（天津专用）2020届高考数学考点规范练20两角和与差的正弦、余弦与正切公式（含解析）新人教A版.docx

考点规范练20两角和与差的正弦、余弦与正切公式一、基础巩固1.sin 35cos 25-cos 145sin 25=()A.-32B.32C.-12D.122.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.453.已知,32,且cos =-45,则tan4-等于()A.7B.17C.-17D.-74.若tan =2tan5,则cos-310sin-5=()A.1B.2C.3D.45.已知cos-6+sin =435,则sin+76的值为()A.12B.32C.-45D.-126.若0yxbcB.bacC.cabD.acb13.已知sin+4=14,-32,-,则cos+712的值为.14.设,0,2,且tan =1+sincos,则2-=.15.(2018浙江,18)已知角的顶点与原点O重合,始边与x轴的非负半轴重合,它的终边过点P-35,-45.(1)求sin(+)的值;(2)若角满足sin(+)=513,求cos 的值.三、高考预测16.在平面直角坐标系中,点P的坐标为35,45,Q是第三象限内一点,|OQ|=1,且POQ=34,则点Q的横坐标为()A.-7210B.-325C.-7212D.-8213考点规范练20两角和与差的正弦、余弦与正切公式1.B解析sin35cos25-cos145sin25=sin35cos25+cos35sin25=sin(35+25)=sin60=32.2.B解析由题意知tan=2,故cos2=cos2-sin2cos2+sin2=1-tan21+tan2=1-221+22=-35.3.B解析因为,32,且cos=-45,所以sin=-35,所以tan=34.所以tan4-=1-tan1+tan=1-341+34=17.4.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=3.5.C解析cos-6+sin=32cos+32sin=435,12cos+32sin=45,即sin+6=45.sin+76=-sin+6=-45.6.B解析0yx2,x-y0,2.又tanx=3tany,tan(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.7.-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-512,12.8.17解析由题意得tan=12,所以tan=tan(-)+=tan(-)+tan1-tan(-)tan=-13+121-1312=17.9.1-33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=1-33.10.解(1),0,2,-2-2.又tan(-)=-130,-2-sin12sin11,acb.故选D.13.-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+38.14.2解析,0,2,且tan=1+sincos,sincos=1+sincos,sincos=cos+cossin.sincos-cossin=cos.sin(-)=cos=sin2-.,0,2,-2,2,2-0,2.函数y=sinx在-2,2内单调递增,由sin(-)=sin2-可得-=2-,即2-=2.15.解(1)由角的终边过点P-35,-45,得sin=-45,所以sin(+)=-sin=45.(2)由角的终边过点P-35,-45,得cos=-35,由sin(+)=513,得cos(+)=1213.由=(+)-,得cos=cos(+)cos+sin(