高中数学第三章三角恒等变换3.1.2两角和与差的正弦余弦正切公式课后习题新人教A版必修4201707.doc

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 B<cos Acos B,则ABC是()A.直角三角形B.钝角三角形C.锐角三角形D.等腰三角形解析:由题意,得cos Acos B-sin Asin B>0,则cos(A+B)>0,所以cos(-C)>0,即cos C<0,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.1-m2B.-1-m2C.m2-1D.-m2-1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 =-1-sin2=-1-m2,故选B.答案:B3.已知sin =12,是第二象限角,且tan(+)=-3,则tan 的值为()A.-3B.3C.-33D.33解析:为第二象限角,cos <0,cos =-32,tan =-33.tan =tan (+)-=tan(+)-tan1+tan(+)tan=-3+331+(-3)×-33=-33.答案:C4.sin47°-sin17°cos30°cos17°=()A.-32B.-12C.12D.32解析:sin 47°=sin (17°+30°)=sin 17°cos 30°+cos 17°·sin 30°,sin47°-sin17°cos30°cos17°=cos17°sin30°cos17°=sin 30°=12.答案:C5.已知tan ,tan 是方程x2+33x+4=0的两根,且-2<<2,-2<<2,则+的值为()A.3B.-23C.-23或3D.无法确定解析:由题意知,tan +tan =-33,tan tan =4,则tan <0,tan <0.所以,-2,0,-2,0.tan (+)=tan+tan1-tantan=-331-4=3.又+(-,0),所以+=-23.答案:B6.函数f(x)=3sin x+3cos x的最小正周期为. 解析:f(x)=3sin x+3cos x=23sinxcos6+cosxsin6=23sinx+6,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)=tanA+tanB1-tanAtanB=1,又0<A+B<,所以A+B=4.答案:48.设角的终边经过点(3,-4),则cos+4的值为. 解析:由三角函数的定义可知,sin =-432+(-4)2=-45,cos =332+(-4)2=35,cos+4=22(cos -sin )=22×75=7210.答案:72109.(2016广东揭阳惠来一中检测)已知函数f(x)=2sin13x-6,xR.(1)求f54的值;(2)设,0,2,f3+2=1013,f(3+2)=65,求cos(+)的值.解:(1)f54=2sin512-6=2sin4=2.(2)f3+2=2sin =1013,sin =513.0,2,cos =1213.又f(3+2)=2sin+2=2cos =65,cos =35.0,2,sin =45.cos(+)=cos cos -sin sin =1213×35-513×45=1665.10.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210,255.(1)求tan(+)的值;(2)求+2的值.解:由条件得cos =210,cos =255.,为锐角,sin =1-cos2=7210,sin =1-cos2=55.因此tan =7,tan =12.(1)tan(+)=tan+tan1-tantan=7+121-7×12=-3.(2)tan(+2)=tan (+)+=tan(+)+tan1-tan(+)tan=-3+121-(-3)×12=-1,又,为锐角,0<+2<32,+2=34.二、B组1.3sin 512-cos 512的值是()A.2B.22C.-2D.sin 712解析:原式=2sin 512cos 6-cos 512sin 6=2sin 512-6=2sin 4=2.答案:A2.(2016山东青岛平度四校联考)已知tan(+)=25,tan-4=14,那么tan+4等于()A.1318B.1322C.322D.16解析:tan (+)=25,tan -4=14,tan +4=tan (+)-4=tan(+)-tan -41+tan(+)tan -4=25-141+25×14=322,故选C.答案:C3.设,都为锐角,且cos =55,sin(+)=35,则sin 等于()A.2525B.11525C.55D.-55或11525解析:为锐角,cos =55,sin =255.,都为锐角,0<+<.sin(+)=35,cos(+)=±45.当cos(+)=-45时,sin =sin(+)-=sin(+)cos -cos(+)sin =35×55+45×255=11525;当cos(+)=45时,sin =sin(+)-=sin(+)cos -cos(+)sin =35×55-45×255=-55,与已知为锐角矛盾.sin =11525.答案:B4.已知tan 4+=2,则12sincos+cos2的值为. 解析:由tan 4+=1+tan1-tan=2,得tan =13,所以12sincos+cos2=sin2+cos22sincos+cos2=tan2+12tan+1=19+123+1=23.答案:235.若cos +4=-7226,32,则cos 的值为. 解析:因为,32,所以+454,74,所以sin +4=-1-72262=-17226.故cos =cos +4-4=-7226×22+-17226×22=-1213.答案:-12136.导学号08720086已知cos-6+sin =453,则sin+76=. 解析:cos-6+sin =cos cos6+sin sin6+sin =453,即32cos +32sin =453,从而12cos +32sin =45,即sin6+=45,所以sin+76=sin+6=-sin+6=-45.答案:-457.已知cos =-55,tan =13,<<32,0<<2,求-的值.解法一:由cos =-55,<<32,得sin =-255,tan =2,又tan =13,于是tan (-)=tan-tan1+tantan=2-131+2×13=1.又由<<32,0<<2,可得-2<-<0,2<-<32,因此-=54.解法二:由cos =-55,<<32,得sin =-255.由tan =13,0<<2,得sin =110,cos =310.所以sin (-)=sin cos -cos sin =-255×310-55×110=-22.又由<<32,0<<2,可得-2<-<0,2<-<32,因此,-=54.8.导学号08720087已知向量a=(cos ,sin ),b=(cos ,sin ),|a-b|=255.(1)求cos(-)的值;(2)若0<<2,-2<<0,且sin =-513,求sin 的值.解:(1)a=(cos ,sin ),b=(cos ,sin ),a-b=(cos -cos ,sin -sin ).又|a-b|=255,(cos-cos)2+(sin-sin)2=255,即2-2cos(-)=45,cos(-)=35.(2)0<<2,-2<<0,0<-<.又cos(-)=35,sin =-513,sin(-)=45,cos =1213.sin =sin (-)+=sin(-)·cos +cos(-)·sin =45×1213+35×-513=3365.8