2025高考數(shù)學(xué)一輪復(fù)習(xí)-4.6-函數(shù)y=sin(ωx+φ)的圖象與性質(zhì)及三角函數(shù)模型的應(yīng)用-專(zhuān)項(xiàng)訓(xùn)練【含答案】

2025高考數(shù)學(xué)一輪復(fù)習(xí)-4.6-函數(shù)y=sin(ωx+φ)的圖象與性質(zhì)及三角函數(shù)模型的應(yīng)用-專(zhuān)項(xiàng)訓(xùn)練【A級(jí)基礎(chǔ)鞏固】1.函數(shù)y=sin(2x-π3)在區(qū)間[-π2.若將函數(shù)y=sin(2x-π4)的圖象上的各點(diǎn)的橫坐標(biāo)伸長(zhǎng)到原來(lái)的2倍(縱坐標(biāo)不變),再向右平移πA.(5π12,0) B.(πC.(π6,0) D.(π3.(多選題)為了得到函數(shù)y=2sin2x的圖象,下列變換正確的是()A.將函數(shù)y=(sinx+cosx)2的圖象向右平移π4B.將函數(shù)y=1-cos2x的圖象向左平移π4C.將函數(shù)y=2sin2(x+π6)的圖象向右平移πD.將函數(shù)y=2sin2(x+π6)的圖象向左平移π4.函數(shù)y=Asin(ωx+)在一個(gè)周期內(nèi)的圖象如圖,則此函數(shù)的解析式為()A.y=2sin(2x+2π3)B.y=2sin(2x+π3C.y=2sin(x2-π3D.y=2sin(2x-π35.已知函數(shù)f(x)=sin3x-3cos3x,則下列結(jié)論錯(cuò)誤的是()A.當(dāng)x∈[0,π2]時(shí),f(x)的取值范圍是[-3B.y=f(x)在[π3,πC.y=f(x)的圖象關(guān)于直線x=-π18D.y=f(x)的圖象可由函數(shù)y=sin3x的圖象向右平移π36.已知函數(shù)f(x)=2cos(ωx-)(ω>0,∈[0,π])的部分圖象如圖所示.若A(π2,2),B(3π2,2),則f(0)=.

7.將函數(shù)f(x)=sin2x的圖象向左平移π6個(gè)單位長(zhǎng)度后得到函數(shù)g(x)的圖象,則函數(shù)g(x)在[0,π]上的零點(diǎn)個(gè)數(shù)為8.如圖,某大風(fēng)車(chē)的半徑為2m,每12s旋轉(zhuǎn)一周,它的最低點(diǎn)O離地面1m,點(diǎn)O在地面上的射影為A.風(fēng)車(chē)圓周上一點(diǎn)從最低點(diǎn)O開(kāi)始,逆時(shí)針?lè)较蛐D(zhuǎn)40s后到達(dá)點(diǎn)P,則點(diǎn)P到地面的距離是m.9.已知函數(shù)f(x)=2sin(2x+π3(1)寫(xiě)出函數(shù)f(x)在x∈R上的單調(diào)遞減區(qū)間;(2)將y=f(x)圖象上所有的點(diǎn)向右平移π3個(gè)單位長(zhǎng)度,橫坐標(biāo)變?yōu)樵瓉?lái)的12,縱坐標(biāo)不變,得到y(tǒng)=g(x)的圖象,求y=g(x)在區(qū)間[0,INCLUDEPICTURE"B組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【B級(jí)能力提升】10.(多選題)如圖1,彈簧掛著的小球做上下運(yùn)動(dòng),它在ts時(shí)相對(duì)于平衡位置的高度h(單位:cm)由關(guān)系式h(t)=Asin(ωt+)(A>0,ω>0)確定.h(t)關(guān)于t的函數(shù)圖象如圖2,則下列敘述正確的是()A.函數(shù)h(t)的周期為2B.函數(shù)h(t)的對(duì)稱(chēng)軸為t=k+14C.函數(shù)h(t)的單調(diào)遞增區(qū)間為[2k-34,2k+1D.函數(shù)h(t)的圖象可由函數(shù)y=2sin(t+π411.將函數(shù)y=sin(2x+π3)的圖象上所有點(diǎn)的縱坐標(biāo)保持不變,橫坐標(biāo)變?yōu)樵瓉?lái)的2ω(ω∈N.

12.若f(x)=3sin2ωx+1(ω>0)在區(qū)間[-3π2,π2]上單調(diào)遞增,則ω的最大值為13.已知函數(shù)f(x)=2sin(ωx+)+1(ω>0,||<π2),函數(shù)f(x)的圖象上兩相鄰對(duì)稱(chēng)軸之間的距離為π2,(1)在①函數(shù)f(x)圖象的一條對(duì)稱(chēng)軸為直線x=-π3;②函數(shù)f(x)圖象的一個(gè)對(duì)稱(chēng)中心為點(diǎn)(5π12,1);③函數(shù)f(x)的圖象經(jīng)過(guò)點(diǎn)((2)若動(dòng)直線x=t,t∈[0,π]與函數(shù)f(x)和函數(shù)g(x)=23sinxcosx的圖象分別交于P,Q兩點(diǎn),求線段PQ長(zhǎng)度的最大值及此時(shí)t的值.INCLUDEPICTURE"B組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【C級(jí)應(yīng)用創(chuàng)新練】14.已知函數(shù)f(x)=Asin(ωx+)(A>0,ω>0,||<π2)的部分圖象如圖所示.(1)求函數(shù)f(x)的解析式及f(x)的單調(diào)遞增區(qū)間;(2)把函數(shù)y=f(x)圖象上各點(diǎn)的橫坐標(biāo)擴(kuò)大到原來(lái)的2倍(縱坐標(biāo)不變),再向左平移π6個(gè)單位長(zhǎng)度,得到函數(shù)y=g(x)的圖象,求關(guān)于x的方程g(x)=m(0<m<2)在x∈[-π3,參考答案【A級(jí)基礎(chǔ)鞏固】1.解析:令x=0得y=sin(-π3)=-32,排除B,D項(xiàng),由f(-π32.解析:將y=sin(2x-π4)的圖象上的各點(diǎn)的橫坐標(biāo)伸長(zhǎng)到原來(lái)的2倍(縱坐標(biāo)不變),可以得到y(tǒng)=sin[2(12x)-π4]=sin(x-π4)的圖象,再向右平移π6個(gè)單位長(zhǎng)度可以得到y(tǒng)=sin[(x-πg(shù)(x)=sin(x-5π12),由g(5π3.解析:將函數(shù)y=(sinx+cosx)2=1+sin2x的圖象向右平移π4個(gè)單位長(zhǎng)度,得到函數(shù)y=1+sin[2(x-π4)]=1+sin(2x-π2)=1-cos2x=2sin2x的圖象,故A正確;將函數(shù)y=1-cos2x的圖象向左平移π4個(gè)單位長(zhǎng)度,得到函數(shù)y=1-cos(2x+π2)=1+sin2x的圖象,故B錯(cuò)誤;將函數(shù)y=2sin2(x+π6)=1-cos(2x+π3)的圖象向右平移π故選AC.4.解析:由已知可得函數(shù)y=Asin(ωx+)的圖象經(jīng)過(guò)點(diǎn)(-π12,2)和點(diǎn)(5π12,-2),則A=2,T=π,所以ω=2,則函數(shù)的解析式為y=2sin(2x+),將(-π12,2)代入得-π6+=π所以=2π3+2kπ,k∈Z,當(dāng)k=0時(shí),=2π3,此時(shí)y=2sin(2x+2π35.解析:f(x)=sin3x-3cos3x=2sin(3x-π3當(dāng)x∈[0,π2]時(shí),3x-π3∈[-π3,7π6],sin(3x-π3當(dāng)x∈[π3,π2]時(shí),3x-π3∈[2π3,正確;當(dāng)x=-π18時(shí),f(-π18)=2sin[3×(-π18)-π3]=2sin(-由函數(shù)y=sin3x的圖象向右平移π3個(gè)單位長(zhǎng)度得到y(tǒng)=sin[3(x-πsin(3x-π)=-sin3x的圖象,D選項(xiàng)錯(cuò)誤.故選D.6.解析:由函數(shù)圖象可知函數(shù)f(x)的周期T=3π2-π2=π,ω=又f(π2)=2cos(π-)=-2·cos=2,則cos=-22.因?yàn)椤蔥0,π],所以=3π4,所以f(x)=2cos(2x-3π4),則f(0)=-2答案:-27.解析:f(x)=sin2x的圖象向左平移π6sin[2(x+π6)]=sin(2x+π3)的圖象,當(dāng)x∈[0,π]時(shí),2x+π3∈[π3,7π3答案:28.解析:以圓心O1為原點(diǎn),以水平方向?yàn)閤軸方向,豎直方向?yàn)閥軸方向建立平面直角坐標(biāo)系,則根據(jù)大風(fēng)車(chē)的半徑為2m,圓上最低點(diǎn)O離地面1m,12s轉(zhuǎn)動(dòng)一周,設(shè)∠OO1P=θ,運(yùn)動(dòng)ts后到地面的距離為f(t).又周期T=12,所以θ=π6則f(t)=3+2sin(θ-π2)=3-2cosπ當(dāng)t=40s時(shí),f(t)=3-2cos(π6×答案:49.解:(1)令2kπ+π2≤2x+π3≤2kπ+解得kπ+π12≤x≤kπ+7π故f(x)在x∈R上的單調(diào)遞減區(qū)間為[kπ+π12,kπ+7π(2)將y=f(x)圖象上所有的點(diǎn)向右平移π3所得函數(shù)的解析式為y=2sin[2(x-π3)+π3]=2sin(2x-再將所得函數(shù)的圖象橫坐標(biāo)變?yōu)樵瓉?lái)的12則g(x)=2sin(4x-π3因?yàn)閤∈[0,π2],則-π3≤4x-π3故當(dāng)4x-π3=π2,即x=5π24當(dāng)4x-π3=3π2,即x=11π24INCLUDEPICTURE"B組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【B級(jí)能力提升】10.解析:由題意和圖象可知,A=2,h(0)=2,h(34)=0,所以2sin=2,即sin=22,所以=2kπ+π4或=2kπ+3π4,k∈Z.因?yàn)閤=0在增區(qū)間內(nèi),所以=2kπ+π4,k∈Z.所以h(t)=2sin(ωt+2kπ+π4)=2sin(ωt+π4).因?yàn)閔(34)=2sin(3所以sin(3ω4+π4)=0,結(jié)合圖象可知,x=34在減區(qū)間內(nèi),所以k∈Z,解得ω=8kπ3+π,k∈Z.根據(jù)圖象可知,T4<34,且T所以32<2πω<3,解得2π3<ω<4π對(duì)于A,T=2ππ對(duì)于B,令πt+π4=kπ+π2,k∈Z,解得t=k+14對(duì)于C,令2kπ-π2≤πt+π4≤2kπ+π2,k∈Z,解得2k-3所以函數(shù)h(t)的單調(diào)遞增區(qū)間為[2k-34,2k+1對(duì)于D,函數(shù)y=2sin(t+π4)圖象上所有點(diǎn)的橫坐標(biāo)伸長(zhǎng)為原來(lái)的π倍得到函數(shù)y=2sin(tπ+故選ABC.11.解析:由題可知g(x)=sin(2·ω2x+π3)=sin(ωx+因?yàn)閤∈(0,π),所以ωx+π3∈(π3,ωπ+y=sinx,x∈(π3要使g(x)的圖象在區(qū)間(0,π)上有且僅有兩條對(duì)稱(chēng)軸和兩個(gè)對(duì)稱(chēng)中心,則2π<ωπ+π3≤5π2,解得53因?yàn)棣亍蔔*,所以ω=2.答案:212.解析:因?yàn)閤∈[-3π2,π2],所以2ωx∈[-3πω,πω],令t=2ωx,則y=3得0<ω≤16所以ω的最大值為16答案:113.解:(1)由函數(shù)f(x)的圖象上兩相鄰對(duì)稱(chēng)軸之間的距離為π2,得該函數(shù)的最小正周期T=2×π2=π,所以ω=2πT此時(shí)f(x)=2sin(2x+)+1.若選①函數(shù)f(x)圖象的一條對(duì)稱(chēng)軸為直線x=-π3,則-2π3+=π2+kπ(k∈Z),解得=7π6+kπ(k∈Z).因?yàn)閨|<π2,所以當(dāng)k=-1時(shí),=π6,此時(shí)f(x)=2sin(2x+若選②函數(shù)f(x)圖象的一個(gè)對(duì)稱(chēng)中心為點(diǎn)(5π12,1),則5π6+=kπ(k∈Z),解得=kπ-5π6(k∈Z).因?yàn)閨|<π2,所以當(dāng)k=1時(shí),=π6此時(shí)f(x)=2sin(2x+π6若選③函數(shù)f(x)的圖象經(jīng)過(guò)點(diǎn)(5π6則f(5π6)=2sin(5π3+得sin(5π3+)=-12因?yàn)閨|<π2,所以7π6<5π3+<13π所以5π3+=11π6,解得=π6此時(shí)f(x)=2sin(2x+π6(2)由(1)可知,函數(shù)f(x)=2sin(2x+π6令h(x)=f(x)-g(x)=2sin(2x+π6)+1-23sinxcosx=2(312cos2x)+1-3當(dāng)2t=0或2t=2π,即當(dāng)t=0或t=π時(shí),線段PQ的長(zhǎng)取到最大值2.INCLUDEPICTURE"B組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【C級(jí)應(yīng)用創(chuàng)新練】14.解:(1)由題中圖象知,A=2,最小正周期T=11π12-(-π12)=π,所以ω=2πT因?yàn)辄c(diǎn)(-π12所以2sin(-2×π12+)=0,即sin(-π6)=0.又因?yàn)?π2<<π2所以-2π3<-π6<π所以-π6=0,從而=π6.故函數(shù)f(x)的解析式為f(x)=2sin(2x+π6令2kπ-π2≤2x+π6≤2kπ+得kπ-π3≤x≤kπ+π故f(x)的單調(diào)遞增區(qū)間為[kπ-π3,kπ+π(2)依題意得g(x)=2