福建專用2019高考數(shù)學一輪復習課時規(guī)范練18同角三角函數(shù)的基本關系及誘導公式理新人教A版.docVIP

課時規(guī)范練18同角三角函數(shù)的基本關系及誘導公式一、基礎鞏固組1.已知sin(+)0,則下列不等關系中必定成立的是()A.sin 0B.sin 0,cos 0,cos 0D.sin 0,cos 02.若cos(3-x)-3cosx+2=0,則tan x等于()A.-12B.-2C.12D.133.已知銳角滿足5的終邊上有一點P(sin(-50),cos 130),則的值為()A.8B.44C.26D.404.1-2sin(+2)cos(-2)等于()A.sin 2-cos 2B.sin 2+cos 2C.(sin 2-cos 2)D.cos 2-sin 25.sin296+cos-293-tan254=()A.0B.12C.1D.-126.已知為銳角,且tan(-)+3=0,則sin 的值是()A.13B.31010C.377D.3557.已知sin(-)=-2sin2+,則sin cos 等于()A.25B.-25C.25或-25D.-158.已知cos512+=13,且-2,則cos12-等于()A.223B.-13C.13D.-223導學號215007189.已知sin +2cos =0,則2sin cos -cos2的值是.10.若f(cos x)=cos 2x,則f(sin 15)=.11.已知為第二象限角,則cos 1+tan2+sin 1+1tan2=.12.已知kZ,則sin(k-)cos(k-1)-sin(k+1)+cos(k+)的值為.二、綜合提升組13.若3sin +cos =0,則1cos2+2sincos的值為()A.103B.53C.23D.-214.已知sin =m-3m+5,cos =4-2mm+5,其中2,則下列結論正確的是()A.3m9B.3mlog2x的解集為.課時規(guī)范練18同角三角函數(shù)的基本關系及誘導公式1.Bsin(+)0,-sin 0.cos(-)0,-cos 0,即cos 0.故選B.2.Dcos(3-x)-3cosx+2=0,-cos x+3sin x=0,tan x=13,故選D.3.B點P(sin(-50),cos 130)化簡為P(cos 220,sin 220),因為090,所以5=220,所以=44.故選B.4.A1-2sin(+2)cos(-2)=1-2sin2cos2=(sin2-cos2)2=|sin 2-cos 2|=sin 2-cos 2.5.A原式=sin4+56+cos-10+3-tan6+4=sin56+cos3-tan4=12+12-1=0.6.B由tan(-)+3=0得tan =3,即sincos=3,sin =3cos ,所以sin2=9(1-sin2),10sin2=9,sin2=910.又因為為銳角,所以sin =31010.7.Bsin(-)=-2sin2+,sin =-2cos ,tan =-2.sin cos =sincossin2+cos2=tan1+tan2=