1、考点规范练21两角和与差的正弦、余弦与正切公式考点规范练A册第14页基础巩固组1.计算cos 42cos 18-cos 48sin 18的结果等于() A.12B.33C.22D.32答案:A解析:原式=sin 48cos 18-cos 48sin 18=sin(48-18)=sin 30=12.2.(2015陕西,理6)“sin =cos ”是“cos 2=0”的()A.充分不必要条件B.必要不充分条件C.充分必要条件D.既不充分也不必要条件答案:A解析:cos 2=cos2-sin2=(cos +sin )(cos -sin ),cos 2=0cos =-sin 或cos =sin ,故选
2、A.3.(2015山西四校联考)已知sin2+=12,-20,则cos-3的值是()A.12B.23C.-12D.1答案:C解析:由已知得cos =12,sin =-32,cos-3=12cos +32sin =-12.4.已知,32,且cos =-45,则tan4-等于()A.7B.17C.-17D.-7答案:B解析:因为,32,且cos =-45,所以sin 0,即sin =-35,所以tan =34.所以tan4-=1-tan1+tan=1-341+34=17.5.(2015重庆,理9)若tan =2tan5,则cos-310sin-5=()A.1B.2C.3D.4导学号92950463
3、答案:C解析:因为tan =2tan5,所以cos-310sin-5=sin-310+2sin-5=sin+5sin-5=sincos5+cossin5sincos5-cossin5=tan+tan5tan-tan5=3tan5tan5=3.6.已知cos-6+sin =435,则sin+76的值为()A.12B.32C.-45D.-12答案:C解析:cos-6+sin =32cos +32sin =435,12cos +32sin =45.sin+76=-sin+6=-32sin+12cos=-45.7.(2015山东潍坊二模)若0,2,且cos2+cos2+2=310,则tan =()A.
4、12B.13C.14D.15答案:B解析:cos2+cos2+2=cos2-sin 2=cos2-2sin cos =cos 2-2sincoscos 2+sin 2=1-2tan1+tan 2=310.整理得3tan2+20tan -7=0,解得tan =13或-7,又0,2,故tan =13.8.(2015四川,理12)sin 15+sin 75的值是.答案:62解析:(方法一)sin 15+sin 75=sin(45-30)+sin(45+30)=sin 45cos 30-cos 45sin 30+sin 45cos 30+cos 45sin 30=2sin 45cos 30=22232
5、=62.(方法二)sin 15+sin 75=sin 15+cos 15=2sin(15+45)=62.9.函数f(x)=sin 2xsin6-cos 2xcos56在-2,2上的单调递增区间为.导学号92950464答案:-512,12解析:f(x)=sin 2xsin6-cos 2xcos56=sin 2xsin6+cos 2xcos6=cos2x-6.当2k-2x-62k(kZ),即k-512xk+12(kZ)时,函数f(x)单调递增.取k=0得-512x12,故函数f(x)在-2,2上的单调递增区间为-512,12.10.(2015浙江,理11)函数f(x)=sin2x+sin xco
6、s x+1的最小正周期是,单调递减区间是.导学号92950465答案:38+k,78+k,kZ解析:f(x)=sin2x+sin xcos x+1=1-cos2x2+12sin 2x+1=12(sin 2x-cos 2x)+32=22sin2x-4+32.故T=22=.令2k+22x-42k+32,kZ,解得k+38xk+78,kZ,故f(x)的单调递减区间为38+k,78+k,kZ.11.已知tan =2.(1)求tan+4的值;(2)求sin2sin2+sincos-cos2-1的值.解:(1)tan+4=tan+tan41-tantan4=tan+11-tan=2+11-2=-3.(2)
7、sin2sin2+sincos-cos2-1=2sincossin2+sincos-(2cos2-1)-1=2sincossin2+sincos-2cos2=2tantan2+tan-2=2222+2-2=1.导学号9295046612.(2015江苏常州一模)已知,均为锐角,且sin =35,tan(-)=-13.(1)求sin(-)的值;(2)求cos 的值.解:(1),0,2,从而-2-2.又tan(-)=-130,-2-0.所以tan =31tan+4tan321tan4tan=34,当且仅当1tan=4tan ,即tan2=14,tan =12时取等号,所以tan 的最大值是34.1
8、4.(2015湖北,理12)函数f(x)=4cos2x2cos2-x-2sin x-|ln(x+1)|的零点个数为.导学号92950468答案:2解析:令f(x)=41+cosx2sin x-2sin x-|ln(x+1)|=sin 2x-|ln(x+1)|=0,即sin 2x=|ln(x+1)|,在同一坐标系作出y=sin 2x与y=|ln(x+1)|的图像.由图像知共2个交点,故f(x)的零点个数为2.15.化简:tan(18-x)tan(12+x)+3tan(18-x)+tan(12+x)=.答案:1解析:tan(18-x)+(12+x)=tan(18-x)+tan(12+x)1-tan
9、(18-x)tan(12+x)=tan 30=33,tan(18-x)+tan(12+x)=331-tan(18-x)tan(12+x),原式=tan(18-x)tan(12+x)+3331-tan(18-x)tan(12+x)=1.16.已知函数f(x)=sin(x+)+acos(x+2),其中aR,-2,2.(1)当a=2,=4时,求f(x)在区间0,上的最大值与最小值;(2)若f2=0,f()=1,求a,的值.解:(1)f(x)=sinx+4+2cosx+2=22(sin x+cos x)-2sin x=22cos x-22sin x=sin4-x,因为x0,从而4-x-34,4.故f(x)在0,上的最大值为22,最小值为-1.(2)由f2=0,f()=1,得cos(1-2asin)=0,2asin2-sin-a=1,又-2,2,知cos 0,解得a=-1,=-6.导学号92950469