高中数学必修4配人教A版-课后习题Word版第一章　1.3　第1课时　诱导公式二、三、四

1、第一章三角函数1.3三角函数的诱导公式第1课时诱导公式二、三、四课后篇巩固探究1.已知sin+=45,则角的终边在()A.第一或第二象限B.第二或第三象限C.第一或第四象限D.第三或第四象限解析由已知得-sin =45,所以sin =-45,故角的终边在第三或第四象限.答案D2.若cos(-)=-12,则cos(-2-)的值为()A.12B.32C.-12D.12解析cos(-)=-cos =-12,cos =12.cos(-2-)=cos(-)=cos =12.答案A3.sin-136-cos-103-tan154的值为()A.-2B.0C.12D.1解析原式=-sin2+6-cos2+43

2、-tan2+74=-sin6-cos+3-tan2-4=-12+cos3+tan4=-12+12+1=1.答案D4.已知tan(-)=12,则sin+cos2sin-cos=()A.14B.-14C.12D.-12解析由已知得-tan =12,所以tan =-12.于是sin+cos2sin-cos=tan+12tan-1=-12+12-12-1=-14.答案B5.若角7-的终边与单位圆的交点坐标是x,35,则cos(-2 018)=()A.45B.35C.45D.-35解析依题意,sin(7-)=35,即sin =35,于是cos =45,故cos(-2 018)=cos =45.答案A6.

3、1-2sin(+2)cos(-2)等于()A.sin 2-cos 2B.sin 2+cos 2C.(sin 2-cos 2)D.cos 2-sin 2解析1-2sin(+2)cos(-2)=1-2sin2cos2=(sin2-cos2)2=|sin 2-cos 2|=sin 2-cos 2.答案A7.记cos(-80)=k,则tan 100等于()A.1-k2kB.-1-k2kC.k1-k2D.-k1-k2解析cos(-80)=cos 80=k,sin 80=1-cos280=1-k2,tan 100=-tan 80=-1-k2k.故选B.答案B8.已知A=sin(k+)sin+cos(k+)

4、cos(kZ),则A的值构成的集合是()A.1,-1,2,-2B.-1,1C.2,-2D.1,-1,0,2,-2解析当k为偶数时,A=sinsin+coscos=2;当k为奇数时,A=-sinsincoscos=-2.故选C.答案C9.已知sin(45+)=513,则sin(135-)=.解析sin(135-)=sin180-(45+)=sin(45+)=513.答案51310.已知tan7+=5,则tan67-=.解析tan67-=tan-7+=-tan7+=-5.答案-511.设tan(5+)=m,则sin(-3)+cos(-)sin(-)-cos(+)=.解析tan(5+)=tan =m

5、,原式=-sin-cos-sin+cos=-tan-1-tan+1=-m-1-m+1=m+1m-1.答案m+1m-112.已知623,cos+3=m(m0),则tan23-=.解析由623,可得+32,.因为cos+3=m0,所以sin+3=1-cos2+3=1-m2,所以tan+3=1-m2m.所以tan23-=tan-+3=-tan+3=-1-m2m.答案-1-m2m13.已知sin(3+)=13,求:sin(180+)cos(720+)tan(540+)sin(-180+)tan(900+)sin(-180-)cos(-180-)的值.解sin(3+)=13,sin =-13.原式=(-

6、sin)costan(-sin)tansin(-cos)=-sin =13.14.(1)已知sin 是方程5x2-7x-6=0的根,求cos(+2)cos(4+)tan2(2+)tan(6+)sin(2+)sin(8+)的值;(2)已知sin(4+)=2sin ,3cos(6+)=2cos(2+),且0,0,求和的值.解(1)因为方程5x2-7x-6=0的两根为2和-35,所以sin =-35.由sin2+cos2=1,得cos =1-sin2=45.当cos =45时,tan =-34;当cos =-45时,tan =34.所以原式=coscostan2tansinsin=tan =34.(2)因为sin(4+)=2sin ,所以sin =2sin .因为3cos(6+)=2cos (2+),所以3cos =2cos .2+2,得