1、5.2.2同角三角函数的基本关系刷新题夯基础 题组一利用同角三角函数的基本关系求值1.(2020福建南平高一下期末)已知为第二象限角,且sin =35,则tan =()A.34B.-43C.-34D.432.已知sin =55,则sin4-cos4的值为() A.-15B.-35C.15D.353.已知是第三象限角,且sin4+cos4=59,则sin cos 的值为()A.23B.-23C.13D.-134.(2020辽宁省实验中学高一下期中)已知cos =45,且为第四象限的角,则tan =.5.已知sin +cos =12,则sin cos =.6.已知cos =-35,且tan 0,则
2、sincos21-sin=.题组二利用同角三角函数的基本关系化简或证明 7.(2020山西长治二中高一下期末)已知sin +cos =2,则tan +cossin的值为()A.-1B.-2C.12 D.2 8.化简:cos-751-sin235=()A.tan 35B.-1tan 35C.1 D.-19.化简sin2+cos4+sin2cos2的结果是()A.14 B.12 C.1 D.3210.求证:2sinxcosx-1cos2x-sin2x=tanx-1tanx+1.题组三正、余弦齐次式的求值问题11.(2021黑龙江哈尔滨六中高一上月考)已知tan =-13,则-2cossin+cos
3、的值为()A.-3B.-34C.-43D.3412.(2020辽宁葫芦岛高一下期末)若3sin+5cossin-2cos=-15,则tan 的值为()A.32B.-32C.2316D.-231613.(2020广东湛江高一下期末)已知tan =3,则sin2-cos2=.14.已知角的始边与x轴的非负半轴重合,顶点与坐标原点重合,终边过点P(3,4),则sin+2cossin-cos=.15.已知tan =23,求下列各式的值:(1)cos-sincos+sin+cos+sincos-sin;(2)1sincos;(3)sin2-2sin cos +4cos2.16.已知2cos2+3cos
4、sin -3sin2=1,-32,-.求:(1)tan ;(2)2sin-3cos4sin-9cos.刷新题培素养 题组一利用同角三角函数的基本关系求值1.(2021四川成都树德中学高一上段测,)是第四象限角,tan =-512,则sin =()A.15B.-15C.513D.-5132.(2020河北石家庄二中高一上期末,)设tan 160=k,则sin 160=()A.-11+k2B.-k1+k2C.k1+k2D.11+k23.()已知02,ln(1+cos )=s,ln 11-cos=t,则ln(sin )=()A.s-tB.s+tC.12(s-t)D.12(s+t)4.(2021江苏扬
5、中高级中学等八校高一上联考,)已知sin +cos =-15.(1)求sin cos 的值;(2)若20,sin cos =23.4.答案-34解析因为cos =45,且为第四象限的角,所以sin =-1-cos2=-1-452=-35,所以tan =sincos=-3545=-34.5.答案-38解析sin +cos =12,(sin +cos )2=14,即sin2+2sin cos +cos2=14,即1+2sin cos =14,sin cos =-38.6.答案-425解析由cos =-350,知是第三象限角,所以sin =-45,故sincos21-sin=-459251+45=-
6、425.7.Dsin +cos =2,(sin +cos )2=2,sin cos =12,tan +cossin=sincos+cossin=1sincos=2.8.D原式=cos-2+35cos235=cos 35cos 35=-1.9.C原式=sin2+cos2(cos2+sin2)=sin2+cos2=1.10.证明证法一:左边=2sinxcosx-(sin2x+cos2x)cos2x-sin2x=-(sin2x-2sinxcosx+cos2x)cos2x-sin2x=(sinx-cosx)2sin2x-cos2x=(sinx-cosx)2(sinx-cosx)(sinx+cosx)=
7、sinx-cosxsinx+cosx=tanx-1tanx+1=右边,原等式成立.证法二:右边=sinxcosx-1sinxcosx+1=sinx-cosxsinx+cosx,左边=1-2sinxcosxsin2x-cos2x=(sinx-cosx)2sin2x-cos2x=(sinx-cosx)2(sinx-cosx)(sinx+cosx)=sinx-cosxsinx+cosx,左边=右边,故原等式成立.11.A因为tan =-13,所以-2cossin+cos=-2tan+1=-2-13+1=-3.故选A.12.D因为3sin+5cossin-2cos=3tan+5tan-2=-15,所以
8、tan =-2316.故选D.13.答案45解析因为tan =3,所以sin2-cos2=sin2-cos2sin2+cos2=sin2cos2-1sin2cos2+1=tan2-1tan2+1=9-19+1=45.故答案为45.14.答案10解析根据角的终边过点P(3,4),利用三角函数的定义,可以求得tan =43,所以sin+2cossin-cos=tan+2tan-1=43+243-1=10313=10.15.解析(1)cos-sincos+sin+cos+sincos-sin=1-tan1+tan+1+tan1-tan,将tan =23代入,得原式=1-231+23+1+231-23
9、=265.(2)1sincos=sin2+cos2sincos=tan2+1tan,将tan =23代入,得原式=136.(3)sin2-2sin cos +4cos2=sin2-2sincos+4cos2sin2+cos2=tan2-2tan+4tan2+1,将tan =23代入,得原式=49-43+449+1=2813.16.解析(1)2cos2+3cos sin -3sin2=2cos2+3cossin-3sin2sin2+cos2=2+3tan-3tan2tan2+1=1,即4tan2-3tan -1=0,解得tan =-14或tan =1.-32,-,为第二象限角,tan 0,tan
10、 =-14.(2)tan =-14,原式=2sincos-3coscos4sincos-9coscos=2tan-34tan-9=2(-14)-34(-14)-9=720.刷新题培素养1.Dcos2=11+tan2=11+-5122=122132,且是第四象限角,cos =1213,sin =tan cos =-513,故选D.2.Btan 160=sin160cos160=k,sin 160=kcos 160.又sin2160+cos2160=1,(kcos 160)2+cos2160=1,cos2160=1k2+1.又160角是第二象限角,cos 1600,cos 160=-11+k2,s
11、in 160=kcos 160=-k1+k2.故选B.3.C依题意得s-t=ln(1+cos )+ln(1-cos )=ln(1-cos2)=ln(sin2),00,s-t=2ln(sin ),即ln(sin )=12(s-t),故选C.4.解析(1)由已知得(sin +cos )2=1+2sin cos =125,sin cos =-1225.(2)1sin-1cos=cos-sinsincos,(cos -sin )2=1-2sin cos =1-2-1225=4925,又2,cos 0,cos -sin 0,cos -sin =-75,原式=-75-1225=3512.5.D为第四象限角,sin 0,cos 0,所以sin -cos =(sin-cos)2=1-2sincos=1713.
