1、3.1.2两角和与差的正弦、余弦、正切公式一、A组1.在ABC中,sin Asin B0,则cos(A+B)0,所以cos(-C)0,即cos C0,所以C是钝角.故ABC是钝角三角形.答案:B2.(2016陕西渭南阶段性测试)若sin(-)cos -cos(-)sin =m,且为第三象限角,则cos 的值为()A.B.-C.D.-解析:sin(-)cos -cos(-)sin =m,sin (-)-=-sin =m,即sin =-m.又为第三象限角,cos 0.由同角三角函数的基本关系可得cos =-=-,故选B.答案:B3.已知sin =,是第二象限角,且tan(+)=-,则tan 的值为
2、()A.-B.C.-D.解析:为第二象限角,cos 0,cos =-,tan =-.tan =tan (+)-=-.答案:C4.=()A.-B.-C.D.解析:sin 47=sin (17+30)=sin 17cos 30+cos 17sin 30,=sin 30=.答案:C5.已知tan ,tan 是方程x2+3x+4=0的两根,且-,-,则+的值为()A.B.-C.-D.无法确定解析:由题意知,tan +tan =-3,tan tan =4,则tan 0,tan 0.所以,.tan (+)=.又+(-,0),所以+=-.答案:B6.函数f(x)=3sin x+cos x的最小正周期为.解析
3、:f(x)=3sin x+cos x=2=2sin,f(x)的最小正周期T=2.答案:27.若A,B是ABC的内角,且(1+tan A)(1+tan B)=2,则A+B等于.解析:由题意知,tan A+tan B+tan Atan B=1,即tan A+tan B=1-tan Atan B.tan (A+B)=1,又0A+B,所以A+B=.答案:8.设角的终边经过点(3,-4),则cos的值为.解析:由三角函数的定义可知,sin =-,cos =,cos(cos -sin )=.答案:9.(2016广东揭阳惠来一中检测)已知函数f(x)=2sin,xR.(1)求f的值;(2)设,f,f(3+2
4、)=,求cos(+)的值.解:(1)f=2sin=2sin.(2)f=2sin =,sin =.,cos =.又f(3+2)=2sin=2cos =,cos =.,sin =.cos(+)=cos cos -sin sin =.10.如图,在平面直角坐标系xOy中,以Ox轴为始边作两个锐角,它们的终边分别与单位圆交于A,B两点,已知A,B的横坐标分别为.(1)求tan(+)的值;(2)求+2的值.解:由条件得cos =,cos =.,为锐角,sin =,sin =.因此tan =7,tan =.(1)tan(+)=-3.(2)tan(+2)=tan (+)+=-1,又,为锐角,0+2,+2=.
5、二、B组1.sin -cos 的值是()A.B.C.-D.sin 解析:原式=2=2sin =2sin .答案:A2.(2016山东青岛平度四校联考)已知tan(+)=,tan,那么tan等于()A.B.C.D.解析:tan (+)=,tan ,tan =tan =,故选C.答案:C3.设,都为锐角,且cos =,sin(+)=,则sin 等于()A.B.C.D.-解析:为锐角,cos =,sin =.,都为锐角,0+.sin(+)=,cos(+)=.当cos(+)=-时,sin =sin(+)-=sin(+)cos -cos(+)sin =;当cos(+)=时,sin =sin(+)-=si
6、n(+)cos -cos(+)sin =-,与已知为锐角矛盾.sin =.答案:B4.已知tan =2,则的值为.解析:由tan =2,得tan =,所以=.答案:5.若cos =-,则cos 的值为.解析:因为,所以+,所以sin =-=-.故cos =cos =-.答案:-6.导学号08720086已知cos+sin =,则sin=.解析:cos+sin =cos cos+sin sin+sin =,即cos +sin =,从而cos +sin =,即sin,所以sin=sin=-sin=-.答案:-7.已知cos =-,tan =,0,求-的值.解法一:由cos =-,得sin =-,t
7、an =2,又tan =,于是tan (-)=1.又由,0,可得-0,-,因此-=.解法二:由cos =-,得sin =-.由tan =,0,得sin =,cos =.所以sin (-)=sin cos -cos sin =-.又由,0,可得-0,-,因此,-=.8.导学号08720087已知向量a=(cos ,sin ),b=(cos ,sin ),|a-b|=.(1)求cos(-)的值;(2)若0,-0,且sin =-,求sin 的值.解:(1)a=(cos ,sin ),b=(cos ,sin ),a-b=(cos -cos ,sin -sin ).又|a-b|=,即2-2cos(-)=,cos(-)=.(2)0,-0,0-.又cos(-)=,sin =-,sin(-)=,cos =.sin =sin (-)+=sin(-)cos +cos(-)sin =.