（天津?qū)Ｓ茫?020屆高考數(shù)學(xué)一輪復(fù)習(xí) 考點(diǎn)規(guī)范練20 兩角和與差的正弦、余弦與正切公式（含解析）新人教A版

考點(diǎn)規(guī)范練20兩角和與差的正弦、余弦與正切公式一、基礎(chǔ)鞏固1.sin 35°cos 25°-cos 145°sin 25°=()A.-32B.32C.-12D.122.已知角的頂點(diǎn)與原點(diǎn)重合,始邊與x軸的正半軸重合,終邊在直線(xiàn)y=2x上,則cos 2=()A.-45B.-35C.35D.453.已知,32,且cos =-45,則tan4-等于()A.7B.17C.-17D.-74.若tan =2tan5,則cos-310sin-5=()A.1B.2C.3D.45.已知cos-6+sin =435,則sin+76的值為()A.12B.32C.-45D.-126.若0<yx<2,且tan x=3tan y,則x-y的最大值為()A.4B.6C.3D.27.函數(shù)f(x)=sin 2xsin6-cos 2xcos56在區(qū)間-2,2上的單調(diào)遞增區(qū)間為. 8.已知tan(3-)=-12,tan(-)=-13,則tan =. 9.在ABC中,C=60°,tanA2+tanB2=1,則tanA2tanB2=. 10.已知,均為銳角,且sin =35,tan(-)=-13.(1)求sin(-)的值;(2)求cos 的值.二、能力提升11.在平面直角坐標(biāo)系中,點(diǎn)(3,t)和(2t,4)分別在以頂點(diǎn)為原點(diǎn),始邊為x軸的非負(fù)半軸的角,+45°的終邊上,則t的值為()A.-6或1B.6或1C.6D.112.設(shè)a=cos 50°cos 127°+cos 40°cos 37°,b=22(sin 56°-cos 56°),c=1-tan239°1+tan239°,則a,b,c的大小關(guān)系是()A.a>b>cB.b>a>cC.c>a>bD.a>c>b13.已知sin+4=14,-32,-,則cos+712的值為. 14.設(shè),0,2,且tan =1+sincos,則2-=. 15.(2018浙江,18)已知角的頂點(diǎn)與原點(diǎn)O重合,始邊與x軸的非負(fù)半軸重合,它的終邊過(guò)點(diǎn)P-35,-45.(1)求sin(+)的值;(2)若角滿(mǎn)足sin(+)=513,求cos 的值.三、高考預(yù)測(cè)16.在平面直角坐標(biāo)系中,點(diǎn)P的坐標(biāo)為35,45,Q是第三象限內(nèi)一點(diǎn),|OQ|=1,且POQ=34,則點(diǎn)Q的橫坐標(biāo)為()A.-7210B.-325C.-7212D.-8213考點(diǎn)規(guī)范練20兩角和與差的正弦、余弦與正切公式1.B解析sin35°cos25°-cos145°sin25°=sin35°cos25°+cos35°sin25°=sin(35°+25°)=sin60°=32.2.B解析由題意知tan=2,故cos2=cos2-sin2cos2+sin2=1-tan21+tan2=1-221+22=-35.3.B解析因?yàn)?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?yàn)閠an=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0<yx<2,x-y0,2.又tanx=3tany,tan(x-y)=tanx-tany1+tanxtany=2tany1+3tan2y=21tany+3tany33=tan6.當(dāng)且僅當(dāng)3tan2y=1時(shí)取等號(hào),x-y的最大值為6,故選B.7.-512,12解析f(x)=sin2xsin6-cos2xcos56=sin2xsin6+cos2xcos6=cos2x-6.當(dāng)2k-2x-62k(kZ),即k-512xk+12(kZ)時(shí),函數(shù)f(x)單調(diào)遞增.取k=0,得-512x12,故函數(shù)f(x)在區(qū)間-2,2上的單調(diào)遞增區(qū)間為-512,12.8.17解析由題意得tan=12,所以tan=tan(-)+=tan(-)+tan1-tan(-)tan=-13+121-13×12=17.9.1-33解析由C=60°,則A+B=120°,即A2+B2=60°.根據(jù)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(-)=-13<0,-2<-<0.sin(-)=-1010.(2)由(1)可得,cos(-)=31010.為銳角,且sin=35,cos=45.cos=cos-(-)=coscos(-)+sinsin(-)=45×31010+35×-1010=91050.11.D解析由題意得tan=t3,tan(+45°)=42t=2t.故tan(+45°)=tan45°+tan1-tan45°tan=1+t31-t3=2t,化簡(jiǎn)得t2+5t-6=0,即(t-1)(t+6)=0,解得t=1或t=-6.若t=-6,則角的終邊在第四象限,+45°的終邊也在第四象限,與點(diǎn)(2t,4)的縱坐標(biāo)矛盾.所以t=-6舍去,故t的值為1,故選D.12.D解析a=sin40°cos127°+cos40°sin127°=sin(40°+127°)=sin167°=sin13°,b=22(sin56°-cos56°)=22sin56°-22cos56°=sin(56°-45°)=sin11°,c=1-tan239°1+tan239°=cos239°-sin239°cos239°cos239°+sin239°cos239°=cos239°-sin239°=cos78°=sin12°.sin13°>sin12°>sin11°,a>c>b.故選D.13.-15+38解析由-32,-,得+4-54,-34.因?yàn)閟in+4=14,所以cos+4=-154.cos+712=cos+4+3=cos+4cos3-sin+4sin3=-154×12-14×32=-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.函數(shù)y=sinx在-2,2內(nèi)單調(diào)遞增,由sin(-)=sin2-可得-=2-,即2-=2.15.解(1)由角的終邊過(guò)點(diǎn)P-35,-45,得sin=-45,所以sin(+)=-sin=45.(2)由角的終邊過(guò)點(diǎn)P-35,-45,得cos=-35,由sin(+)=513,得cos(+)=±1213.由=(+)-,得cos=cos(+)cos+sin(+)sin,所以cos=-5665或cos=1665.16.A解析設(shè)xOP=,則cos=35,sin=45.因?yàn)槭堑谌笙迌?nèi)一點(diǎn),所以xQ=cos+34=35×-22-45×22=-7210,故選A.7