（浙江專用）2020版高考數(shù)學(xué)大一輪復(fù)習(xí) 第四章 三角函數(shù)、解三角形 考點(diǎn)規(guī)范練16 同角三角函數(shù)的基本關(guān)系及誘導(dǎo)公式

考點(diǎn)規(guī)范練16同角三角函數(shù)的基本關(guān)系及誘導(dǎo)公式基礎(chǔ)鞏固組1.sin 600°的值為()A.-12B.-32C.12D.32答案B解析sin600°=sin(360°+240°)=sin240°=sin(180°+60°)=-sin60°=-32.2.已知sin2+=-35,2,則tan =()A.34B.-34C.-43D.43答案C解析sin2+=-35,sin2+=cos,cos=-35,又2,sin=1-cos2=45,tan=sincos=-43.故選C.3.若cos(3-x)-3cosx+2=0,則tan x等于()A.-12B.-2C.12D.13答案D解析cos(3-x)-3cosx+2=0,-cosx+3sinx=0.tanx=13.故選D.4.1+2sin(-3)cos(+3)化簡(jiǎn)的結(jié)果是()A.sin 3-cos 3B.cos 3-sin 3C.±(sin 3-cos 3)D.以上都不對(duì)答案A解析sin(-3)=sin3,cos(+3)=-cos3,原式=1-2sin3·cos3=(sin3-cos3)2=|sin3-cos3|.2<3<,sin3>0,cos3<0.原式=sin3-cos3.故選A.5.已知tan =3,則1+2sincossin2-cos2的值是()A.12B.2C.-12D.-2答案B解析原式=sin2+cos2+2sincossin2-cos2=(sin+cos)2(sin+cos)(sin-cos)=sin+cossin-cos=tan+1tan-1=3+13-1=2.6.sin(-1 071°)sin 99°+sin(-171°)sin(-261°)+tan(-1 089°)tan(-540°)=. 答案0解析原式=(-sin1071°)·sin99°+sin171°·sin261°+tan1089°·tan540°=-sin(3×360°-9°)sin(90°+9°)+sin(180°-9°)sin(270°-9°)+tan(3×360°+9°)tan(360°+180°)=sin9°cos9°-sin9°cos9°+tan9°tan180°=0.7.(2018浙江紹興3月模擬)已知sin(30°+)=35,60°<<150°,則cos(30°+)=;cos =. 答案-453-4310解析60°<<150°,90°<30°+<180°.sin(30°+)=35,cos(30°+)=-1-sin2(30°+)=-1-(35) 2=-45.cos=cos(30°+-30°)=cos(30°+)cos30°+sin(30°+)·sin30°=-45×32+35×12=3-4310,故答案為-45,3-4310.8.已知R,sin2+4sin cos +4cos2=52,則tan =. 答案3或-13解析由sin2+4sincos+4cos2=52可得52=sin2+4sincos+4cos2sin2+cos2=tan2+4tan+4tan2+1,解得tan=3或tan=-13.能力提升組9.已知:sin(-1 000°);cos(-2 200°);tan(-10);sin710costan179.其中符號(hào)為負(fù)的是()A.B.C.D.答案C解析sin(-1000°)=sin80°>0;cos(-2200°)=cos(-40°)=cos40°>0,tan(-10)=tan(3-10)<0;sin710·costan179=-sin710tan179,sin710>0,tan179<0.故選C.10.已知傾斜角為的直線與直線x-3y+1=0垂直,則23sin2-cos2=()A.103B.-103C.1013D.-1013答案C解析直線x-3y+1=0的斜率為13,因此與此直線垂直的直線的斜率k=-3,tan=-3,23sin2-cos2=2(sin2+cos2)3sin2-cos2=2(tan2+1)3tan2-1,把tan=-3代入得,原式=2×(-3)2+13×(-3)2-1=1013.故選C.11.已知cos512+=13,且-<<-2,則cos12-等于()A.223B.13C.-13D.-223答案D解析因?yàn)?12+12-=2,所以cos12-=sin2-12-=sin512+.因?yàn)?<<-2,所以-712<+512<-12.又cos512+=13>0,所以-2<+512<-12,所以sin512+=-1-cos2512+=-1-132=-223.12.若1sin+1cos=3,則sin cos =()A.-13B.13C.-13或1D.13或-1答案A解析由1sin+1cos=3,可得sin+cos=3sincos,兩邊平方,得1+2sincos=3sin2cos2,解得sincos=-13或sincos=1.由題意,知-1<sin<1,-1<cos<1,且sin0,cos0,所以sincos1,故選A.13.(2018浙江溫州模擬)已知sin x-sin y=-23,cos x-cos y=23,且x,y為銳角,則tan (x-y)的值是()A.2145B.-2145C.±2145D.-1414答案B解析sinx-siny=-23,cosx-cosy=23,兩式平方相加得,cos(x-y)=59,又x,y為銳角,sinx-siny<0,x<y,sin(x-y)=-1-cos2(x-y)=-2149,tan(x-y)=sin(x-y)cos(x-y)=-214959=-2145.14.(2018浙江金華十校)已知sin 2=2425,0<<2,則2cos4-的值為. 答案75解析sin2=2sincos=2425,0<<2,sin和cos均為正數(shù),又(sin+cos)2=sin2+cos2+2sincos=1+2425=4925,所以sin+cos=75,2cos4-=222cos+22sin=sin+cos=75.15.(2018浙江杭州二中6月熱身)已知R,sin +2cos =102,則tan =;tan 2=. 答案3或-13-34解析由sin+2cos=102可以得到sin2+4sincos+4cos2=52,即sin2+4sincos+4cos2sin2+cos2=52,化簡(jiǎn)得tan2+4tan+4tan2+1=52,整理得3tan2-8tan-3=0,解得tan=3或tan=-13.當(dāng)tan=3時(shí),tan2=2×31-32=-34;當(dāng)tan=-13時(shí),tan2=2×(-13)1-132=-34.16.當(dāng)0<x<4時(shí),函數(shù)f(x)=cos2xcosxsinx-sin2x的最小值是. 答案4解析當(dāng)0<x<4時(shí),0<tanx<1,f(x)=cos2xcosxsinx-sin2x=1tanx-tan2x,設(shè)t=tanx,則0<t<1,y=1t-t2=1t(1-t)4,當(dāng)且僅當(dāng)t=1-t,即t=12時(shí)等號(hào)成立.17.(2018浙江平湖中學(xué)模擬)已知tan =-34,求下列各式的值:(1)sin(+32)+cos(-2)2sin(+)+cos(-);(2)2+sin cos -cos2.解(1)原式=-cos+sin-2sin-cos=-1+tan-2tan-1=-1-3464-1=-72;(2)原式=2+sincos-cos2sin2+cos2=2+tan-1tan2+1=2+-34-1916+1=2225.18.已知f()=sin(-)cos(2-)tan-+32tan2+·sin(-).(1)化簡(jiǎn)f();(2)若是第三象限角,且cos-32=15,求f()的值.解(1)f()=sin·cos·tan-+32-2tan2+·sin=sin·cos·-tan2+tan2+·sin=-cos.(2)cos-32=-sin=15,sin=-15,又是第三象限角,cos=-1-sin2=-265,故f()=265.6