(A+C)+cos(A-C),将cosA+C2=cos60=12,cos(A+C)=cos120=-12代入上式,得cosA-C2=22-2cos(A-C),因为cos(A-C)=cosA-C2+A-C2=cosA-C2cosA-C2-sinA-C2sinA-C2=cos2A-C2-sin2A-C2=cos2A-C2-1-cos2A-C2=2cos2A-C2-1,代入上式并整理得42cos2A-C2+2cosA-C2-32=0,即2cosA-C2-222cosA-C2+3=0.因为22cosA-C2+30,所以2cosA-C2-2=0.所以cosA-C2=22.