1、第4節(jié)三角函數(shù)的圖象與性質(zhì)知識(shí)點(diǎn)、方法基礎(chǔ)鞏固練綜合運(yùn)用練應(yīng)用創(chuàng)新練三角函數(shù)的定義域與值域1,212三角函數(shù)的單調(diào)性4,5,811三角函數(shù)的周期性、奇偶性、對(duì)稱性3,6,7,910,1518綜合問(wèn)題13,14,16,171.函數(shù)y=tan(4-x)的定義域是(D)A.xx4B.xx-4C.xxk+4,kZD.xxk+34,kZ解析:y=tan(4-x)=-tan(x-4),由x-42+k,kZ,得xk+34,kZ.故選D.2.若函數(shù)f(x)=cos(x-3)(0)的最小正周期為2,則f(x)在0,4上的值域?yàn)?B)A.-32,12 B.-12,1C.-32,1 D.12,1解析:因?yàn)門=2=2

2、,所以=4,f(x)=cos(4x-3).因?yàn)閤0,4,所以4x-3-3,23,所以-12f(x)=cos(4x-3)1,所以f(x)-12,1.故選B.3.(多選題)已知函數(shù)f(x)=32sin 2x-12cos 2x,則下列判斷正確的是(AC)A.關(guān)于直線x=3對(duì)稱B.關(guān)于直線x=6對(duì)稱C.關(guān)于點(diǎn)(12,0)對(duì)稱D.關(guān)于點(diǎn)(3,0)對(duì)稱解析:f(x)=32sin 2x-12cos 2x=sin(2x-6),則f(3)=sin(23-6)=sin 2=1,即函數(shù)關(guān)于直線x=3對(duì)稱,故A正確,D錯(cuò)誤;f(6)=sin(2 6-6)=sin 6=12,則函數(shù)不關(guān)于直線x=6對(duì)稱,故B錯(cuò)誤;f(1

3、2)=sin(2 12-6)=0,即f(x)關(guān)于點(diǎn)(12,0)對(duì)稱,故C正確.故選AC.4.函數(shù)y=2sin(6-2x)(x0,)的單調(diào)遞增區(qū)間是(C)A.0,3B.12,712C.3,56D.56,解析:因?yàn)閥=2sin(6-2x)=-2sin(2x-6),由2+2k2x-632+2k,kZ,解得3+kx56+k,kZ,即函數(shù)在R上的單調(diào)遞增區(qū)間為3+k,56+k,kZ,所以函數(shù)在0,上的單調(diào)遞增區(qū)間為3,56.故選C.5.函數(shù)f(x)=sin x(0)的圖象向右平移12個(gè)單位長(zhǎng)度得到函數(shù)y=g(x)的圖象,并且函數(shù)g(x)在區(qū)間6,3上單調(diào)遞增,在區(qū)間3,2上單調(diào)遞減,則實(shí)數(shù)的值為(C)A

4、.74 B.32 C.2D.54解析:因?yàn)閷⒑瘮?shù)f(x)=sin x(0)的圖象向右平移12個(gè)單位長(zhǎng)度得到函數(shù)y=g(x)的圖象,所以g(x)=sin(x-12),又函數(shù)g(x)在區(qū)間6,3上單調(diào)遞增,在區(qū)間3,2上單調(diào)遞減,所以g(3)=sin 4=1,且23,所以=8k+2(kZ),06, 所以=2.故選C.6.設(shè)函數(shù)f(x)=cos(x+6)在-,上的圖象大致如圖,則f(x)的最小正周期為(C)A.109B.76C.43D.32解析:由圖可得,函數(shù)圖象過(guò)點(diǎn)(-49,0),將它代入函數(shù)f(x)可得cos(-49+6)=0,又(-49,0)是函數(shù)f(x)圖象與x軸負(fù)半軸的第一個(gè)交點(diǎn),所以-4

5、9+6=-2,解得=32,所以函數(shù)f(x)的最小正周期為T=2=232=43.故選C.7.已知函數(shù)f(x)=sin x+acos x的圖象關(guān)于直線x=53對(duì)稱,則實(shí)數(shù)a的值為.解析:由x=53是f(x)圖象的對(duì)稱軸,可得f(0)=f(103),即sin 0+acos 0=sin 103+acos 103,解得a=-33.答案:-338.若函數(shù)f(x)=sin x(02)在區(qū)間0,3上單調(diào)遞增,在區(qū)間3,2上單調(diào)遞減,則等于.解析:根據(jù)題意知f(x)在x=3處取得最大值1,所以sin 3=1,所以3=2k+2,kZ,即=6k+32,kZ.又00),將f(x)圖象上的所有點(diǎn)的橫坐標(biāo)向右平移3個(gè)單位

6、長(zhǎng)度,縱坐標(biāo)不變,所得函數(shù)圖象的一個(gè)對(duì)稱中心為(-4,-1),則的最小值為(B)A.27 B.107C.127D.227解析:因?yàn)楹瘮?shù)f(x)=cos(x+3)-1(0),將f(x)圖象上的所有點(diǎn)的橫坐標(biāo)向右平移3個(gè)單位長(zhǎng)度,縱坐標(biāo)不變,可得函數(shù)y=cos(x-3+3)-1的圖象,所得函數(shù)圖象的一個(gè)對(duì)稱中心為(-4,-1),則cos(-4)-3+3-1=-1,所以cos(-4)-3+3=0,即(-4)-3+3=k+2,kZ,即=-127k-27,kZ,所以的最小值為107.故選B.11.函數(shù)f(x)=Asin(2x+)(|,A0)的部分圖象如圖所示,則f(x)(B)A.在(-512,12)上是

7、減函數(shù)B.在(-512,12)上是增函數(shù)C.在(3,56)上是減函數(shù)D.在(3,56)上是增函數(shù)解析:由圖象可知A=2,函數(shù)的圖象過(guò)點(diǎn)(3,0),所以有2sin(23+)=023+=k(kZ)=k-23(kZ).因?yàn)閨,所以=3或=-23,當(dāng)=-23時(shí),f(x)=2sin(2x-23),此時(shí)f(0)0,不符合題意,所以=3.所以f(x)=2sin(2x+3).當(dāng)x(-512,12)時(shí),2x+3(-2,2),所以f(x)單調(diào)遞增;當(dāng)x(3,56)時(shí),2x+3(,2),函數(shù)f(x)不具有單調(diào)性.故選B.12.已知函數(shù)f(x)=2sin x+sin 2x,則f(x)的最小值是.解析:f(x)=2co

8、s x+2cos 2x=4cos2x+2cos x-2=4(cos x+1)(cos x-12),所以當(dāng)cos x12時(shí),函數(shù)f(x)單調(diào)遞增,從而得到函數(shù)f(x)的單調(diào)遞減區(qū)間為2k-53,2k-3(kZ),函數(shù)f(x)的單調(diào)遞增區(qū)間為2k-3,2k+3(kZ),所以當(dāng)x=2k-3,kZ時(shí),函數(shù)f(x)取得最小值,此時(shí)sin x=-32,sin 2x=-32,所以f(x)min=2(-32)-32=-332.答案:-33213.函數(shù)y=2sin 2x+sin2x(xR)的最小正周期是 ,值域是.解析:函數(shù)y=2sin 2x+sin2x=2sin 2x+1-cos2x2=172sin(2x-)

9、+12,cos =41717,sin =1717,xR,因?yàn)?1sin(2x-)1,所以1-172172sin(2x-)+121+172,即函數(shù)的值域?yàn)?-172,1+172,最小正周期T=22=.答案:1-172,1+17214.(2021浙江金華三模)已知函數(shù)f(x)=sin x(cos x-sin x)+12,則函數(shù)f(x)的最小值為,若f()=26,(0,2),則cos 2的值為.解析:由題意,函數(shù)f(x)=sin x(cos x-sin x)+12=sin xcos x-sin2x+12=12sin 2x-1-cos2x2+12=12sin 2x+12cos 2x=22sin(2x+

10、4),當(dāng)2x+4=-2+2k,kZ,即x=-38+k,kZ時(shí),函數(shù)f(x)取得最小值,最小值為-22.由f()=26,即22sin(2+4)=26,即sin(2+4)=13,因?yàn)?0,2),所以2+4(4,54),又因?yàn)閟in(2+4)=1322,所以2+4(34,),所以cos(2+4)=-223,則cos 2=cos(2+4)-4=cos(2+4)cos 4+sin(2+4)sin 4=-22322+1322=-4+26.答案:-22-4+2615.關(guān)于函數(shù)f(x)=sin x+1sinx有如下四個(gè)命題:f(x)的圖象關(guān)于y軸對(duì)稱;f(x)的圖象關(guān)于原點(diǎn)對(duì)稱;f(x)的圖象關(guān)于直線x=2對(duì)

11、稱;f(x)的最小值為2.屬于真命題的序號(hào)是.解析:對(duì)于命題,f(6)=12+2=52,f(-6)=-12-2=-52,則f(-6)f(6),所以函數(shù)f(x)的圖象不關(guān)于y軸對(duì)稱,命題錯(cuò)誤;對(duì)于命題,函數(shù)f(x)的定義域?yàn)閤|xk,kZ,定義域關(guān)于原點(diǎn)對(duì)稱,f(-x)=sin(-x)+1sin(-x)=-sin x-1sinx=-(sin x+1sinx)=-f(x),所以函數(shù)f(x)的圖象關(guān)于原點(diǎn)對(duì)稱,命題正確;對(duì)于命題,因?yàn)閒(2-x)=sin(2-x)+1sin(2-x)=cos x+1cosx,f(2+x)=sin(2+x)+1sin(2+x)=cos x+1cosx,則f(2-x)=

12、f(2+x),所以函數(shù)f(x)的圖象關(guān)于直線x=2對(duì)稱,命題正確;對(duì)于命題,當(dāng)-x0時(shí),sin x0,則f(x)=sin x+1sinx02,命題錯(cuò)誤.答案:16.已知函數(shù)f(x)=2sin(2x+4).(1)求函數(shù)f(x)圖象的對(duì)稱軸方程;(2)求函數(shù)f(x)的單調(diào)遞增區(qū)間;(3)當(dāng)x4,34時(shí),求函數(shù)f(x)的最大值和最小值.解:(1)令2x+4=k+2,kZ,得x=k2+8,kZ.所以函數(shù)f(x)圖象的對(duì)稱軸方程是x=k2+8,kZ.(2)令2k-22x+42k+2,kZ,得k-38xk+8,kZ.故函數(shù)f(x)的單調(diào)遞增區(qū)間為k-38,k+8,kZ.(3)當(dāng)x4,34時(shí),342x+47

13、4,所以-1sin(2x+4)22,所以-2f(x)1,所以當(dāng)x4,34時(shí),函數(shù)f(x)的最大值為1,最小值為-2.17.設(shè)函數(shù)f(x)=2sin(2x-6)+m的圖象關(guān)于直線x=對(duì)稱,其中012.(1)求函數(shù)f(x)的最小正周期;(2)若函數(shù)y=f(x)的圖象過(guò)點(diǎn)(,0),求函數(shù)f(x)在0,32上的值域.解:(1)由直線x=是y=f(x)圖象的一條對(duì)稱軸,可得sin(2-6)=1,所以2-6=k+2(kZ),即=k2+13(kZ).又00,-122),給出以下四個(gè)論斷:f(x)的最小正周期為;f(x)在區(qū)間(-6,0)上是增函數(shù);f(x)的圖象關(guān)于點(diǎn)(3,0)對(duì)稱;f(x)的圖象關(guān)于直線x=12對(duì)稱.以其中兩個(gè)論斷作為條件,另兩個(gè)論斷作為結(jié)論,寫出你認(rèn)為正確的一個(gè)命題(寫成“pq”的形式) .(用到的論斷都用序號(hào)表示)解析:若f