1、考点规范练22两角和与差的正弦、余弦与正切公式基础巩固1.sin 20cos 10-cos 160sin 10=()A.-32B.32C.-12D.12答案:D解析:sin20cos10-cos160sin10=sin20cos10+cos20sin10=sin(10+20)=sin30=12.2.已知角的顶点与原点重合,始边与x轴的正半轴重合,终边在直线y=2x上,则cos 2=()A.-45B.-35C.35D.45答案:B解析:由题意知tan=2,故cos2=cos2-sin2cos2+sin2=1-tan21+tan2=1-221+22=-35.3.已知(0,),且3cos 2-8co
2、s =5,则sin =()A.53B.23C.13D.59答案:A解析:原式化简得3cos2-4cos-4=0,解得cos=-23或cos=2(舍去).(0,),sin=1-cos2=53.4.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=-45.5.若0yx2,且tan x=3tan y,则x-y的最大值为()A.4B.6C.3D.2答案:B解析:0yx2,x-y0,2.又tanx=3tany,t
3、an(x-y)=tanx-tany1+tanxtany=2tany1+3tan2y=21tany+3tany33=tan6.当且仅当3tan2y=1时取等号,x-y的最大值为6,故选B.6.函数f(x)=sin 2xsin6-cos 2xcos56在区间-2,2上的单调递增区间为.答案:-512,12解析:f(x)=sin2xsin6-cos2xcos56=sin2xsin6+cos2xcos6=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)在区间-2,2上的单调递增区间为-512,12.7.在A
4、BC中,C=60,tanA2+tanB2=1,则tanA2tanB2=.答案:1-33解析:由C=60,则A+B=120,即A2+B2=60.根据tanA2+B2=tanA2+tanB21-tanA2tanB2,tanA2+tanB2=1,得3=11-tanA2tanB2,解得tanA2tanB2=1-33.8.函数f(x)=sin2x+sin xcos x+1的最小正周期是,单调递减区间是.答案:38+k,78+k,kZ解析:f(x)=sin2x+sinxcosx+1=1-cos2x2+12sin2x+1=12(sin2x-cos2x)+32=22sin2x-4+32.故T=22=.令2k+
5、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.9.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,-2-2.又tan(-)=-130,-2-bcB.bacC.cabD.acb答案:D解析:a=sin40cos127+cos40sin127=sin(40+127)=sin167=sin13,b=22(sin56-cos56)=22sin56-22cos56=sin(56-45)=sin11,c=1-tan2391+tan239=cos239-sin239co
6、s239cos239+sin239cos239=cos239-sin239=cos78=sin12.sin13sin12sin11,acb.故选D.12.已知sin+4=14,-32,-,则cos+712的值为.答案:-15+38解析:由-32,-得+4-54,-34.因为sin+4=14,所以cos+4=-154.cos+712=cos+4+3=cos+4cos3-sin+4sin3=-15412-1432=-15+38.13.已知sin 10+mcos 10=2cos 140,则m=.答案:-3解析:由sin10+mcos10=2cos140可得,m=2cos140-sin10cos10=
7、-2cos40-sin10cos10=-2cos(30+10)-sin10cos10=-3cos10cos10=-3.14.设函数f(x)=sinx-6+sinx-2,其中03.已知f6=0.(1)求.(2)将函数y=f(x)的图象上各点的横坐标伸长为原来的2倍(纵坐标不变),再将得到的图象向左平移4个单位,得到函数y=g(x)的图象,求g(x)在区间-4,34上的最小值.解:(1)因为f(x)=sinx-6+sinx-2,所以f(x)=32sinx-12cosx-cosx=32sinx-32cosx=312sinx-32cosx=3sinx-3.由题设知f6=0,所以6-3=k,kZ.故=6k+2,kZ,又03,所以=2.(2)由(1)得f(x)=3sin2x-3,所以g(x)=3sinx+4-3=3sinx-12.因为x-4,34,所以x-12-3,23,当x-12=-3,即x=-4时,g(x)取得最小值-32.高考预测15.已知sin3-=23,则cos3+2=()A.-59B.-23C.23D.59答案:A解析:依题意有cos23-2=cos23-=1-2sin23-=59,故cos3+2=cos-23-2=-cos23-2=-59.6
