第1講三角函數(shù)的圖象與性質(zhì)、三角恒等變換(限時(shí):45分鐘)【選題明細(xì)表】知識(shí)點(diǎn)、方法題號(hào)三角函數(shù)圖象4,5,9三角函數(shù)性質(zhì)1,6,7,8,10,11三角恒等變換2,3,12一、選擇題1.(2018廣西桂林市一模)下列函數(shù)中,最小正周期為,且圖象關(guān)于原點(diǎn)對(duì)稱的函數(shù)是(A)(A)y=cos(2x+2) (B)y=sin(2x+2)(C)y=sin 2x+cos 2x(D)y=sin x+cos x解析:對(duì)于選項(xiàng)A,y=-sin 2x,T=22=,且圖象關(guān)于原點(diǎn)對(duì)稱.故選A.2.(2018甘肅模擬)已知cos(4-2)=23,則sin 等于(C)(A)79(B)19(C)-19(D)-79解析:因?yàn)閏os(4-2)=23,所以cos(2-)=2cos2(4-2)-1=-19=sin ,即sin =-19,故選C.3.(2018佛山一模)已知tan +1tan=4,則cos2(+4)等于(C)(A)12(B)13(C)14(D)15解析:由tan +1tan=4,得sincos+cossin=4,即sin2+cos2sincos=4,所以sin cos =14,所以cos2(+4)=1+cos(2+2)2=1-sin22=1-2sincos2=1-2142=14.故選C.4.(2018江西省六校聯(lián)考)設(shè)>0,函數(shù)y=sin(x+3)-1的圖象向左平移23個(gè)單位后與原圖象重合,則的最小值是(D)(A)23(B)43(C)32(D)3解析:因?yàn)閳D象向左平移23個(gè)單位后與原圖象重合,所以23是一個(gè)周期的整數(shù)倍,即23=2k,=3k,kZ.的最小值是3.選D.5.(2018遼寧葫蘆島二模)已知函數(shù)f(x)=Asin(x+)(A>0,>0,0<<)的圖象如圖所示,則下列說法正確的是(B)(A)函數(shù)f(x)的周期為(B)函數(shù)y=f(x-)為奇函數(shù)(C)函數(shù)f(x)在-,2上單調(diào)遞增(D)函數(shù)f(x)的圖象關(guān)于點(diǎn)(34,0)對(duì)稱解析:觀察圖象可得,函數(shù)的最小值為-2,所以A=2,又由圖象可知函數(shù)圖象過(0,3),(54,-2),即3=2sin,-2=2sin(54+),結(jié)合>0,0<<可得=1415,=3,或=23,=23,又12T=>54,即<45,所以f(x)=2sin(23x+23),顯然A選項(xiàng)錯(cuò)誤;對(duì)于B,f(x-)=2sin23(x-)+23=2sin23x,是奇函數(shù);對(duì)于C,x-,2,則23x+230,f(x)不單調(diào);對(duì)于D,當(dāng)x=34時(shí),f(x)=2sin(2334+23)=2cos230,不正確.故選B.6.(2018陜西西工大附中七模)已知f(x)=sin(2 017x+6)+cos(2 017x-3)的最大值為A,若存在實(shí)數(shù)x1,x2,使得對(duì)任意實(shí)數(shù)x總有f(x1)f(x)f(x2)成立,則A|x1-x2|的最小值為(B)(A)2 017(B)22 017(C)42 017(D)4 034解析:f(x)=sin(2 017x+6)+cos(2 017x-3)=32sin 2 017x+12cos 2 017x+12cos 2 017x+32sin 2 017x=2sin(2 017x+6),所以A=2,|x1-x2|T2=222 017,所以A|x1-x2|22 017.選B.7.(2018河南洛陽聯(lián)考)已知函數(shù)f(x)=sin(sin x)+cos(sin x),xR,則下列說法正確的是(C)(A)函數(shù)f(x)是周期函數(shù)且最小正周期為(B)函數(shù)f(x)是奇函數(shù)(C)函數(shù)f(x)在區(qū)間0,2上的值域?yàn)?,2(D)函數(shù)f(x)在4,2上是增函數(shù)解析:A中,f(x+)=sinsin(x+)+cossin(x+)=sin(-sin x)+cos(-sin x)=-sin(sin x)+cos(sin x)f(x),A不對(duì);B中,f(-x)=sinsin(-x)+cossin(-x)=-sin(sin x)+cos(sin x)-f(x),B不對(duì);C中,令t=sin x,因?yàn)閤0,2,所以t0,1,則y=sin t+cos t=2sin(t+4),t0,1,所以t+44,1+4,所以sin(t+4)22,1,所以y1,2,C正確;D中,f(x)=2sin(sin x+4),令t=sin x+4,則y=2sin t,內(nèi)層函數(shù)t=sin x+4在4,2上單調(diào),而x4,2時(shí),t22+4,1+4,此時(shí)外層函數(shù)y=2sin t不單調(diào),D不對(duì).故選C.二、填空題8.(2018東北三校二模)函數(shù)f(x)=cos xsin(x+3)-3cos2x+34在閉區(qū)間-4,4上的最小值是.解析:f(x)=cos x(12sin x+32cos x)-3cos2x+34=14sin 2x-32cos2x+34=14sin 2x-34(cos 2x+1)+34=12(12sin 2x-32cos 2x)=12sin(2x-3),由x-4,4,所以2x-3-56,6,所以當(dāng)2x-3=-2時(shí)f(x)min=-12.答案:-129.(2018云南玉溪模擬)函數(shù)y=Asin(x+)+kA>0,>0,|<2,xR的部分圖象如圖所示,則該函數(shù)表達(dá)式為.解析:根據(jù)函數(shù)y=Asin(x+)+kA>0,>0,|<2,xR的部分圖象,可得k=-1+32=1,A=3-(-1)2=2,342=132-2,所以=3.再根據(jù)五點(diǎn)法作圖可得32+=2,所以=-6,故該函數(shù)的解析式為y=2sin(3x-6)+1.答案:y=2sin(3x-6)+110.(2018吉林大學(xué)附中四模)已知定義域?yàn)镽的函數(shù)f(x)既是奇函數(shù),又是周期為3的周期函數(shù),當(dāng)x(0,32)時(shí),f(x)=sin x,則函數(shù)f(x)在區(qū)間0,6上的零點(diǎn)個(gè)數(shù)是.解析:因?yàn)楹瘮?shù)f(x)的定義域?yàn)镽,周期為3,所以f(0)=f(32)=f(92)=0,如圖所示,畫出函數(shù)的圖象,由圖象可知在0,6上的零點(diǎn)為0,1,32,2,3,4,92,5,6,所以共有9個(gè)零點(diǎn).答案:9三、解答題11.(2018合肥三模)已知函數(shù)f(x)=3sin xcos x-12cos(2x-3).(1)求函數(shù)f(x)圖象的對(duì)稱軸方程;(2)將函數(shù)f(x)圖象向右平移4個(gè)單位,所得圖象對(duì)應(yīng)的函數(shù)為g(x).當(dāng)x0,2時(shí),求函數(shù)g(x)的值域.解:(1)f(x)=3sin xcos x-12cos(2x-3)=34sin 2x-14cos 2x=12sin(2x-6).令2x-6=2+k,kZ,解得x=3+k2,所以函數(shù)f(x)圖象的對(duì)稱軸方程為x=3+k2,kZ.(2)把f(x)=12sin(2x-6)的圖象向右平移4個(gè)單位,可得g(x)=12sin(2x-23),因?yàn)閤0,2,所以2x-23-23,3,所以sin(2x-23)-1,32,所以g(x)=12sin(2x-23)-12,34,即當(dāng)x0,2時(shí),函數(shù)g(x)的值域?yàn)?12,34.12.(2018湖南省永州市一模)已知函數(shù)f(x)=Asin (x+)A>0,>0,|<2的部分圖象如圖所示.(1)求f(x)的解析式;(2)方程f(x)=32在0,2上的兩解分別為x1,x2,求sin (x1+x2),cos (x1-x2)的值.解:(1)由圖象可知A=2,T=76-6=,因?yàn)門=2|,所以=2,因?yàn)閒(x)的圖象過點(diǎn)(6,2),即2sin(26+)=2,3+=2k+2(kZ),即=2k+6(kZ),又因?yàn)閨<2,所以=6,所以f(x)=2sin(2x+6).(2)因?yàn)閒(x)的圖象在y軸右側(cè)的第一個(gè)波峰的橫坐標(biāo)為6,圖象f(x)=32在0,2上的兩解x1,x2關(guān)于直線x=6對(duì)稱,所以x1+x2=3,所以sin (x1+x2)=32,因?yàn)閏os (x1-x2)=cos(2x1-3)=sin(2x1+6),f(x1)=2sin(2x1+6)=32,所以cos (x1-x2)=34.