新教材北师大版高中数学必修二 2.4积化和差与和差化积公式 课时练（课后作业设计）

1、精品文档 精心整理精品文档 可编辑的精品文档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tan

2、x-tan y=sinxcosx-sinycosy=sinxcosy-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cos 20-cos 50=-2sin20+502sin20-502=-2sin 35sin(-15)=2sin 15sin 35.答案C4.sin220+cos250+sin 20cos 50=()A.

3、-1B.2C.43D.34解析原式=-12cos(20+20)-cos(20-20)+12cos(50+50)+cos(50-50)+12(sin 70-sin 30)=12(1-cos 40)+12(1+cos 100)+12(sin 70-sin 30)=1-12cos 40+12cos 100+12sin 70-12sin 30=34+12sin 70+12(cos 100-cos 40)=34+12sin 70-sin100+402sin100-402=34+12sin 70-sin 30sin 70=34.答案D5.cos(x+2 020)-cos(x-2 020)=.解析原式=-2

4、sinx+2 020+x-2 0202sinx+2 020-(x-2 020)2=-2sin xsin 2 020.答案-2sin xsin 2 0206.cos 37.5cos 22.5=.解析cos 37.5cos 22.5=12(cos 60+cos 15)=14+12cos 15=2+6+28.答案2+6+287.cos 15cos 60cos 75=.解析原式=12cos 15cos 75=14cos 90+cos(-60)=18.答案188.求值:sin 42-cos 12+sin 54.解原式=sin 42-sin 78+sin 54=-2cos 60sin 18+sin 54=

5、sin 54-sin 18=2cos 36sin 18=22cos36sin18cos182cos18=2cos36(sin18cos18+sin18cos18)2cos18=2sin36cos362cos18=sin36cos36+sin36cos362cos18=sin722cos18=12.9.求证:2cos 20+2sin 20-12cos 20-2sin 20-1tan 25=cos15sin15.证明左边=2cos 20sin 25+2sin 20sin 25-sin 252cos 20cos 25-2sin 20cos 25-cos 25=sin45-sin(-5)-cos45+

6、cos(-5)-sin25cos45+cos(-5)-sin45-sin(-5)-cos25=sin5+cos5-sin25sin5+cos5-cos25=sin5+sin85-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(+)

7、=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.直角三角形解析由已知得sin Asin B=-12cos(A+B)-cos(A-B)=12(1+cos C).又A+B=-C,所以cos(A-B)-cos(-C)=1+cos C.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.-22si

8、n 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=-2sin xsin 3+2cos xsin 3=-2sin 3(sin x-cos x)=-22sin 3sinx-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cos 80+

9、2cos 40sin 80=cos 80+cos 40+cos 40sin 80=2cos 60cos 20+cos 40sin 80=cos 20+cos 40sin 80=2cos 30cos 10sin 80=3sin 80sin 80=3.答案36.计算:sin 20cos 70+sin 10sin 50.解sin 20cos 70+sin 10sin 50=12sin(20+70)+sin(20-70)-12cos(10+50)-cos(10-50)=12(sin 90-sin 50)-12(cos 60-cos 40)=12-12sin 50-14+12cos 40=12-12si

10、n 50-14+12sin 50=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=2-2cosB,|b|=2,ab=2sin B.由夹角公式,得cos3=2sinB22-2cosB,整理得2sin2B+cos B-1=0,即2cos2B-cos B-1=0.所以cos B=1(舍去)或cos B=-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-C2

11、6.所以sin A+sin C=2sinA+C2cosA-C2=2sin6cosA-C2=cosA-C2.所以sin A+sin C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+sin 4x解析2cos2x+3sin2x-3=sin2x+3+2x-3-sin2x+3-2x-3=sin 4x+sin23=32+sin 4x.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cos A+cos C=-22cos Acos C,则2cosA+C2cosA-C2=-2cos(A+C)+cos(A-C),将cosA+C2=cos 60=12,cos(A+C)=cos 120=-12代入上式,得cosA-C2=22-2cos(A-C),因为cos(A-C)=cosA-C2+A-C2