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1、考点规范练16同角三角函数的基本关系及诱导公式考点规范练第18页基础巩固组1.sin 600的值为()A.-12B.-32C.12D.32答案B解析sin 600=sin(360+240)=sin 240=sin(180+60)=-sin 60=-32.2.已知sin2+=-35,2,则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.故选C.3.若cos(3-x)-3cosx+2=0,则tan x等于()A.-12B.-2C.12D.13答案D解析cos(
2、3-x)-3cosx+2=0,-cos x+3sin x=0.tan x=13.故选D.4.1+2sin(-3)cos(+3)化简的结果是()A.sin 3-cos 3B.cos 3-sin 3C.(sin 3-cos 3)D.以上都不对答案A解析sin(-3)=sin 3,cos(+3)=-cos 3,原式=1-2sin3cos3=(sin3-cos3)2=|sin 3-cos 3|.230,cos 30.原式=sin 3-cos 3.故选A.5.已知tan =3,则1+2sincossin2-cos2的值是()A.12B.2C.-12D.-2答案B解析原式=sin2+cos2+2sinco
3、ssin2-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=(-sin 1 071)sin 99+sin 171sin 261+tan 1 089tan 540=-sin(3360-9)sin(90+9)+sin(180-9)sin(270-9)+tan(3360+9)tan(360+180)=sin 9cos 9-sin 9cos 9+tan 9tan 180=0.7
4、.(2018浙江绍兴3月模拟)已知sin(30+)=35,60150,则cos(30+)=;cos =.答案-453-4310解析60150,9030+0;cos(-2 200)=cos(-40)=cos 400,tan(-10)=tan(3-10)0,tan1790.故选C.10.已知倾斜角为的直线与直线x-3y+1=0垂直,则23sin2-cos2=()A.103B.-103C.1013D.-1013答案C解析直线x-3y+1=0的斜率为13,因此与此直线垂直的直线的斜率k=-3,tan =-3,23sin2-cos2=2(sin2+cos2)3sin2-cos2=2(tan2+1)3ta
5、n2-1,把tan =-3代入得,原式=2(-3)2+13(-3)2-1=1013.故选C.11.已知cos512+=13,且-2,则cos12-等于()A.223B.13C.-13D.-223答案D解析因为512+12-=2,所以cos12-=sin2-12-=sin512+.因为-2,所以-712+5120,所以-2+512-12,所以sin512+=-1-cos2512+=-1-132=-223.12.若1sin+1cos=3,则sin cos =()A.-13B.13C.-13或1D.13或-1答案A解析由1sin+1cos=3,可得sin +cos=3sin cos ,两边平方,得1
6、+2sin cos =3sin2cos2,解得sin cos =-13或sin cos =1.由题意,知-1sin 1,-1cos 1,且sin 0,cos 0,所以sin cos 1,故选A.13.(2018浙江温州模拟)已知sin x-sin y=-23,cos x-cos y=23,且x,y为锐角,则tan (x-y)的值是()A.2145B.-2145C.2145D.-1414答案B解析sin x-sin y=-23,cos x-cos y=23,两式平方相加得,cos (x-y)=59,又x,y为锐角,sin x-sin y0,xy,sin(x-y)=-1-cos2(x-y)=-21
7、49,tan(x-y)=sin(x-y)cos(x-y)=-214959=-2145.14.(2018浙江金华十校)已知sin 2=2425,02,则2cos4-的值为.答案75解析sin 2=2sin cos =2425,02,sin 和cos 均为正数,又(sin +cos )2=sin2+cos2+2sin cos =1+2425=4925,所以sin +cos =75,2cos4-=222cos +22sin =sin +cos =75.15.(2018浙江杭州二中6月热身)已知R,sin +2cos =102,则tan =;tan 2=.答案3或-13-34解析由sin +2cos
8、=102可以得到sin2+4sin cos +4cos2=52,即sin2+4sincos+4cos2sin2+cos2=52,化简得tan2+4tan+4tan2+1=52,整理得3tan2-8tan -3=0,解得tan =3或tan =-13.当tan =3时,tan 2=231-32=-34;当tan =-13时,tan 2=2(-13)1-132=-34.16.当0x4时,函数f(x)=cos2xcosxsinx-sin2x的最小值是.答案4解析当0x4时,0tan x1,f(x)=cos2xcosxsinx-sin2x=1tanx-tan2x,设t=tan x,则0t1,y=1t-
9、t2=1t(1-t)4,当且仅当t=1-t,即t=12时等号成立.17.(2018浙江平湖中学模拟)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)化简f();(2)若是第三象限角,且cos-32=15,求f()的值.解(1)f()=sincostan-+32-2tan2+sin=sincos-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.
