1、5.2.2同角三角函数的基本关系课后篇巩固提升合格考达标练1.已知cos =45,且322,则1tan的值为()A.34B.-34C.43D.-43答案D解析因为cos=45,且320,cos0,所以cos1|cos|+sin1|sin|=-1+1=0,即原式等于0.7.(2021福建泉州质检)已知为第四象限角,sin +3cos =1,则tan =.答案-43解析由题意知(sin+3cos)2=sin2+cos2,得6sincos=-8cos2,又因为为第四象限角,所以cos0,所以6sin=-8cos,所以tan=-43.8.已知tan =23,求下列各式的值:(1)cos-sincos+
2、sin+cos+sincos-sin;(2)1sincos.解(1)cos-sincos+sin+cos+sincos-sin=1-tan1+tan+1+tan1-tan=1-231+23+1+231-23=265.(2)1sincos=sin2+cos2sincos=tan2+1tan=136.等级考提升练9.化简11+tan2160的结果为()A.-cos 160B.cos 160C.1cos160D.1-cos160答案A解析原式=11+sin2160cos2160=1cos2160+sin2160cos2160=11cos2160=cos2160=|cos160|=-cos160.故选
3、A.10.(2021河南郑州高一月考)已知是第三象限角,且sin4+cos4=59,则sin cos 的值为()A.23B.-23C.13D.-13答案A解析为第三象限角,则sin0,cos0,sincos=23.11.若是ABC的一个内角,且sin cos =-18,则sin -cos 的值为()A.-32B.32C.-52D.52答案D解析由题意知2,所以sin-cos0,sin-cos=(sin-cos)2=1-2sincos=52.12.若cos +2sin =-5,则tan 等于()A.12B.2C.-12D.-2答案B解析(方法一)由cos+2sin=-5,cos2+sin2=1联
4、立消去cos,得(-5-2sin)2+sin2=1.化简得5sin2+45sin+4=0,(5sin+2)2=0,sin=-255.cos=-5-2sin=-55.tan=sincos=2.(方法二)cos+2sin=-5,cos2+4sincos+4sin2=5.cos2+4sincos+4sin2cos2+sin2=5.1+4tan+4tan21+tan2=5,tan2-4tan+4=0.(tan-2)2=0,tan=2.13.(多选题)化简cos1-sin2+2sin1-cos2的值为()A.-1B.1C.-3D.2答案ABC解析原式=cos|cos|+2sin|sin|,当为第一象限角
5、时,上式值为3;当为第二象限角时,上式值为1;当为第三象限角时,上式值为-3;当为第四象限角时,上式值为-1.14.(多选题)已知tan2x-2tan2y-1=0,则下列式子成立的是()A.sin2y=2sin2x+1B.sin2y=-2sin2x-1C.sin2y=2sin2x-1D.sin2y=1-2cos2x答案CD解析tan2x-2tan2y-1=0,sin2xcos2x-2sin2ycos2y-1=0,整理得sin2xcos2y-2sin2ycos2x=cos2ycos2x,(1-cos2x)(1-sin2y)-sin2ycos2x=(cos2y+sin2y)cos2x,即1-cos
6、2x-sin2y+sin2ycos2x-sin2ycos2x=cos2x,即sin2y=1-2cos2x=2sin2x-1,C,D正确.15.已知cos+4=13,02,则sin+4=.答案223解析sin2+4+cos2+4=1,sin2+4=1-19=89.02,4+40,且a1,若loga(sin x-cos x)=0,则sin8x+cos8x=.答案1解析设a0且a1,若loga(sinx-cosx)=0,所以sinx-cosx=a0=1,所以(sinx-cosx)2=sin2x+cos2x-2sinxcosx=1,又sin2x+cos2x=1,所以sinxcosx=0,又由(sin2
7、x+cos2x)2=sin4x+cos4x+2sin2xcos2x=1,则sin4x+cos4x=1,所以sin8x+cos8x=(sin4x+cos4x)2-2sin4xcos4x=(sin4x+cos4x)2=1.17.若322,化简:1-cos1+cos+1+cos1-cos.解322,sin0.原式=(1-cos)2(1+cos)(1-cos)+(1+cos)2(1-cos)(1+cos)=(1-cos)2sin2+(1+cos)2sin2=|1-cos|sin|+|1+cos|sin|=-1-cossin-1+cossin=-2sin.新情境创新练18.已知(0,),且sin ,co
8、s 是方程25x2-5x-12=0的两个根,求sin3+cos3和tan -1tan的值.解(方法一)由题意得sin+cos=15,sincos=-1225,易知2.sin3+cos3=(sin+cos)(sin2-sincos+cos2)=(sin+cos)(1-sincos)=151+1225=37125.tan-1tan=sincos-cossin=sin2-cos2sincos=(sin+cos)(sin-cos)sincos.(0,),sincos0,cos0.sin-cos=(sin-cos)2=1-2sincos=1+21225=4925=75.tan-1tan=1575-1225=-712.(方法二)方程25x2-5x-12=0的两根分别为45和-35.(0,),且sincos=-12250,cos0,则sin=45,cos=-35,sin3+cos3=453+-353=64125-27125=37125,tan-1tan=sincos-cossin=45-35-3545=-43+34=-712.
