專題3-1三角函數(shù)圖像與性質(zhì)目錄TOC\o"1-1"\h\u題型01三角函數(shù)單調(diào)性 1題型02求周期 3題型03非同名函數(shù)平移 6題型04對(duì)稱軸最值應(yīng)用 8題型05對(duì)稱中心最值應(yīng)用 11題型06輔助角最值 14題型07正余弦換元型最值 17題型08一元二次型換元最值 20題型09分式型最值 21題型10最值型綜合 23題型11恒等變形：求角 25題型12恒等變形：拆角求值（分式型） 27題型13恒等變形：拆角求值（復(fù)合型） 29題型14恒等變形：拆角求值（正切型對(duì)偶） 31高考練場(chǎng) 33題型01三角函數(shù)單調(diào)性【解題攻略】A，ω，φ對(duì)函數(shù)y＝Asin（ωx＋φ）圖象的影響（1）函數(shù)y＝Asin（ωx＋φ）（A＞0，ω＞0）中參數(shù)A．φ、ω的作用參數(shù)作用AA決定了函數(shù)的值域以及函數(shù)的最大值和最小值，通常稱A為振幅.φφ決定了x=0時(shí)的函數(shù)值，通常稱φ為初相，ωx＋φ為相位.ωω決定了函數(shù)的周期T＝SKIPIF1<0.（2）圖象的變換（1）振幅變換要得到函數(shù)y＝Asinx（A＞0，A≠1）的圖象，只要將函數(shù)y＝sinx的圖象上所有點(diǎn)的縱坐標(biāo)伸長(zhǎng)（當(dāng)A＞1時(shí)）或縮短（當(dāng)0＜A＜1時(shí)）到原來(lái)的A倍（橫坐標(biāo)不變）即可得到．（2）平移變換要得到函數(shù)y＝sin（x＋φ）的圖象，只要將函數(shù)y＝sinx的圖象上所有點(diǎn)向左（當(dāng)φ＞0時(shí)）或向右（當(dāng)φ＜0時(shí)）平行移動(dòng)|φ|個(gè)單位長(zhǎng)度即可得到．（3）周期變換要得到函數(shù)y＝sinωx（x∈R）（其中ω＞0且ω≠1）的圖象，可以把函數(shù)y＝sinx上所有點(diǎn)的橫坐標(biāo)縮短（當(dāng)ω＞1時(shí)）或伸長(zhǎng)（當(dāng)0＜ω＜1時(shí)）到原來(lái)的SKIPIF1<0_倍（縱坐標(biāo)不變）即可得到．【典例1-1】（2023·全國(guó)·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，則使得SKIPIF1<0和SKIPIF1<0都單調(diào)遞增的一個(gè)區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】利用復(fù)合函數(shù)的單調(diào)性，判斷各選項(xiàng)是否正確.【詳解】當(dāng)SKIPIF1<0從SKIPIF1<0增加到SKIPIF1<0時(shí)，SKIPIF1<0從0遞減到SKIPIF1<0，SKIPIF1<0從SKIPIF1<0遞增到1，所以SKIPIF1<0從SKIPIF1<0遞減到SKIPIF1<0，SKIPIF1<0從SKIPIF1<0遞減到SKIPIF1<0，A錯(cuò)誤；當(dāng)SKIPIF1<0從SKIPIF1<0增加到SKIPIF1<0時(shí)，SKIPIF1<0從SKIPIF1<0遞減到SKIPIF1<0，SKIPIF1<0從1遞減到SKIPIF1<0，所以SKIPIF1<0從SKIPIF1<0遞增到SKIPIF1<0，SKIPIF1<0從SKIPIF1<0遞減到SKIPIF1<0，B錯(cuò)誤；當(dāng)SKIPIF1<0從SKIPIF1<0增加到SKIPIF1<0時(shí)，SKIPIF1<0從SKIPIF1<0遞減到SKIPIF1<0，SKIPIF1<0從SKIPIF1<0遞減到SKIPIF1<0，所以SKIPIF1<0從SKIPIF1<0遞增到SKIPIF1<0，SKIPIF1<0從SKIPIF1<0遞減到SKIPIF1<0，C錯(cuò)誤；當(dāng)SKIPIF1<0從SKIPIF1<0增加到SKIPIF1<0時(shí)，SKIPIF1<0從-1遞增到SKIPIF1<0，SKIPIF1<0從SKIPIF1<0遞減到0，所以SKIPIF1<0從SKIPIF1<0遞增到SKIPIF1<0，SKIPIF1<0從SKIPIF1<0遞增到SKIPIF1<0，D正確；故選：D【典例1-2】已知函數(shù)SKIPIF1<0，則f（x）（

）A．在（0，SKIPIF1<0）單調(diào)遞減 B．在（0，π）單調(diào)遞增C．在（—SKIPIF1<0，0）單調(diào)遞減 D．在（—SKIPIF1<0，0）單調(diào)遞增【答案】D【分析】先用誘導(dǎo)公式化簡(jiǎn)得到SKIPIF1<0，再將選項(xiàng)代入檢驗(yàn)，求出正確答案.【詳解】SKIPIF1<0，故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0不單調(diào)，AB錯(cuò)誤；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故D正確故選：D【變式1-1】（2022上·福建莆田·高三校考）函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】先換元,求定義域再結(jié)合復(fù)合函數(shù)的單調(diào)性,最后根據(jù)正弦函數(shù)的單調(diào)性求解即可.【詳解】設(shè)SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，取SKIPIF1<0單調(diào)增的部分，所以可得：SKIPIF1<0，即SKIPIF1<0，解得：SKIPIF1<0答案：A.【變式1-2】（2023·全國(guó)·模擬預(yù)測(cè)）函數(shù)SKIPIF1<0在下列某個(gè)區(qū)間上單調(diào)遞增，這個(gè)區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】由二倍角公式結(jié)合輔助角公式化簡(jiǎn)可得SKIPIF1<0的表達(dá)式，求出其單調(diào)增區(qū)間，結(jié)合選項(xiàng)，即可判斷出答案.【詳解】∵SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增.故選：A【變式1-3】（2023·黑龍江齊齊哈爾·統(tǒng)考二模）“SKIPIF1<0”是“函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增”的（

）A．充分不必要條件 B．必要不充分條件 C．充要條件 D．既不充分也不必要條件【答案】A【分析】根據(jù)正弦函數(shù)的單調(diào)性和參數(shù)范圍即可求解.【詳解】若函數(shù)SKIPIF1<0區(qū)間SKIPIF1<0上單調(diào)遞增，則令SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，結(jié)合SKIPIF1<0是區(qū)間，所以SKIPIF1<0，解得SKIPIF1<0.“SKIPIF1<0”是“SKIPIF1<0”的充分不必要條件,故選：A..題型02求周期【解題攻略】求周期方法直接法：形如y＝Asin（ωx＋φ）或者y＝Acos（ωx＋φ）函數(shù)的周期T＝SKIPIF1<0.y＝Atan（ωx＋φ）的周期是T＝SKIPIF1<0觀察法：形如

SKIPIF1<0

SKIPIF1<0

SKIPIF1<0等等諸如此類的帶絕對(duì)值型，可以通過簡(jiǎn)圖判定是否有周期，以及最小正周期的值3.恒等變形轉(zhuǎn)化法。4.定義證明法【典例1-1】（2023下·湖南長(zhǎng)沙·高三長(zhǎng)沙一中校考階段練習(xí)）設(shè)函數(shù)SKIPIF1<0，則SKIPIF1<0的最小正周期（

）A．與SKIPIF1<0有關(guān)，且與SKIPIF1<0有關(guān) B．與SKIPIF1<0有關(guān)，但與SKIPIF1<0無(wú)關(guān)C．與SKIPIF1<0無(wú)關(guān)，且與SKIPIF1<0無(wú)關(guān) D．與SKIPIF1<0無(wú)關(guān)，但與SKIPIF1<0有關(guān)【答案】D【分析】根據(jù)三角函數(shù)的周期性，結(jié)合周期成倍數(shù)關(guān)系的兩個(gè)函數(shù)之和，其周期為這兩個(gè)函數(shù)的周期的最小公倍數(shù)這一結(jié)論，解答即可.【詳解】SKIPIF1<0，對(duì)于SKIPIF1<0，其最小正周期為SKIPIF1<0，對(duì)于SKIPIF1<0，其最小正周期為SKIPIF1<0，所以對(duì)于任意SKIPIF1<0，SKIPIF1<0的最小正周期都為SKIPIF1<0，對(duì)于SKIPIF1<0，其最小正周期為SKIPIF1<0，故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，其最小正周期為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，其最小正周期為SKIPIF1<0，所以SKIPIF1<0的最小正周期與SKIPIF1<0無(wú)關(guān)，但與SKIPIF1<0有關(guān).故選：D.【典例1-2】（2023上·福建廈門·高三福建省廈門第二中學(xué)校考階段練習(xí)）以下函數(shù)中最小正周期為SKIPIF1<0的個(gè)數(shù)是（

）SKIPIF1<0

SKIPIF1<0

SKIPIF1<0

SKIPIF1<0A．1 B．2 C．3 D．4【答案】A【分析】對(duì)于A，直接畫出函數(shù)圖象驗(yàn)證即可；對(duì)于BCD，舉出反例推翻即可.【詳解】畫出函數(shù)SKIPIF1<0的圖象如圖所示：

由圖可知函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0，滿足題意；對(duì)于SKIPIF1<0而言，SKIPIF1<0，即函數(shù)SKIPIF1<0的最小正周期不是SKIPIF1<0，不滿足題意；對(duì)于SKIPIF1<0而言，SKIPIF1<0，即函數(shù)SKIPIF1<0的最小正周期不是SKIPIF1<0，不滿足題意；對(duì)于SKIPIF1<0而言，SKIPIF1<0，即函數(shù)SKIPIF1<0的最小正周期不是SKIPIF1<0，不滿足題意；綜上所述，滿足題意的函數(shù)的個(gè)數(shù)有1個(gè).故選：A.【變式1-1】（2023·全國(guó)·高三專題練習(xí)）下列函數(shù)中是奇函數(shù)，且最小正周期是SKIPIF1<0的函數(shù)是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】確定SKIPIF1<0和SKIPIF1<0，SKIPIF1<0為偶函數(shù)，排除，驗(yàn)證D選項(xiàng)滿足條件，得到答案.【詳解】對(duì)選項(xiàng)A：SKIPIF1<0，函數(shù)定義域?yàn)镾KIPIF1<0，SKIPIF1<0，函數(shù)為偶函數(shù)，排除；對(duì)選項(xiàng)B：SKIPIF1<0，函數(shù)定義域?yàn)镾KIPIF1<0，SKIPIF1<0，函數(shù)為偶函數(shù)，排除；對(duì)選項(xiàng)C：SKIPIF1<0，函數(shù)定義域?yàn)镾KIPIF1<0，SKIPIF1<0，函數(shù)為偶函數(shù)，排除；對(duì)選項(xiàng)D：SKIPIF1<0，函數(shù)定義域?yàn)镾KIPIF1<0，SKIPIF1<0，函數(shù)為奇函數(shù)，SKIPIF1<0，滿足條件；故選：D.【變式1-2】（2023·廣東·統(tǒng)考二模）已知函數(shù)SKIPIF1<0，SKIPIF1<0的定義域?yàn)镽，則“SKIPIF1<0，SKIPIF1<0為周期函數(shù)”是“SKIPIF1<0為周期函數(shù)”的（

）A．充分不必要條件 B．必要不充分條件C．充要條件 D．既不充分也不必要條件【答案】D【分析】根據(jù)通過反例和周期的性質(zhì)判斷即可.【詳解】?jī)蓚€(gè)周期函數(shù)之和是否為周期函數(shù)，取決于兩個(gè)函數(shù)的周期的比是否為有理數(shù)，若為有理數(shù)，則有周期，若不為有理數(shù)，則無(wú)周期.SKIPIF1<0的周期為SKIPIF1<0，SKIPIF1<0的周期為SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)，只有周期的整數(shù)倍才是函數(shù)的周期，則不是充分條件；若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0為周期函數(shù)，但SKIPIF1<0，SKIPIF1<0為周期函數(shù)不正確，故不是必要條件；因此為不充分不必要條件.故選：D【變式1-3】（2023上·江蘇·高三專題練習(xí)）在函數(shù)①SKIPIF1<0，②SKIPIF1<0，③SKIPIF1<0，④SKIPIF1<0中，最小正周期為π的函數(shù)有（）A．①③ B．①④C．③④ D．②③【答案】D【分析】根據(jù)函數(shù)圖象的翻折變換和周期公式可得.【詳解】①由余弦函數(shù)的奇偶性可知，SKIPIF1<0，最小值周期為SKIPIF1<0；②由翻折變換可知，函數(shù)SKIPIF1<0的圖象如圖：由圖知SKIPIF1<0的最小值周期為SKIPIF1<0；③由周期公式得SKIPIF1<0，所以SKIPIF1<0的最小值周期為SKIPIF1<0；④SKIPIF1<0的最小值周期為SKIPIF1<0.故選：D題型03非同名函數(shù)平移【解題攻略】平移變換：1.基本法：提系數(shù)（就是直接換x，其余的都不動(dòng)）；2.正到余，余到正：方法一：誘導(dǎo)公式化為同名（盡量化正為余，因?yàn)橛嘞沂桥己瘮?shù)，可以解決系數(shù)是負(fù)的）；方法二：直接第極大值法（通過快速畫圖，正弦對(duì)應(yīng)第一極大值軸處。余弦即五點(diǎn)第一點(diǎn)處，本方法是重點(diǎn)）【典例1-1】（2023秋·山東·高三山東省實(shí)驗(yàn)中學(xué)校考期末）要得到函數(shù)SKIPIF1<0的圖象，只需將函數(shù)SKIPIF1<0的圖象（

）A．先向右平移SKIPIF1<0個(gè)單位長(zhǎng)度，再向下平移1個(gè)單位長(zhǎng)度B．先向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，再向上平移1個(gè)單位長(zhǎng)度C．先向右平移SKIPIF1<0個(gè)單位長(zhǎng)度，再向下平移1個(gè)單位長(zhǎng)度D．先向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，再向上平移1個(gè)單位長(zhǎng)度【答案】B【解析】根據(jù)SKIPIF1<0，SKIPIF1<0可判斷.【詳解】SKIPIF1<0，所以SKIPIF1<0先向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，再向上平移1個(gè)單位長(zhǎng)度可得到SKIPIF1<0的圖象.故選：B.【典例1-2】（2021春·河南許昌·高三許昌實(shí)驗(yàn)中學(xué)校考）要得到函數(shù)SKIPIF1<0的圖象，只需將函數(shù)SKIPIF1<0的圖象（

）A．向左平移SKIPIF1<0個(gè)單位長(zhǎng)度 B．向右平移SKIPIF1<0個(gè)單位長(zhǎng)度C．向左平移1個(gè)單位長(zhǎng)度 D．向右平移1個(gè)單位長(zhǎng)度【答案】C【分析】把SKIPIF1<0化成SKIPIF1<0可得平移的發(fā)現(xiàn)及其長(zhǎng)度.【詳解】因?yàn)镾KIPIF1<0，所以要得到函數(shù)SKIPIF1<0的圖象，只需把函數(shù)SKIPIF1<0的圖象上所有的點(diǎn)向左平移1個(gè)單位長(zhǎng)度.故選：C.【變式1-1】（2020春·全國(guó)·高三校聯(lián)考階段練習(xí)）要得到函數(shù)SKIPIF1<0的圖象，只需將函數(shù)SKIPIF1<0的圖象（

）A．向左平移SKIPIF1<0個(gè)單位 B．向右平移SKIPIF1<0個(gè)單位C．向左平移SKIPIF1<0個(gè)單位 D．向右平栘SKIPIF1<0個(gè)單位【答案】C【解析】由題意利用函數(shù)SKIPIF1<0的圖象變換規(guī)律，得出結(jié)論．【詳解】解：要得到函數(shù)SKIPIF1<0的圖象，只需將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位即可，故選：C．【變式1-2】（2022·全國(guó)·高三專題練習(xí)）為得到函數(shù)SKIPIF1<0的圖象，只需將函數(shù)SKIPIF1<0圖象上所有的點(diǎn)（

）A．向左平移SKIPIF1<0個(gè)單位長(zhǎng)度 B．向右平移SKIPIF1<0個(gè)單位長(zhǎng)度C．向左平移SKIPIF1<0個(gè)單位長(zhǎng)度 D．向右平移SKIPIF1<0個(gè)單位長(zhǎng)度【答案】D【分析】先得到SKIPIF1<0，再利用平移變換求解.【詳解】解：因?yàn)镾KIPIF1<0，將其圖象上所有的點(diǎn)向右平移SKIPIF1<0個(gè)單位長(zhǎng)度，得到函數(shù)SKIPIF1<0的圖象.A，B，C都不滿足.故選：D【變式1-3】（2022·河南鶴壁·鶴壁高中校考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，為了得到函數(shù)SKIPIF1<0的圖象只需將y=f（x）的圖象（

）A．向右平移SKIPIF1<0個(gè)單位 B．向右平移SKIPIF1<0個(gè)單位C．向左平移SKIPIF1<0個(gè)單位 D．向左平移SKIPIF1<0個(gè)單位【答案】C【分析】根據(jù)誘導(dǎo)公式SKIPIF1<0，SKIPIF1<0即可得到平移方法.【詳解】函數(shù)SKIPIF1<0，SKIPIF1<0，所以為了得到函數(shù)SKIPIF1<0的圖象只需將y=f（x）的圖象向左平移SKIPIF1<0個(gè)單位.故選：C題型04對(duì)稱軸最值應(yīng)用【解題攻略】正余弦對(duì)稱軸：最值處，令sin(ωx＋φ)＝1，則ωx＋φ＝kπ＋eq\f(π,2)(k∈Z)，可求得對(duì)稱軸方程；對(duì)稱軸代入，三角函數(shù)部分必為正負(fù)1，還可以理解為輔助角那個(gè)整體取得最大值或者最小值SKIPIF1<0【典例1-1】已知函數(shù)SKIPIF1<0的最大值為SKIPIF1<0，若存在實(shí)數(shù)SKIPIF1<0，使得對(duì)任意實(shí)數(shù)SKIPIF1<0，總有SKIPIF1<0成立，則SKIPIF1<0的最小值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0湖北省荊州市沙市中學(xué)2021-2022學(xué)年高三上學(xué)期數(shù)學(xué)試題【答案】B【分析】結(jié)合三角恒等變換求得SKIPIF1<0的最大值和最小值，并求得SKIPIF1<0的最小值.【詳解】SKIPIF1<0SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0取得最大值為SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)SKIPIF1<0取得最小值為SKIPIF1<0.依題意，存在實(shí)數(shù)SKIPIF1<0，使得對(duì)任意實(shí)數(shù)SKIPIF1<0，總有SKIPIF1<0成立，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0是整數(shù)，SKIPIF1<0為奇數(shù)，所以SKIPIF1<0的最小值為SKIPIF1<0.故選：B【典例1-2】（2022屆湘贛十四校高三聯(lián)考第二次考試?yán)頂?shù)試題=）已知函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度得到SKIPIF1<0的圖象，若SKIPIF1<0為SKIPIF1<0的一條對(duì)稱軸，則SKIPIF1<0__________．【答案】SKIPIF1<0【分析】直接利用三角函數(shù)關(guān)系式的恒等變換和平移變換得SKIPIF1<0，SKIPIF1<0，再利用三角函數(shù)對(duì)稱性列方程求解即可．【詳解】設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0是SKIPIF1<0的一條對(duì)稱軸，∴SKIPIF1<0SKIPIF1<0，即SKIPIF1<0.故答案為SKIPIF1<0【變式1-1】已知把函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度，再把橫坐標(biāo)縮小到原來(lái)一半，縱坐標(biāo)不變，得到函數(shù)SKIPIF1<0的圖象，若SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的最大值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】先化簡(jiǎn)函數(shù)SKIPIF1<0，然后根據(jù)圖像的變換得函數(shù)SKIPIF1<0的解析式，通過判斷得SKIPIF1<0，SKIPIF1<0同時(shí)令SKIPIF1<0取得最大值或最小值時(shí)，SKIPIF1<0，再結(jié)合函數(shù)SKIPIF1<0的圖像，即可求得SKIPIF1<0的最大值.【詳解】SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0．將圖象向右平移至SKIPIF1<0個(gè)單位長(zhǎng)度，再把橫坐標(biāo)縮小到原來(lái)一半，縱坐標(biāo)不變，得到函數(shù)SKIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0同時(shí)令SKIPIF1<0取得最大值或最小值時(shí)，SKIPIF1<0．當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0，根據(jù)函數(shù)的圖象可知SKIPIF1<0的最大值為SKIPIF1<0個(gè)周期的長(zhǎng)度，即SKIPIF1<0故選：C.【變式1-2】（河南省三門峽市2021-2022學(xué)年高三上學(xué)期階段性檢測(cè)理科數(shù)學(xué)試題）.將函數(shù)SKIPIF1<0的圖象上的所有點(diǎn)的橫坐標(biāo)縮短為原來(lái)的SKIPIF1<0，縱坐標(biāo)不變，再把所得的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，然后再把所得的圖象向下平移1個(gè)單位長(zhǎng)度，得到函數(shù)SKIPIF1<0的圖象，若SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最大值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】根據(jù)三角函數(shù)平移變換,先求得SKIPIF1<0的解析式.根據(jù)SKIPIF1<0,可知SKIPIF1<0,即SKIPIF1<0.根據(jù)SKIPIF1<0可分別求得SKIPIF1<0的最大值和SKIPIF1<0的最小值,即可求得SKIPIF1<0的最大值.【詳解】根據(jù)平移變換將函數(shù)SKIPIF1<0的圖象上的所有點(diǎn)的橫坐標(biāo)縮短為原來(lái)的SKIPIF1<0,縱坐標(biāo)不變,再把所得的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度,然后再把所得的圖象向下平移1個(gè)單位長(zhǎng)度,可得SKIPIF1<0由SKIPIF1<0,可知SKIPIF1<0即SKIPIF1<0SKIPIF1<0所以SKIPIF1<0SKIPIF1<0的最大值為SKIPIF1<0,SKIPIF1<0的最小值為SKIPIF1<0則SKIPIF1<0的最大值為SKIPIF1<0,SKIPIF1<0的最小值為SKIPIF1<0所以SKIPIF1<0的最大值為SKIPIF1<0故選:A【變式1-3】（2021屆安徽省馬鞍山二中高三下學(xué)期4月高考模擬數(shù)學(xué)試題）將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0（SKIPIF1<0）個(gè)單位長(zhǎng)度后得到函數(shù)SKIPIF1<0的圖象，若使SKIPIF1<0成立的a、b有SKIPIF1<0，則下列直線中可以是函數(shù)SKIPIF1<0圖象的對(duì)稱軸的是A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】根據(jù)三角函數(shù)平移關(guān)系求出SKIPIF1<0的解析式，結(jié)合SKIPIF1<0成立的SKIPIF1<0有SKIPIF1<0，求出SKIPIF1<0的關(guān)系，結(jié)合最小值建立方程求出SKIPIF1<0的值即可.解：將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0（SKIPIF1<0）個(gè)單位長(zhǎng)度后得到函數(shù)SKIPIF1<0的圖象，

即SKIPIF1<0，若SKIPIF1<0成立，即SKIPIF1<0，即SKIPIF1<0，

則SKIPIF1<0與SKIPIF1<0一個(gè)取最大值1，一個(gè)取最小值?1，不妨設(shè)SKIPIF1<0，

則SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0，

∵SKIPIF1<0，∴當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，

SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0（舍），

即SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，

當(dāng)SKIPIF1<0時(shí)，對(duì)稱軸方程為SKIPIF1<0.故選：D.題型05對(duì)稱中心最值應(yīng)用【解題攻略】正余弦對(duì)稱中心：零點(diǎn)處，令sin(ωx＋φ)＝0，ωx＋φ＝kπ(k∈Z)，可求得對(duì)稱中心的橫坐標(biāo)對(duì)稱中心橫坐標(biāo)代入，三角函數(shù)那部分必為0【典例1-1】（2022·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0的兩條相鄰對(duì)稱軸之間的距離為SKIPIF1<0，則下列點(diǎn)的坐標(biāo)為SKIPIF1<0的對(duì)稱中心的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】根據(jù)相鄰對(duì)稱軸之間距離可得最小正周期為SKIPIF1<0，由此可求得SKIPIF1<0，得到SKIPIF1<0解析式；利用正弦型函數(shù)對(duì)稱中心的求法可求得對(duì)稱中心，對(duì)比選項(xiàng)可得結(jié)果.【詳解】SKIPIF1<0兩條相鄰對(duì)稱軸之間的距離為SKIPIF1<0，SKIPIF1<0最小正周期SKIPIF1<0，解得：SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，解得：SKIPIF1<0，此時(shí)SKIPIF1<0，SKIPIF1<0的對(duì)稱中心為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的一個(gè)對(duì)稱中心為SKIPIF1<0.故選：C.【典例1-2】（2022·天津南開·二模）函數(shù)SKIPIF1<0SKIPIF1<0，其圖象的一個(gè)最低點(diǎn)是SKIPIF1<0，距離SKIPIF1<0點(diǎn)最近的對(duì)稱中心為SKIPIF1<0，則（

）A．SKIPIF1<0B．SKIPIF1<0是函數(shù)SKIPIF1<0圖象的一條對(duì)稱軸C．SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0單調(diào)遞增D．SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位后得到SKIPIF1<0的圖象，若SKIPIF1<0是奇函數(shù)，則SKIPIF1<0的最小值是SKIPIF1<0【答案】C【分析】由函數(shù)的圖像的頂點(diǎn)坐標(biāo)求出SKIPIF1<0，由周期求出SKIPIF1<0，由最低點(diǎn)求出SKIPIF1<0的值，可得函數(shù)的解析式，再利用三角函數(shù)的圖像和性質(zhì)，得出結(jié)論．【詳解】解：SKIPIF1<0函數(shù)SKIPIF1<0，SKIPIF1<0的圖象的一個(gè)最低點(diǎn)是SKIPIF1<0，距離SKIPIF1<0點(diǎn)最近的對(duì)稱中心為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0，所以函數(shù)SKIPIF1<0，故A錯(cuò)誤；SKIPIF1<0，故函數(shù)關(guān)于SKIPIF1<0對(duì)稱，故B錯(cuò)誤；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞增，故C正確；把SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位后得到SKIPIF1<0的圖象，若SKIPIF1<0是奇函數(shù)，則SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0的最小值是SKIPIF1<0，故D錯(cuò)誤，故選：C【變式1-1】．（2022·四川涼山·三模（理））將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位，再將縱坐標(biāo)伸長(zhǎng)為原來(lái)的4倍（橫坐標(biāo)不變）得到函數(shù)SKIPIF1<0的圖象，且SKIPIF1<0的圖象上一條對(duì)稱軸與一個(gè)對(duì)稱中心的最小距離為SKIPIF1<0，對(duì)于函數(shù)SKIPIF1<0有以下幾個(gè)結(jié)論：（1）SKIPIF1<0；（2）它的圖象關(guān)于直線SKIPIF1<0對(duì)稱；（3）它的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱；（4）若SKIPIF1<0，則SKIPIF1<0；則上述結(jié)論正確的個(gè)數(shù)為（

）A．1 B．2 C．3 D．4【答案】C【分析】先根據(jù)圖像平移的性質(zhì)求出SKIPIF1<0的函數(shù)解析式，逐項(xiàng)代入分析即可.【詳解】解：由題意得：SKIPIF1<0，向左平移SKIPIF1<0個(gè)單位，再將縱坐標(biāo)伸長(zhǎng)為原來(lái)的4倍（橫坐標(biāo)不變）得到函數(shù)SKIPIF1<0.對(duì)于選項(xiàng)A：由SKIPIF1<0的圖象上一條對(duì)稱軸與一個(gè)對(duì)稱中心的最小距離為SKIPIF1<0，最小正周期SKIPIF1<0，即SKIPIF1<0SKIPIF1<0，解得SKIPIF1<0，故SKIPIF1<0，所以（1）錯(cuò)誤；當(dāng)SKIPIF1<0時(shí)，代入SKIPIF1<0可知SKIPIF1<0，故圖像的一條對(duì)稱軸是SKIPIF1<0，故（2）正確；當(dāng)SKIPIF1<0時(shí)，代入SKIPIF1<0可知SKIPIF1<0，故圖像的一個(gè)對(duì)稱點(diǎn)是SKIPIF1<0，故（3）正確；若SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0因此SKIPIF1<0在SKIPIF1<0上的取值范圍是SKIPIF1<0，故（4）正確；由上可知（2）（3）（4）正確，正確的個(gè)數(shù)為SKIPIF1<0個(gè).故選：C【變式1-2】（2023·全國(guó)·高三專題練習(xí)）將函數(shù)SKIPIF1<0的圖象分別向左、向右各平移SKIPIF1<0個(gè)單位長(zhǎng)度后，所得的兩個(gè)圖象對(duì)稱中心重合，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．2 C．3 D．6【答案】A【分析】根據(jù)三角函數(shù)的圖象變換求得SKIPIF1<0和SKIPIF1<0，根據(jù)函數(shù)SKIPIF1<0與SKIPIF1<0的對(duì)稱中心重合，得到SKIPIF1<0，即可求解.【詳解】解：將函數(shù)SKIPIF1<0的圖象分別向左平移SKIPIF1<0個(gè)單位長(zhǎng)度后，可得SKIPIF1<0將函數(shù)SKIPIF1<0的圖象分別向右各平移SKIPIF1<0個(gè)單位長(zhǎng)度后，可得SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0與SKIPIF1<0的對(duì)稱中心重合，所以SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0的最小值為SKIPIF1<0.故選：A.【變式1-3】（2021·安徽·六安一中高三階段練習(xí)（理））已知SKIPIF1<0的一個(gè)對(duì)稱中心為SKIPIF1<0，把SKIPIF1<0的圖像向右平移SKIPIF1<0個(gè)單位后，可以得到偶函數(shù)SKIPIF1<0的圖象，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】利用輔助角公式將函數(shù)化簡(jiǎn)，即可求出函數(shù)的對(duì)稱中心坐標(biāo)，再根據(jù)三角函數(shù)的平移變換規(guī)則得到SKIPIF1<0的解析式，結(jié)合函數(shù)的奇偶性，求出SKIPIF1<0的取值，從而計(jì)算可得；【詳解】解：因?yàn)镾KIPIF1<0，令SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，即函數(shù)的對(duì)稱中心坐標(biāo)為SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，把SKIPIF1<0的圖像向右平移SKIPIF1<0個(gè)單位得到SKIPIF1<0，因?yàn)镾KIPIF1<0為偶函數(shù)，所以SKIPIF1<0，解得SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0且SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)SKIPIF1<0；故選：D題型06輔助角最值【解題攻略】SKIPIF1<0輔助角范圍滿足：SKIPIF1<0【典例1-1】（江西省上饒市（天佑中學(xué)、余干中學(xué)等）六校2021屆高三下學(xué)期第一次聯(lián)考數(shù)學(xué)試題已知函數(shù)SKIPIF1<0，SKIPIF1<0的最小值為SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】由SKIPIF1<0可得出不等式SKIPIF1<0對(duì)任意的SKIPIF1<0恒成立，化簡(jiǎn)得出SKIPIF1<0，分SKIPIF1<0、SKIPIF1<0兩種情況討論，結(jié)合SKIPIF1<0可求得實(shí)數(shù)SKIPIF1<0的取值范圍.【詳解】SKIPIF1<0且SKIPIF1<0，由題意可知，對(duì)任意的SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，可得SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0成立；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0，此時(shí)SKIPIF1<0.綜上所述，實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故選：C.【典例1-2】已知函數(shù)SKIPIF1<0在SKIPIF1<0上的值域?yàn)镾KIPIF1<0，則SKIPIF1<0的取值范圍為______.【答案】SKIPIF1<0【分析】化簡(jiǎn)得SKIPIF1<0，其中SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，再結(jié)合三角函數(shù)的性質(zhì)可求解.【詳解】由題意得SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，其中SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0.因?yàn)镾KIPIF1<0，SKIPIF1<0，故SKIPIF1<0，因?yàn)镾KIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0，則SKIPIF1<0.又當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減，且SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0.【變式1-1】.已知函數(shù)SKIPIF1<0，周期SKIPIF1<0，SKIPIF1<0，且在SKIPIF1<0處取得最大值，則使得不等式SKIPIF1<0恒成立的實(shí)數(shù)SKIPIF1<0的最小值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】先根據(jù)三角恒等變換和三角形函數(shù)的性質(zhì)，以及同角的三角函數(shù)的關(guān)系可得SKIPIF1<0，①，再根據(jù)SKIPIF1<0，可得SKIPIF1<0，②，通過①②求出SKIPIF1<0的值，再根據(jù)三角函數(shù)的性質(zhì)可得SKIPIF1<0，SKIPIF1<0，求出SKIPIF1<0，根據(jù)不等式SKIPIF1<0恒成立，則SKIPIF1<0，即可求出答案．【詳解】SKIPIF1<0，其中SKIPIF1<0，SKIPIF1<0SKIPIF1<0處取得最大值，SKIPIF1<0SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，①，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，②，①SKIPIF1<0②得SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0（舍去），由①得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0在第一象限，SKIPIF1<0取SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，使SKIPIF1<0最小，則SKIPIF1<0，即SKIPIF1<0，若不等式SKIPIF1<0恒成立，則SKIPIF1<0，故選：B【變式1-2】（浙江省紹興市諸暨市海亮高級(jí)中學(xué)2021-2022學(xué)年高三上學(xué)期12月選考數(shù)學(xué)試題）已知當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0取到最大值，則SKIPIF1<0是（）A．奇函數(shù)，在SKIPIF1<0時(shí)取到最小值； B．偶函數(shù)，在SKIPIF1<0時(shí)取到最小值；C．奇函數(shù)，在SKIPIF1<0時(shí)取到最小值； D．偶函數(shù)，在SKIPIF1<0時(shí)取到最小值；【答案】B【分析】由輔助角公式可得SKIPIF1<0，根據(jù)SKIPIF1<0時(shí)SKIPIF1<0有最大值可得SKIPIF1<0，求出SKIPIF1<0，再根據(jù)奇偶性并計(jì)算SKIPIF1<0、SKIPIF1<0可得答案.【詳解】SKIPIF1<0SKIPIF1<0，取SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有最大值SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，SKIPIF1<0為偶函數(shù)，SKIPIF1<0，SKIPIF1<0，故B正確，故選：B.【變式1-3】（江蘇省淮安市淮陰中學(xué)2020屆高三下學(xué)期5月高考模擬數(shù)學(xué)試題）若存在正整數(shù)m使得關(guān)于x的方程SKIPIF1<0在SKIPIF1<0上有兩個(gè)不等實(shí)根，則正整數(shù)n的最小值是______．【答案】4【分析】化簡(jiǎn)SKIPIF1<0，因?yàn)镾KIPIF1<0，則SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上有兩個(gè)不等實(shí)根，轉(zhuǎn)化為SKIPIF1<0在SKIPIF1<0上有兩個(gè)不等實(shí)根，故SKIPIF1<0，即可得出答案．【詳解】SKIPIF1<0，其中SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，則SKIPIF1<0，+SKIPIF1<0在SKIPIF1<0上有兩個(gè)不等實(shí)根，SKIPIF1<0在SKIPIF1<0上有兩個(gè)不等實(shí)根，則SKIPIF1<0，所以SKIPIF1<0①對(duì)任意SKIPIF1<0，SKIPIF1<0，恒成立．由②得SKIPIF1<0，存在SKIPIF1<0，成立，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0．故答案為：4題型07正余弦換元型最值【解題攻略】SKIPIF1<0與SKIPIF1<0在同一函數(shù)中一般可設(shè)SKIPIF1<0進(jìn)行換元．換元時(shí)注意新元的取值范圍．SKIPIF1<0之間的互化關(guān)系1.SKIPIF1<02.SKIPIF1<0【典例1-1】（2021下·上海徐匯·高三南洋中學(xué)校考階段練習(xí)）已知函數(shù)SKIPIF1<0，則SKIPIF1<0的值域?yàn)椋敬鸢浮縎KIPIF1<0【分析】利用換元法，令SKIPIF1<0，進(jìn)而可得SKIPIF1<0，再利用函數(shù)的單調(diào)性即可求解.【詳解】由SKIPIF1<0令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，又對(duì)勾函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0，SKIPIF1<0；單調(diào)遞增區(qū)間為SKIPIF1<0，SKIPIF1<0，結(jié)合對(duì)勾函數(shù)的圖象，如下：所以SKIPIF1<0，所以SKIPIF1<0，所以函數(shù)的值域?yàn)镾KIPIF1<0.故答案為：SKIPIF1<0.【典例1-2】（2022·高三單元測(cè)試）函數(shù)SKIPIF1<0的值域?yàn)椋敬鸢浮縎KIPIF1<0【分析】利用SKIPIF1<0通過換元將原函數(shù)轉(zhuǎn)化為含未知量SKIPIF1<0的函數(shù)SKIPIF1<0，再解出函數(shù)SKIPIF1<0的值域即為函數(shù)SKIPIF1<0的值域.【詳解】令SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0．故答案為：SKIPIF1<0.【變式1-1】已知函數(shù)SKIPIF1<0，則SKIPIF1<0的最大值為___________.【答案】SKIPIF1<