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4.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 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.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-sin17cos30cos17=()a.-32b.-12c.12d.32解析:sin 47=sin (17+30)=sin 17cos 30+cos 17sin 30,sin47-sin17cos30cos17=cos17sin30cos17=sin 30=12.答案:c5.已知tan ,tan 是方程x2+33x+4=0的两根,且-22,-22,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0a+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 )=2275=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 =121335-51345=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+121-712=-3.(2)tan(+2)=tan (+)+=tan(+)+tan1-tan(+)tan=-3+121-(-3)12=-1,又,为锐角,0+232,+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+2514=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 =3555+45255=11525;当cos(+)=45时,sin =sin(+)-=sin(+)cos -cos(+)sin =3555-45255=-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=-722622+-1722622=-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 =13,32,02,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+213=1.又由32,02,可得-2-0,2-32,因此-=54.解法二:由cos =-55,32,得sin =-255.由tan =13,02,得sin =110,cos =310.所以sin (-)=sin cos -cos sin =-255310-55110=-22.又由32,02,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02,-20,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,