2020版数学（理）新攻略总复习课标通用练习：第四章 -第三节　两角和与差的正弦、余弦、正切公式 WORD版含解析.docx

1、第三节两角和与差的正弦、余弦、正切公式A组基础题组1.计算-sin 133cos 197-cos 47cos 73的结果为() A.12B.33C.22D.32答案A-sin 133cos 197-cos 47cos 73=-sin 47(-cos 17)-cos 47sin 17=sin(47-17)=sin 30=12.2.已知cos+3=sin-3,则tan 的值为()A.-1B.1 C.3D.-3答案B由已知得12cos -32sin =12sin -32cos ,整理得,12+32sin =12+32cos ,则sin =cos ,故tan =1.3.tan 18+tan 12+33

2、tan 18tan 12=()A.3B.2C.22D.33答案Dtan 30=tan(18+12)=tan18+tan121-tan18tan12=33,tan 18+tan 12=33(1-tan 18tan 12),原式=33.故选D.4.(2019广东惠州质检)已知tan +1tan=4,则cos2+4=()A.12 B.13C.14 D.15答案C由tan +1tan=4,得sincos+cossin=4,即sin2+cos2sincos=4,sin cos =14,cos2+4=1+cos2+22=1-sin22=1-2sincos2=1-2142=14.故选C.5.已知sin =5

3、5,sin(-)=-1010,、均为锐角,则cos 2=()A.-32B.-1C.0 D.1答案C由题意知cos =1-552=255,cos(-)=1-10102=31010,所以cos =cos-(-),=cos cos(-)+sin sin(-)=22,所以cos 2=2cos2-1=212-1=0.6.已知sin2+=12,-2,0,则cos-3的值为.答案-12解析由已知得cos =12,sin =-32,所以cos-3=12cos +32sin =-12.7.(2018课标全国,15,5分)已知sin +cos =1,cos +sin =0,则sin(+)=.答案-12解析由sin

4、 +cos =1,cos +sin =0,两式平方相加,得2+2sin cos +2cos sin =1,整理得sin(+)=-12.8.已知sin(-)cos -cos(-)sin =35,是第三象限角,则sin+54=.答案7210解析依题意可将已知条件变形为sin(-)-=-sin =35,sin =-35,又是第三象限角,因此有cos =-45,所以sin+54=-sin+4=-sin cos4-cos sin4=7210.9.已知tan =2.(1)求tan+4的值;(2)求sin2sin2+sincos-cos2-1的值.解析(1)因为tan =2,所以tan+4=tan+tan4

5、1-tantan4=2+11-21=-3.(2)因为tan =2,所以sin2sin2+sincos-cos2-1=2sincossin2+sincos-(cos2-sin2)-(sin2+cos2)=2sincossin2+sincos-2cos2=2tantan2+tan-2=2222+2-2=1.10.已知0,2,tan =12,求tan 2和sin2+3的值.解析tan =12,tan 2=2tan1-tan2=2121-14=43.sincos=12,即cos =2sin ,又sin2+cos2=1,5sin2=1,0,2,sin =55,cos =255,sin 2=2sin co

6、s =255255=45,cos 2=cos2-sin2=45-15=35,sin2+3=sin 2cos 3+cos 2sin 3=4512+3532=4+3310.B组提升题组1.已知sin-4=7210,cos 2=725,则sin =() A.45B.-45C.35D.-35答案C由sin-4=7210得sin -cos =75,由cos 2=725得cos2-sin2=725,即(cos -sin )(cos +sin )=725,由可得cos +sin =-15,由可得sin =35.2.(2018广西贵港联考)已知cos-6+sin =435,0,2,则cos+23=.答案-45

7、解析因为cos-6+sin =32cos +32sin =3sin+6=435,所以sin+6=45,又+23-+6=2,所以cos+23=cos2+6=-sin+6=-45.3.(2018浙江,18,14分)已知角的顶点与原点O重合,始边与x轴的非负半轴重合,它的终边过点P-35,-45.(1)求sin(+)的值;(2)若角满足sin(+)=513,求cos 的值.解析(1)由角的终边过点P-35,-45得sin =-45,所以sin(+)=-sin =45.(2)由角的终边过点P-35,-45得cos =-35,由sin(+)=513得cos(+)=1213.由=(+)-得cos =cos

8、(+)cos +sin(+)sin ,所以cos =-5665或cos =1665.4.已知sin +cos =355,0,4,sin-4=35,4,2.(1)求sin 2和tan 2的值;(2)求cos(+2)的值.解析(1)sin +cos =355,两边同时平方,得(sin +cos )2=95,即1+sin 2=95,所以sin 2=45,又0,4,则20,2,所以cos 2=1-sin22=35,所以tan 2=sin2cos2=43.(2)因为4,2,所以-40,4,又sin-4=35,所以cos-4=45,于是sin 2-4=2sin-4cos-4=2425,又sin 2-4=-cos 2,所以cos 2=-2425,又22,所以sin 2=725,又cos2=1+cos22=45,0,4,所以cos =255,sin =55,所以cos(+2)=cos cos 2-sin sin 2=255-2425-55725=-11525.