1、1東風高級中學高一數(shù)學限時訓練、選擇題A.1答案3n33.已知COS(+a)=，且a是第四象限角，貝卩cos(3n+a)()254443A.匚B.C+ D555.5答案 B解析3n331.3誘導公式習題課1.cos(20n）等于（解析cos(20nV20n)=cos 3-cos(6n2n2n+2)=cos(n)=ncos312.2.若sin(31，則7ncos(2a）等于（B.D.解析由已知，得sina=17n2，則cos(2a)= sin12.2-cos(2+a)=-sina =5，3答案Cn解析f(cosx)=fsin( x)=3cos(n 2x)=3+cos2x6.若角A,B, C是厶A

2、BC的三個內(nèi)角，則下列等式中一定成立的是（）答案D解析TA+B+C=n，A+B=nC,cos(A+B)= cosC, sin(A+B)=sinC4.若sin( 31a)=35n貝 Scos( 6 a）的值為（)1 AB1.一二C.2.2D333答案 B解析5ncos(6na)=cos +(n3 a)2.23=sin(35.若f(sinx)=3cos2x，貝Uf(cosx)等于()A.3cos2xB. 3si n2xC. 3+cos2xD. 3+si n2xA.cos(A+B=cosCAC.cos( 2+C)=sinBB. sin(A+E)= sinCB+ CAD. sin -=cos2-COS

3、( 3n + a)= COSa = 1Sina =14所以A, B都不正確；同理，B+C=nA,5B+CnA A所以sin = sin( q)=cos，因此D是正確的.、rtn7.設tan(5n + a)=m( a半kn+ 2,kZ),a5n)COs(3n a)=sin(5n a)cos(2n + n a=sin(n a)cos(n a)=sina (COsa)=Sina COsaSinaCOsa 2 2Sina +cosa則+ COS COS的值為（m+1m-1m1m+1.1 D.1答案A解析sina COSatan(5n+a)=m,.tana=m,原式=Sina +cosatana 1ta

4、na +1m-1m+1m+1m- 1,故選A.8若sinasina +COsa=2,COsa則sin(a5n)COs(3na)等于()答案解析3103_10丄sinasina cosa =+COsa2, 得tanoc3.則sin(6tanatan2a +17=109.已知n為整數(shù)，化簡倉n;；a所得結果是（）A.tan(na)B.tan(na)C. tanD.tana答案解析若n=2k(kZ)，則nn + a=sinann+akn+aCOSatana；若nn + an=2k+1(kZ),則亦S1112 kn+n+asinatana.n + a COSa、na為銳角，2tan(na)=3cos(

5、+B)= 5,tan(n + a)+6sin(B)=1,則sina =()3.5AA.5B.罕C.3, 1010答案C解析由已知可得，一2ta na +3si ntana3,擠.3小0故sina選C._ 、填空題已知an311.(0,2),tan(n aJ=4,答案35解析由于tan(na)=-tana =10.D貝卩sina3,則tanaB +5=0,tana 6sinB =1解得3=4,89sina3解方程組COSa4.2 2Sina+COSa=13冗得sina=，又a(0,)，所以sina0.52答案答案答案12.所以sina=3=5.n、右COS( +0)=f，則COS(0)=13.解

6、析已知sin(5nCOS(0)=cosn(nn36+ 0)= COS(-+ 0)=虧.3=4,則sin(nn14.解析sin(已知sin(a)=COS4)=5,)=COSa=3=4.3=4,那么cos(a+4)的值是10解析nn+才)-(a-匸)11nnnn3cos(a+才)=cos +(a4)= sin(a才)=5.15.設f(x)=asin(nx+a)+bcos(nx+3),其中a、b、a、B為非零常數(shù).若f(2014)=1,貝 Sf(2015)等于_ .答案1解析Tf(2015)=asin(2015n+a)+bcos(2015n+3)=asin(n+2014n + a)+bcos(n +

7、2014n + 3)=asin(2014n + a)bcos(2014n +3)= f(2014)，又f(2014)=1,三、解答題16.求值:(1)sin1 320 ;(2)cos(31n5n5n(2)cos( 6)=cos(6n +6 )=cosg=cos(n 百)=cos=A.-2B鳥C.-1+3D.扌+3答案B解析sin600+tan240 =sin(360 +240) +tan(18 0 +60)sin240 +tan60=sin(180+60 )+tan60 =sin60+tan60 =于+33=2 . f(2015)=1.31n_).解析(1)sin1 320=sin(3=sin

8、(180+60)= sin60 x360+240 )=sin240二2 .12)(3)sin600+tan240 的值是(1315nn-+ asin 答案-COS寺+a17.化簡a COS號+a解析原式=.nsin8n + a cossin4n +n2acos n +COSaSinacosasin=-1.5COS尹-a 3n+aCOS - a21n a答案-1COS2n +解析原式=nsinn +2+ aCOSCOS a Sill專+ a COS專-aCOSaSinaCOSaCOSasina=1.7na214SIH n a _ SIH n asinasinaCOS2 asina sina CO

9、S aCOSa sin asin1ET=sit 180+ a+ a13L1-sina = Ig=lg10=3肩318已知COS180+ a_a + 36(Ta 180 cos 180sinsin+ aIg,310求cosaCOSCOS解析COS_ a2n1 cosa COS n a1S1T + a sin a+36(Tsina 180=cos+ COS a 2nsin刊廣+ a180的值.cosa SlKla sinasina cosa=sinaIg丄,(3)SIH 2 n+ a COS n a COSCOSCOS兀 一a SI JJn a SIH n + a解析原式=sina cosasin

10、acos2n +nn + 2 acosasin2n+ n asinsin2 n +sinasinacosnn+2 asina sinasina S=tana151= +COSa +1 1COSaCOSa +1 +COSa21COSaS HI n abcin a + n SI：(1)化簡f(解析(1)f(a)sinaCOSatanatanasinaCOSa+_COS a-2n(OS n a 1 COSa COS n a + COS a 2nCuS n+aCOSaCOSaCOSa COSaCOSa COSa +COSa=_2sin2 =18.a19.已知a是第四象限角,若sina=5,求f(a)；

11、31n求f(a).f(-f(a)=COSa4=5.(2)vsina =31n亍=COS()nn=COS(3)=COS_1=2.20.若sin(180+a)=-10!?，0 a90.COS 2 n a d-H=COSa.16103 10a =,COsa =一，10 10 sina COsaCOsa +sinaViQ3鎖7010=2.COS扌-a亦亍+a1227n*22已知tan八硏是關于X的方程x-kx+k-3=的兩實根，且3n0,cosB0,cosC0，從而ABC定是銳角三角形.tana+tana=k，tantana= -3,L=k2_4k2- 3sinacosa =k，即k2-3=1,k20,故k=2.7n21SinaCOsa =sina +cosa = a +cosa=1+2sinacosa =2.n + a)= (cosa +sin即SinaCOsa=2