1、PAGE PAGE 6考點(diǎn)規(guī)范練20同角三角函數(shù)的基本關(guān)系及誘導(dǎo)公式基礎(chǔ)鞏固1.已知sin(+)0,則下列不等關(guān)系中必定成立的是()A.sin0B.sin0,cos0,cos0D.sin0,cos0且a1)的圖象過(guò)定點(diǎn)P,且角的始邊與x軸的非負(fù)半軸重合,終邊過(guò)點(diǎn)P,則cos112-sin92+sin2cos2+sin-=()A.23B.-23C.32D.-326.已知sin =3cos ,則sin2+sin cos+1=()A.4+34B.7+34C.1D.37.已知sinx+6=14,則sin56-x+cos3-x的值為()A.0B.14C.12D.-128.(2021寧夏銀川一中高三月考)

2、已知sin cos=-18,434,則sin +cos的值等于()A.32B.-32C.34D.-349.(2021上海楊浦一模)已知sin =-55,-2,2,則sin+2=.10.已知tan(-)=2,則sin+cossin-cos=.11.(2021廣西崇左高三檢測(cè))化簡(jiǎn):sin2+cos2-cos(+)+sin(-)cos2+sin(+)=.12.已知kZ,則sin(k-)cos(k-1)-sin(k+1)+cos(k+)的值為.能力提升13.(2021海南?？谀M)已知tan +1tan=4,則sin4+cos4=()A.38B.12C.34D.7814.已知sin1+cos=2,則

3、tan =()A.-43B.34C.43D.215.已知在ABC中,3sin2-A=3sin(-A),且cosA=-3cos(-B),則ABC為()A.等腰三角形B.直角三角形C.等腰直角三角形D.等邊三角形16.已知cos512+=13,且-2,則cos12-等于()A.223B.-13C.13D.-22317.(2021上海衛(wèi)育中學(xué)高三月考)若sin ,cos是關(guān)于x的方程x2-ax+a=0的兩個(gè)根,則a=.高考預(yù)測(cè)18.(2021河北唐山模擬)已知tan =-3,則sin2+cos32-=()A.110B.310C.12D.35答案：1.B解析sin(+)0,-sin0.cos(-)0,

4、-cos0,即cos0且a1)的圖象過(guò)定點(diǎn)P(2,3),則由題意可得tan=32.于是cos112-sin92+sin2cos2+sin(-)=cos32-sin2+sin2-sinsin=-sincos+2sincos-sin2=sincos-sin2=-cossin=-1tan=-23.6.B解析sin=3cos,tan=3,sin2+sincos+1=2sin2+sincos+cos2sin2+cos2=2tan2+tan+1tan2+1=23+3+13+1=7+34.7.C解析因?yàn)閟inx+6=14,所以sin56-x+cos3-x=sin-x+6+cos2-x+6=2sinx+6=2

5、14=12.故選C.8.A解析sincos=-180,434,20,(sin+cos)2=sin2+cos2+2sincos=1-14=34,sin+cos=32.9.255解析因?yàn)閟in=-55,-2,2,所以-2,0,可得cos0,所以cos=1-sin2=1-552=255,所以sin+2=cos=255.10.13解析tan(-)=2,tan=-2,sin+cossin-cos=tan+1tan-1=-2+1-2-1=13.11.0解析原式=cossin-cos+sin(-sin)-sin=-sin+sin=0.12.-1解析當(dāng)k=2n(nZ)時(shí),原式=sin(2n-)cos(2n-1

6、)-sin(2n+1)+cos(2n+)=sin(-)cos(-)sin(+)cos=-sin(-cos)-sincos=-1.當(dāng)k=2n+1(nZ)時(shí),原式=sin(2n+1)-cos(2n+1-1)-sin(2n+1+1)+cos(2n+1)+=sin(-)cossincos(+)=sincossin(-cos)=-1.綜上,原式=-1.13.D解析tan+1tan=sincos+cossin=sin2+cos2sincos=1sincos=4,則sincos=14.故sin4+cos4=(sin2+cos2)2-2sin2cos2=1-2116=78.14.A解析sin1+cos=2,s

7、in=2+2cos,兩邊平方,得sin2=4+8cos+4cos2,即1-cos2=4+8cos+4cos2,整理得,5cos2+8cos+3=0,解得cos=-1,或cos=-35.當(dāng)cos=-1時(shí),1+cos=0,sin1+cos無(wú)意義,cos=-35,此時(shí),sin=45,tan=sincos=-43.15.B解析由3sin2-A=3sin(-A),得3cosA=3sinA,所以tanA=33,又因?yàn)锳是ABC的內(nèi)角,所以A=6.由cosA=-3cos(-B),得cos6=3cosB,所以cosB=12,所以B=3.所以C=-A-B=2.故ABC為直角三角形.16.D解析sin12-=cos512+=13,-2,71212-1312.cos12-=-1-sin212-=-223.17.1-2解析由題意,=a2-4a0,sin+cos=a,sincos=a,所以a4或a0,且sin+cos=sincos,所以(sin+cos)2=(sincos)2,所以1+2s