紅對勾新課標2016高考數(shù)學大一輪復習第三章三角函數(shù)、解三角形課時作業(yè)19理新人教A版

1、課時作業(yè)19同角三角函數(shù)基本關(guān)系式與誘導公式一、選擇題1sineq f(29,6)coseq blc(rc)(avs4alco1(f(29,3)taneq f(25,4)()A0 B.eq f(1,2)C1Deq f(1,2)解析：原式sin(4eq f(5,6)cos(10eq f(,3)tan(6eq f(,4)sineq f(5,6)coseq f(,3)taneq f(,4)eq f(1,2)eq f(1,2)10.答案：A2若sin是5x27x60的根，則eq f(sinblc(rc)(avs4alco1(f(3,2)sinblc(rc)(avs4alco1(f(3,2)tan22,

2、cosblc(rc)(avs4alco1(f(,2)cosblc(rc)(avs4alco1(f(,2)sin)()A.eq f(3,5) B.eq f(5,3)C.eq f(4,5) D.eq f(5,4)解析：由5x27x60，得xeq f(3,5)或2.sineq f(3,5).原式eq f(coscostan2,sinsinsin)eq f(1,sin)eq f(5,3).答案：B3已知tanx2，則sin2x1()A0 B.eq f(9,5)C.eq f(4,3) D.eq f(5,3)解析：sin2x1eq f(2sin2xcos2x,sin2xcos2x)eq f(2tan2x1

3、,tan2x1)eq f(9,5).答案：B4已知eq blc(rc)(avs4alco1(，f(3,2)，coseq f(4,5)，則taneq blc(rc)(avs4alco1(f(,4)等于()A7 B.eq f(1,7)Ceq f(1,7)D7解析：eq blc(rc)(avs4alco1(，f(3,2)且coseq f(4,5)，sineq f(3,5)，taneq f(3,4).taneq blc(rc)(avs4alco1(f(,4)eq f(tanf(,4)tan,1tanf(,4)tan)eq f(1f(3,4),1f(3,4)eq f(1,7).答案：B5已知eq f(s

4、in3cos,3cossin)5，則sin2sincos的值是()A.eq f(2,5)Beq f(2,5)C2D2解析：由eq f(sin3cos,3cossin)5得eq f(tan3,3tan)5即tan2，所以sin2sincoseq f(sin2sincos,sin2cos2)eq f(tan2tan,tan21)eq f(2,5).答案：A6已知sincoseq f(4,3)eq blc(rc)(avs4alco1(0f(,4)，則sincos的值為()A.eq f(r(2),3)Beq f(r(2),3)C.eq f(1,3)Deq f(1,3)解析：sincoseq f(4,3

5、)，(sincos)21sin2eq f(16,9)，sin2eq f(7,9)，又0eq f(,4)，sincos，sincoseq r(sincos2)eq r(1sin2)eq f(r(2),3).答案：B二、填空題7(tanxeq f(1,tanx)cos2x化簡的結(jié)果是_解析：(tanxeq f(1,tanx)cos2x(eq f(sinx,cosx)eq f(cosx,sinx)cos2xeq f(sin2xcos2x,sinxcosx)cos2xeq f(cosx,sinx)eq f(1,tanx).答案：eq f(1,tanx)8已知角終邊上一點P(4,3)，則eq f(cos

6、f(,2)sin,cosf(11,2)sinf(9,2)的值為_解析：taneq f(y,x)eq f(3,4)，eq f(cosf(,2)sin,cosf(11,2)sinf(9,2)eq f(sinsin,sincos)taneq f(3,4).答案：eq f(3,4)9已知sineq blc(rc)(avs4alco1(f(,12)eq f(1,3)，且eq f(,2)，則coseq blc(rc)(avs4alco1(f(,12)_.解析：sineq blc(rc)(avs4alco1(f(,12)eq f(1,3)，又eq f(,2)，eq f(7,12)eq f(,12)eq f(

7、13,12)，coseq blc(rc)(avs4alco1(f(,12)eq r(1sin2blc(rc)(avs4alco1(f(,12)eq f(2r(2),3).答案：eq f(2r(2),3)三、解答題10已知sin(3)eq f(1,3)，求eq f(cos,coscos1)eq f(cos2,sinf(3,2)cossinf(3,2)的值解：sin(3)sineq f(1,3)，sineq f(1,3)，原式eq f(cos,coscos1)eq f(cos2,sinf(3,2)coscos)eq f(1,1cos)eq f(cos,cos2cos)eq f(1,1cos)eq

8、f(1,1cos)eq f(2,1cos2)eq f(2,sin2)eq f(2,f(1,3)2)18.11已知在ABC中，sinAcosAeq f(1,5).(1)求sinAcosA的值；(2)判斷ABC是銳角三角形還是鈍角三角形；(3)求tanA的值解：(1)sinAcosAeq f(1,5)，兩邊平方得12sinAcosAeq f(1,25)，sinAcosAeq f(12,25)，(2)由sinAcosAeq f(12,25)0，且0A，可知cosA0，cosA0，sinAcosAeq f(7,5)，由，可得sinAeq f(4,5)，cosAeq f(3,5)，tanAeq f(si

9、nA,cosA)eq f(f(4,5),f(3,5)eq f(4,3).1已知sineq f(1,3)，(eq f(,2)，eq f(,2)，則sin(5)sin(eq f(3,2)的值是()A.eq f(2r(2),9)Beq f(2r(2),9)Ceq f(1,9) D.eq f(1,9)解析：sineq f(1,3)，(eq f(,2)，eq f(,2)，coseq r(1sin2)eq f(2r(2),3).原式sin()(cos)sincoseq f(1,3)eq f(2r(2),3)eq f(2r(2),9).答案：B2當0 xeq f(,4)時，函數(shù)f(x)eq f(cos2x,

10、cosxsinxsin2x)的最小值是()A.eq f(1,4) B.eq f(1,2)C2D4解析：當0 xeq f(,4)時，0tanx1，f(x)eq f(cos2x,cosxsinxsin2x)eq f(1,tanxtan2x)，設(shè)ttanx，則0t1，yeq f(1,tt2)eq f(1,t1t)eq f(1,f(t1t,2)2)4.當且僅當t1t，即teq f(1,2)時等號成立答案：D3已知coseq blc(rc)(avs4alco1(f(,6)a(|a|1)，則coseq blc(rc)(avs4alco1(f(5,6)sineq blc(rc)(avs4alco1(f(2,

11、3)的值是_解析：coseq blc(rc)(avs4alco1(f(5,6)coseq blcrc(avs4alco1(blc(rc)(avs4alco1(f(,6)coseq blc(rc)(avs4alco1(f(,6)a.sineq blc(rc)(avs4alco1(f(2,3)sineq blcrc(avs4alco1(f(,2)blc(rc)(avs4alco1(f(,6)coseq blc(rc)(avs4alco1(f(,6)a，coseq blc(rc)(avs4alco1(f(5,6)sineq blc(rc)(avs4alco1(f(2,3)0.答案：04在平面直角坐標

12、系xOy中，鈍角eq f(,4)的頂點為坐標原點，始邊與x軸的非負半軸重合若角eq f(,4)的終邊與單位圓x2y21交于點Aeq blc(rc)(avs4alco1(f(3,5)，t).(1)求sin的值；(2)設(shè)f(x)coseq blc(rc)(avs4alco1(f(,2)x)，求f(1)f(2)f(9)解：(1)由三角函數(shù)的定義，得coseq blc(rc)(avs4alco1(f(,4)eq f(3,5)，sineq blc(rc)(avs4alco1(f(,4)eq f(4,5).sinsineq blcrc(avs4alco1(blc(rc)(avs4alco1(f(,4)f(,4)sineq blc(rc)(avs4alco1(f(,4)coseq f(,4)coseq blc(rc)(avs4alco1(f(,4)sineq f(,4)eq f(4,5)eq f(r(2),2)eq blc(rc)(avs4alco1(f(3,5)eq f(r(2),2)eq f(7r(2),10).(2)f(1)coseq blc(rc)(avs4alco1(f(,2)sin，f(2)coseq blc(rc)(avs4alco1(2f(,2)cos，f(3)coseq blc(rc)(avs4a