人教A版高中數(shù)學(xué)（必修第一冊）同步講義 5.4三角函數(shù)圖象（教師版）

第第頁第05講5.4.1正弦函數(shù)、余弦函數(shù)的圖象課程標(biāo)準(zhǔn)學(xué)習(xí)目標(biāo)①理解并掌握用單位圓作正弦函數(shù)以及作余弦函數(shù)的圖象的方法.掌握數(shù)形結(jié)合的優(yōu)勢。②通過兩類函數(shù)圖象認(rèn)識函數(shù)圖象的特點(diǎn)，并能通過兩類圖象的形狀掌握兩類函數(shù)的性質(zhì)。會作正弦函數(shù)、余弦函數(shù)的圖象的同時(shí)，能認(rèn)識圖象與三角函數(shù)的密切關(guān)系，并能解決與圖象有關(guān)的三角函數(shù)問題知識點(diǎn)01：正弦函數(shù)的圖象正弦函數(shù)SKIPIF1<0,SKIPIF1<0的圖象叫做正弦曲線.知識點(diǎn)02：正弦函數(shù)圖象的畫法(1)幾何法:①在單位圓上,將點(diǎn)SKIPIF1<0繞著點(diǎn)SKIPIF1<0旋轉(zhuǎn)SKIPIF1<0弧度至點(diǎn)SKIPIF1<0,根據(jù)正弦函數(shù)的定義,點(diǎn)SKIPIF1<0的縱坐標(biāo)SKIPIF1<0.由此,以SKIPIF1<0為橫坐標(biāo),SKIPIF1<0為縱坐標(biāo)畫點(diǎn),即得到函數(shù)圖象上的點(diǎn)SKIPIF1<0.②將函數(shù)SKIPIF1<0,SKIPIF1<0的圖象不斷向左、向右平行移動(dòng)(每次移動(dòng)SKIPIF1<0個(gè)單位長度).(2)“五點(diǎn)法”:在函數(shù)SKIPIF1<0,SKIPIF1<0的圖象上,以下五個(gè)點(diǎn):SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在確定圖象形狀時(shí)起關(guān)鍵作用.描出這五個(gè)點(diǎn),函數(shù)SKIPIF1<0,SKIPIF1<0的圖象形狀就基本確定了.因此,在精確度要求不高時(shí),常先找出這五個(gè)關(guān)鍵點(diǎn),再用光滑的曲線將它們連接起來,得到正弦函數(shù)的簡圖.【即學(xué)即練1】（2023春·陜西西安·高二西北工業(yè)大學(xué)附屬中學(xué)?？茧A段練習(xí)）用五點(diǎn)法作出函數(shù)SKIPIF1<0的大致圖象．【答案】圖象見解析【詳解】解：因?yàn)镾KIPIF1<0，列表：SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0描點(diǎn)、連線，函數(shù)圖象如下圖所示：知識點(diǎn)03：余弦函數(shù)的圖象余弦函數(shù)SKIPIF1<0,SKIPIF1<0的圖象叫做余弦曲線.知識點(diǎn)04：余弦函數(shù)圖象的畫法(1)要得到SKIPIF1<0,SKIPIF1<0的圖象,只需把SKIPIF1<0,SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長度即可,這是因?yàn)镾KIPIF1<0.(2)用“五點(diǎn)法”:畫余弦函數(shù)SKIPIF1<0在SKIPIF1<0上的圖象時(shí),所取的五個(gè)關(guān)鍵點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0再用光滑的曲線連接起來.【即學(xué)即練1】（2023·全國·高三專題練習(xí)）作出函數(shù)SKIPIF1<0的圖象【答案】見解析【詳解】SKIPIF1<0，SKIPIF1<0，作出函數(shù)SKIPIF1<0圖象后，將SKIPIF1<0軸下方的部分沿SKIPIF1<0軸翻折到SKIPIF1<0軸上方，即為函數(shù)SKIPIF1<0的圖象，如圖

題型01用“五點(diǎn)法”作三角函數(shù)的圖象【典例1】（2023·全國·高三專題練習(xí)）用“五點(diǎn)法”在給定的坐標(biāo)系中，畫出函數(shù)SKIPIF1<0在SKIPIF1<0上的大致圖像.

【答案】答案見解析【詳解】列表：SKIPIF1<00SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0120SKIPIF1<001描點(diǎn)，連線，畫出SKIPIF1<0在SKIPIF1<0上的大致圖像如圖：【典例2】（2023·全國·高一課堂例題）（1）作出函數(shù)SKIPIF1<0的簡圖；（2）作出函數(shù)SKIPIF1<0的簡圖．【答案】（1）答案見解析；（2）答案見解析【詳解】（1）列表：SKIPIF1<00SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0010SKIPIF1<00SKIPIF1<0020SKIPIF1<00描點(diǎn)并用光滑的曲線連接起來，可得函數(shù)SKIPIF1<0的圖象，如圖所示：

（2）列表：SKIPIF1<00SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<001SKIPIF1<001SKIPIF1<001210描點(diǎn)并用光滑的曲線連接起來，可得函數(shù)SKIPIF1<0的圖象，如圖所示：

【變式1】（2023春·北京·高一北京市第三十五中學(xué)?？茧A段練習(xí)）用五點(diǎn)法畫出函數(shù)SKIPIF1<0一個(gè)周期的圖象.【答案】答案見解析【詳解】令SKIPIF1<0，則SKIPIF1<0.列表：SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0函數(shù)SKIPIF1<0在一個(gè)周期內(nèi)的圖象如下圖所示：【變式2】（2023·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0，SKIPIF1<0．在用“五點(diǎn)法”作函數(shù)SKIPIF1<0的圖象時(shí)，列表如下：SKIPIF1<0xSKIPIF1<0完成上述表格，并在坐標(biāo)系中畫出函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的圖象；

【答案】填表見解析；作圖見解析【詳解】由題意列出以下表格：SKIPIF1<0SKIPIF1<00SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0x0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0020SKIPIF1<0SKIPIF1<0函數(shù)圖象如圖所示：

【變式3】（2023·全國·高一隨堂練習(xí)）用五點(diǎn)法分別畫下列函數(shù)在SKIPIF1<0上的圖象:(1)SKIPIF1<0;(2)SKIPIF1<0.【答案】（1）見解析（2）見解析【詳解】解:xSKIPIF1<0SKIPIF1<00SKIPIF1<0SKIPIF1<0SKIPIF1<0010-10SKIPIF1<032123

題型02利用圖象解三角不等式【典例1】（2023·全國·高一假期作業(yè)）不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】解：SKIPIF1<0SKIPIF1<0SKIPIF1<0函數(shù)圖象如下所示：

SKIPIF1<0SKIPIF1<0，SKIPIF1<0不等式的解集為：SKIPIF1<0．故選：SKIPIF1<0．【典例2】（2023秋·江西撫州·高二黎川縣第二中學(xué)校考開學(xué)考試）不等式SKIPIF1<0的解集為．【答案】SKIPIF1<0【詳解】畫出SKIPIF1<0時(shí)，SKIPIF1<0的圖象．

令SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0又SKIPIF1<0的周期為SKIPIF1<0，所以SKIPIF1<0的解集為SKIPIF1<0．用SKIPIF1<0代替SKIPIF1<0解出SKIPIF1<0．可得SKIPIF1<0則SKIPIF1<0的解集為SKIPIF1<0.故答案為：SKIPIF1<0.【變式1】（2023·全國·高一假期作業(yè)）不等式SKIPIF1<0SKIPIF1<0的解集是（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】A【詳解】如圖所示，不等式SKIPIF1<0，SKIPIF1<0的解集為SKIPIF1<0故選：A【變式2】（2023春·高一課時(shí)練習(xí)）在(0，2π)內(nèi)使sinx>|cosx|的x的取值范圍是（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】A【詳解】因?yàn)閟inx>|cosx|且x∈(0，2π)，所以sinx>0，所以x∈(0，π)，在同一平面直角坐標(biāo)系中畫出y=sinx，x∈(0，π)與y=|cosx|，x∈(0，π)的圖象，觀察圖象易得x∈SKIPIF1<0．故選：A.【變式3】（2023春·上海嘉定·高一?？计谥校┎坏仁絊KIPIF1<0的解集為.【答案】SKIPIF1<0【分析】畫出SKIPIF1<0的圖象，由圖象即可求解.【詳解】

畫出SKIPIF1<0的圖象，如圖所示，由圖可知，不等式SKIPIF1<0的解集為SKIPIF1<0.故答案為：SKIPIF1<0題型03利用圖象求方程的解或函數(shù)零點(diǎn)的個(gè)數(shù)問題【典例1】（2023·全國·高一假期作業(yè)）函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)為.【答案】3【詳解】由SKIPIF1<0，則函數(shù)SKIPIF1<0零點(diǎn)個(gè)數(shù)為SKIPIF1<0圖象交點(diǎn)個(gè)數(shù)，在同一坐標(biāo)系中畫出兩函數(shù)圖象如下，則交點(diǎn)有3個(gè)，即SKIPIF1<0有3個(gè)零點(diǎn).故答案為：3【典例2】（2023秋·河南新鄉(xiāng)·高一校聯(lián)考期末）已知函數(shù)SKIPIF1<0在SKIPIF1<0上恰有2個(gè)零點(diǎn)，則SKIPIF1<0的取值范圍為．【答案】SKIPIF1<0【詳解】由SKIPIF1<0且SKIPIF1<0，得SKIPIF1<0，若SKIPIF1<0在SKIPIF1<0上無零點(diǎn)，則SKIPIF1<0在SKIPIF1<0上恰有2個(gè)零點(diǎn)，則SKIPIF1<0，無解；若SKIPIF1<0在SKIPIF1<0上恰有1個(gè)零點(diǎn)，則SKIPIF1<0在SKIPIF1<0上恰有1個(gè)零點(diǎn)，則SKIPIF1<0，解得SKIPIF1<0；若SKIPIF1<0在SKIPIF1<0上恰有2個(gè)零點(diǎn)，則SKIPIF1<0在SKIPIF1<0上無零點(diǎn)，則SKIPIF1<0，無解，所以SKIPIF1<0的取值范圍為SKIPIF1<0.故答案為：SKIPIF1<0【典例3】（2023春·新疆塔城·高一塔城地區(qū)第一高級中學(xué)?？茧A段練習(xí)）函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)為．【答案】7【詳解】依題意求函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)，可以轉(zhuǎn)化為求函數(shù)SKIPIF1<0與SKIPIF1<0的交點(diǎn)個(gè)數(shù)，SKIPIF1<0，如圖，對于函數(shù)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；所以在SKIPIF1<0軸非負(fù)半軸上兩個(gè)函數(shù)圖像有4個(gè)交點(diǎn)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；所以在SKIPIF1<0軸負(fù)半軸上兩個(gè)函數(shù)圖像有3個(gè)交點(diǎn)，

綜上，函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)為7.故答案為：7.【典例4】（2023春·貴州遵義·高一統(tǒng)考期中）已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上有且僅有10個(gè)零點(diǎn)，則ω的取值范圍是.【答案】SKIPIF1<0【詳解】由SKIPIF1<0，則SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0在區(qū)間SKIPIF1<0上有且僅有10個(gè)零點(diǎn)，所以由余弦函數(shù)的性質(zhì)可知：SKIPIF1<0解得SKIPIF1<0，故答案為：SKIPIF1<0.【典例5】（2023春·高一單元測試）方程SKIPIF1<0的解的個(gè)數(shù)是．【答案】7【詳解】由正弦函數(shù)值域可得SKIPIF1<0，又因?yàn)楫?dāng)SKIPIF1<0時(shí)，SKIPIF1<0；所以，分別畫出SKIPIF1<0和SKIPIF1<0在SKIPIF1<0上的圖象如下圖所示：

根據(jù)圖像并根據(jù)其對稱性可知，在SKIPIF1<0上兩函數(shù)圖象共有7個(gè)交點(diǎn)；由函數(shù)與方程可知，方程SKIPIF1<0有7個(gè)解.故答案為：7【變式1】（2023·四川綿陽·統(tǒng)考模擬預(yù)測）已知函數(shù)SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上的零點(diǎn)個(gè)數(shù)為．【答案】2【詳解】令SKIPIF1<0，可得SKIPIF1<0，原題意等價(jià)于求SKIPIF1<0與SKIPIF1<0在SKIPIF1<0上的交點(diǎn)個(gè)數(shù)，∵SKIPIF1<0，則SKIPIF1<0，且SKIPIF1<0，有余弦函數(shù)可知SKIPIF1<0與SKIPIF1<0在SKIPIF1<0上有2個(gè)交點(diǎn)所以SKIPIF1<0與SKIPIF1<0在SKIPIF1<0上有2個(gè)交點(diǎn).故答案為：2.【變式2】（2023春·湖北武漢·高一華中科技大學(xué)附屬中學(xué)校聯(lián)考期中）已知函數(shù)SKIPIF1<0），若方程SKIPIF1<0在SKIPIF1<0上恰有5個(gè)實(shí)數(shù)解，則實(shí)數(shù)SKIPIF1<0的取值范圍為．【答案】SKIPIF1<0【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0在區(qū)間SKIPIF1<0上恰好有5個(gè)x，使得SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上恰有5條對稱軸．令SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上恰有5條對稱軸，如圖：所以SKIPIF1<0，解得SKIPIF1<0．故答案為：SKIPIF1<0.【變式3】（2023春·江西南昌·高一校考階段練習(xí)）已知函數(shù)SKIPIF1<0，若存在實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0SKIPIF1<0互不相等SKIPIF1<0，則SKIPIF1<0的取值范圍是.【答案】SKIPIF1<0【詳解】函數(shù)SKIPIF1<0的圖象如下圖所示：存在實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0SKIPIF1<0互不相等SKIPIF1<0，不妨設(shè)SKIPIF1<0，則由圖可知SKIPIF1<0關(guān)于SKIPIF1<0對稱，所以SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，此時(shí)SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，因?yàn)镾KIPIF1<0解得SKIPIF1<0或SKIPIF1<0，故而SKIPIF1<0，SKIPIF1<0，且由圖可得SKIPIF1<0，即SKIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0，所以SKIPIF1<0，綜上所述SKIPIF1<0；故答案為：SKIPIF1<0.【變式4】（2023春·四川廣安·高一?？茧A段練習(xí)）已知關(guān)于x的方程SKIPIF1<0在SKIPIF1<0上有兩個(gè)不同的實(shí)數(shù)根，則m的取值范圍是．【答案】SKIPIF1<0【詳解】由題設(shè)SKIPIF1<0在SKIPIF1<0上有兩個(gè)不同的實(shí)數(shù)根，又SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0的圖象如下，只需SKIPIF1<0與SKIPIF1<0在給定區(qū)間內(nèi)有兩個(gè)交點(diǎn)即可，如圖，SKIPIF1<0，則SKIPIF1<0.故答案為：SKIPIF1<0【變式5】（2023秋·河北衡水·高二衡水市第二中學(xué)?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0，令SKIPIF1<0在區(qū)間SKIPIF1<0上恰有2個(gè)零點(diǎn)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0.【答案】SKIPIF1<0SKIPIF1<0【詳解】由題意得SKIPIF1<0在SKIPIF1<0上恰有2個(gè)零點(diǎn)，即SKIPIF1<0在SKIPIF1<0上恰有2個(gè)零點(diǎn)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，畫出SKIPIF1<0在SKIPIF1<0時(shí)的函數(shù)圖象，

SKIPIF1<0關(guān)于SKIPIF1<0對稱，故SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0故答案為：SKIPIF1<0，SKIPIF1<0A夯實(shí)基礎(chǔ)1．（2023秋·高一課時(shí)練習(xí)）函數(shù)SKIPIF1<0的圖象中與y軸最近的最高點(diǎn)的坐標(biāo)為（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】B；【詳解】用五點(diǎn)法畫出函數(shù)SKIPIF1<0的部分圖象如圖所示，由圖易知與y軸最近的最高點(diǎn)的坐標(biāo)為SKIPIF1<0.

故選：B2．（2023·全國·高三專題練習(xí)）用“五點(diǎn)法”作SKIPIF1<0的圖象，首先描出的五個(gè)點(diǎn)的橫坐標(biāo)是（）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】B【詳解】由“五點(diǎn)法”作圖知：令SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，即為五個(gè)關(guān)鍵點(diǎn)的橫坐標(biāo).故選：B.3．（2023·全國·高三專題練習(xí)）三角函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的圖像為（）A． B．C． D．【答案】C【詳解】解：∵SKIPIF1<0為奇函數(shù)，∴SKIPIF1<0的圖像關(guān)于原點(diǎn)對稱，故排除A、D選項(xiàng)，三角函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值為SKIPIF1<0，故排除B選項(xiàng).故選：C.4．（2023·全國·高三專題練習(xí)）從函數(shù)SKIPIF1<0的圖象來看，當(dāng)SKIPIF1<0時(shí)，對于SKIPIF1<0的x有（

）A．0個(gè) B．1個(gè) C．2個(gè) D．3個(gè)【答案】C【詳解】先畫出SKIPIF1<0，SKIPIF1<0的圖象，即A與D之間的部分，再畫出SKIPIF1<0的圖象，如下圖：由圖象可知它們有2個(gè)交點(diǎn)B、C，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的x的值有2個(gè).故選：C5．（2023·全國·高三專題練習(xí)）函數(shù)SKIPIF1<0與SKIPIF1<0圖像交點(diǎn)的個(gè)數(shù)為（

）A．0 B．1 C．2 D．3【答案】C【詳解】作出函數(shù)SKIPIF1<0在SKIPIF1<0上的圖象，并作出直線SKIPIF1<0，如圖：觀察圖形知：函數(shù)SKIPIF1<0在SKIPIF1<0上的圖象與直線SKIPIF1<0有兩個(gè)公共點(diǎn)，所以函數(shù)SKIPIF1<0與SKIPIF1<0圖像交點(diǎn)的個(gè)數(shù)為2.故選：C6．（2023·全國·高三專題練習(xí)）函數(shù)SKIPIF1<0的簡圖是（

）A．B．C． D．【答案】B【詳解】由SKIPIF1<0知，其圖象和SKIPIF1<0的圖象相同，故選B．7．（2023秋·安徽合肥·高一校聯(lián)考期末）函數(shù)SKIPIF1<0，SKIPIF1<0的圖象在區(qū)間SKIPIF1<0的交點(diǎn)個(gè)數(shù)為（

）A．3 B．4 C．5 D．6【答案】A【詳解】分別作出SKIPIF1<0，SKIPIF1<0在區(qū)間SKIPIF1<0上的圖象，如圖所示，

由圖象可知：SKIPIF1<0，SKIPIF1<0的圖象在區(qū)間SKIPIF1<0的交點(diǎn)個(gè)數(shù)為3.故選：A.8．（2023春·遼寧·高一鳳城市第一中學(xué)校聯(lián)考階段練習(xí)）華羅庚說：“數(shù)缺形時(shí)少直觀，形少數(shù)時(shí)難入微，數(shù)形結(jié)合百般好，隔離分家萬事休.”所以研究函數(shù)時(shí)往往要作圖，那么函數(shù)SKIPIF1<0的部分圖像可能是（

）A．

B．

C．

D．

【答案】B【詳解】因?yàn)镾KIPIF1<0，所以ACD錯(cuò)誤.故選：B二、多選題9．（2023秋·高一課時(shí)練習(xí)）（多選）函數(shù)SKIPIF1<0與SKIPIF1<0有一個(gè)交點(diǎn)，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．0C．1 D．SKIPIF1<0【答案】BD【詳解】畫出SKIPIF1<0的圖象.如圖：直線SKIPIF1<0和SKIPIF1<0與SKIPIF1<0的圖象只有一個(gè)交點(diǎn)，

故SKIPIF1<0或SKIPIF1<0.故選：BD.10．（2023·全國·高一假期作業(yè)）函數(shù)SKIPIF1<0，SKIPIF1<0的圖像與直線SKIPIF1<0（t為常數(shù)，SKIPIF1<0）的交點(diǎn)可能有（

）A．0個(gè) B．1個(gè) C．2個(gè) D．3個(gè)【答案】ABC【詳解】作出SKIPIF1<0，SKIPIF1<0的圖像觀察可知，

當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0的圖像與直線SKIPIF1<0的交點(diǎn)個(gè)數(shù)為0；當(dāng)SKIPIF1<0或SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0的圖像與直線SKIPIF1<0的交點(diǎn)個(gè)數(shù)為l；當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0的圖像與直線SKIPIF1<0的交點(diǎn)個(gè)數(shù)為2.故選：ABC.三、填空題11．（2023秋·湖南邵陽·高三湖南省邵東市第一中學(xué)校考階段練習(xí)）若函數(shù)SKIPIF1<0在SKIPIF1<0上有且僅有3個(gè)零點(diǎn)，則SKIPIF1<0的最小值為.【答案】SKIPIF1<0【詳解】令SKIPIF1<0，得SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，依題意，SKIPIF1<0在SKIPIF1<0上有且僅有3個(gè)零點(diǎn)，則SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0的最小值為SKIPIF1<0.故答案為：SKIPIF1<0.12．（2023春·上海嘉定·高一?？计谥校┎坏仁絊KIPIF1<0的解集為.【答案】SKIPIF1<0【詳解】

畫出SKIPIF1<0的圖象，如圖所示，由圖可知，不等式SKIPIF1<0的解集為SKIPIF1<0.故答案為：SKIPIF1<0四、解答題13．（2023·全國·高三專題練習(xí)）用“五點(diǎn)法”在給定的坐標(biāo)系中，畫出函數(shù)SKIPIF1<0在SKIPIF1<0上的大致圖像.

【答案】答案見解析【詳解】列表：SKIPIF1<00SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0120SKIPIF1<001描點(diǎn)，連線，畫出SKIPIF1<0在SKIPIF1<0上的大致圖像如圖：14．（2023·全國·高三專題練習(xí)）函數(shù)SKIPIF1<0，用五點(diǎn)作圖法畫出函數(shù)SKIPIF1<0在SKIPIF1<0上的圖象；（先列表，再畫圖）【答案】答案見解析【詳解】SKIPIF1<0，按五個(gè)關(guān)鍵點(diǎn)列表：SKIPIF1<00SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0010SKIPIF1<00SKIPIF1<003010描點(diǎn)并將它們用光滑的曲線連接起來如下圖所示：15．（2023·高一課時(shí)練習(xí)）作函數(shù)SKIPIF1<0的圖象.【答案】圖象見解析.【詳解】SKIPIF1<0SKIPIF1<0故SKIPIF1<0的圖象實(shí)際就是SKIPIF1<0的圖象在x軸下方的部分翻折到x軸上方后得到的圖象，如圖16．（2023秋·山東泰安·高一泰山中學(xué)?？计谀┮阎瘮?shù)SKIPIF1<0（1）作出該函數(shù)的圖象；（2）若SKIPIF1<0，求SKIPIF1<0的值；（3）若SKIPIF1<0，討論方程SKIPIF1<0的解的個(gè)數(shù)．【答案】（1）圖見解析；（2）SKIPIF1<0或SKIPIF1<0或SKIPIF1<0；（3）當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，解的個(gè)數(shù)為0；當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，解的個(gè)數(shù)為1；當(dāng)SKIPIF1<0時(shí)，解的個(gè)數(shù)為3．【詳解】（1）SKIPIF1<0的函數(shù)圖象如下：（2）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，解得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，綜上，SKIPIF1<0或SKIPIF1<0或SKIPIF1<0；（3）方程SKIPIF1<0的解的個(gè)數(shù)等價(jià)于SKIPIF1<0與SKIPIF1<0的圖象的交點(diǎn)個(gè)數(shù)，則由（1）中函數(shù)圖象可得，當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，解的個(gè)數(shù)為0；當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，解的個(gè)數(shù)為1；當(dāng)SKIPIF1<0時(shí)，解的個(gè)數(shù)為3．B能力提升1．（2023春·江蘇揚(yáng)州·高一統(tǒng)考期中）設(shè)函數(shù)SKIPIF1<0的定義域?yàn)镽，SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上零點(diǎn)的個(gè)數(shù)為（

）A．4 B．5 C．6 D．7【答案】D【詳解】由SKIPIF1<0，得SKIPIF1<0的圖象關(guān)于y軸對稱，由SKIPIF1<0，得SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對稱，令SKIPIF1<0，得SKIPIF1<0，函數(shù)SKIPIF1<0是周期為1的偶函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，在同一坐標(biāo)系內(nèi)作出函數(shù)SKIPIF1<0在SKIPIF1<0上的圖象，函數(shù)SKIPIF1<0在SKIPIF1<0上的圖象，如圖，

觀察圖象知，函數(shù)SKIPIF1<0與SKIPIF1<0的圖象在SKIPIF1<0上的交點(diǎn)有7個(gè)，所以函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上零點(diǎn)的個(gè)數(shù)為7.故選：D2．（2023春·四川瀘州·高一四川省瀘縣第一中學(xué)?？茧A段練習(xí)）若SKIPIF1<0(SKIPIF1<0)在SKIPIF1<0上有且只有兩個(gè)零點(diǎn)，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】∵SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上有且只有兩個(gè)零點(diǎn)，則SKIPIF1<0﹒解得SKIPIF1<0.故選：A3．（2023春·高一課時(shí)練習(xí)）函數(shù)SKIPIF1<0的圖象與直線SKIPIF1<0有且僅有兩個(gè)不同的交點(diǎn)，求實(shí)數(shù)SKIPIF1<0的取值范圍．【答案】SKIPIF1<0．【詳解】SKIPIF1<0，其圖象如圖所示．

若使SKIPIF1<0的圖象與直線SKIPIF1<0有且僅有兩個(gè)不同的交點(diǎn)，根據(jù)圖象，可得實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0．

第06講5.4.2正弦函數(shù)、余弦函數(shù)的性質(zhì)課程標(biāo)準(zhǔn)學(xué)習(xí)目標(biāo)①結(jié)合正弦函數(shù)、余弦函數(shù)的圖象掌握正、余弦函數(shù)的性質(zhì)。②會求正、余弦函數(shù)的周期，單調(diào)區(qū)間、對稱點(diǎn)、對稱軸及最值，及結(jié)合函數(shù)的圖象會求函數(shù)的解析式，并能求出相關(guān)的基本量。會求正、余弦函數(shù)的最小正周期，單調(diào)區(qū)間，對稱點(diǎn)，對稱區(qū)間，會求兩類函數(shù)的最值.知識點(diǎn)01：函數(shù)的周期性1.周期函數(shù)的定義一般地,設(shè)函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0,如果存在一個(gè)非零常數(shù)SKIPIF1<0,使得對每一個(gè)SKIPIF1<0,都有SKIPIF1<0,且SKIPIF1<0,那么函數(shù)SKIPIF1<0就叫做周期函數(shù).非零常數(shù)SKIPIF1<0叫做這個(gè)函數(shù)的周期.2.最小正周期的定義如果在周期函數(shù)f(x)的所有周期中存在一個(gè)最小的正數(shù),那么這個(gè)最小正數(shù)就叫做SKIPIF1<0的最小正周期.

【即學(xué)即練1】（2023春·天津紅橋·高一統(tǒng)考期末）函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0，則SKIPIF1<0.【答案】SKIPIF1<0【詳解】SKIPIF1<0的最小正周期SKIPIF1<0，SKIPIF1<0.故答案為：SKIPIF1<0.知識點(diǎn)02：正弦函數(shù)、余弦函數(shù)的周期性和奇偶性函數(shù)奇偶性SKIPIF1<0奇函數(shù)SKIPIF1<0偶函數(shù)SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0為奇函數(shù)；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0為偶函數(shù)；SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0為奇函數(shù)；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0為偶函數(shù)；【即學(xué)即練2】（2023秋·高一課時(shí)練習(xí)）已知函數(shù)SKIPIF1<0SKIPIF1<0的最小正周期為SKIPIF1<0，則SKIPIF1<0.【答案】12【詳解】由于SKIPIF1<0，依題意可知SKIPIF1<0.故答案為：SKIPIF1<0知識點(diǎn)03：正弦、余弦型函數(shù)的常用周期函數(shù)最小正周期SKIPIF1<0或SKIPIF1<0（SKIPIF1<0）SKIPIF1<0SKIPIF1<0或SKIPIF1<0SKIPIF1<0SKIPIF1<0或SKIPIF1<0（SKIPIF1<0）SKIPIF1<0SKIPIF1<0無周期SKIPIF1<0SKIPIF1<0【即學(xué)即練3】（2023秋·湖北荊州·高三沙市中學(xué)?？茧A段練習(xí)）函數(shù)SKIPIF1<0的最小正周期為.【答案】SKIPIF1<0/SKIPIF1<0【詳解】由誘導(dǎo)公式可知，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0與SKIPIF1<0不恒相等，故SKIPIF1<0的最小正周期為SKIPIF1<0，故答案為：SKIPIF1<0知識點(diǎn)04：正弦函數(shù)、余弦函數(shù)的圖象和性質(zhì)函數(shù)SKIPIF1<0SKIPIF1<0圖象定義域定義域SKIPIF1<0SKIPIF1<0值域SKIPIF1<0SKIPIF1<0周期性SKIPIF1<0SKIPIF1<0奇偶性奇函數(shù)偶函數(shù)單調(diào)性在每一個(gè)閉區(qū)間SKIPIF1<0(SKIPIF1<0)上都單調(diào)遞增;在每一個(gè)閉區(qū)間SKIPIF1<0(SKIPIF1<0上都單調(diào)遞減在每一個(gè)閉區(qū)間SKIPIF1<0(SKIPIF1<0)上都單調(diào)遞增;在每一個(gè)閉區(qū)間SKIPIF1<0(SKIPIF1<0)上都單調(diào)遞減最值當(dāng)SKIPIF1<0(SKIPIF1<0)時(shí)，SKIPIF1<0;當(dāng)SKIPIF1<0(SKIPIF1<0)時(shí)，SKIPIF1<0;當(dāng)SKIPIF1<0(SKIPIF1<0)時(shí)，SKIPIF1<0;當(dāng)SKIPIF1<0(SKIPIF1<0)時(shí)，SKIPIF1<0;圖象的對稱性對稱中心為SKIPIF1<0(SKIPIF1<0),對稱軸為直線SKIPIF1<0(SKIPIF1<0)對稱中心為SKIPIF1<0(SKIPIF1<0),對稱軸為直線SKIPIF1<0(SKIPIF1<0)【即學(xué)即練4】（2023·全國·高三專題練習(xí)）y＝cosSKIPIF1<0的單調(diào)遞減區(qū)間為.【答案】SKIPIF1<0【詳解】因?yàn)镾KIPIF1<0，所以由SKIPIF1<0得，SKIPIF1<0，SKIPIF1<0，即所求單調(diào)遞減區(qū)間為SKIPIF1<0.故答案為：SKIPIF1<0.題型01三角函數(shù)的周期問題及簡單應(yīng)用【典例1】（2023秋·高一課時(shí)練習(xí)）下列函數(shù)，最小正周期為SKIPIF1<0的是（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】C【詳解】函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0，故A不符合；函數(shù)SKIPIF1<0，其最小正周期為SKIPIF1<0，故B不符合；因?yàn)楹瘮?shù)SKIPIF1<0的最小正周期為SKIPIF1<0，所以函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0，故C符合；因?yàn)楹瘮?shù)SKIPIF1<0的最小正周期為SKIPIF1<0，所以函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0，故D不符合.故選：C.【典例2】（多選）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】AC【詳解】對于A，函數(shù)SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0的定義域?yàn)镾KIPIF1<0關(guān)于原點(diǎn)對稱，即SKIPIF1<0是奇函數(shù)，且注意到其周期為SKIPIF1<0，故A正確；對于B：函數(shù)SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0的定義域?yàn)镾KIPIF1<0關(guān)于原點(diǎn)對稱，所以SKIPIF1<0是偶函數(shù)，不是奇函數(shù)，故B錯(cuò)誤；對于C：SKIPIF1<0，由A選項(xiàng)分析易知SKIPIF1<0是奇函數(shù)，同時(shí)也是最小正周期是SKIPIF1<0的周期函數(shù)，故C正確；對于D：函數(shù)SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0的定義域?yàn)镾KIPIF1<0關(guān)于原點(diǎn)對稱，所以SKIPIF1<0是偶函數(shù)，不是奇函數(shù)，故D錯(cuò)誤．故選：AC．【典例3】（2023秋·高一課時(shí)練習(xí)）求下列函數(shù)的最小正周期．(1)SKIPIF1<0；(2)SKIPIF1<0.【答案】(1)SKIPIF1<0(2)SKIPIF1<0.【詳解】（1）因?yàn)镾KIPIF1<0，所以自變量SKIPIF1<0至少要增加到SKIPIF1<0，函數(shù)SKIPIF1<0，SKIPIF1<0的值才能重復(fù)出現(xiàn)，所以函數(shù)SKIPIF1<0的最小正周期是SKIPIF1<0.（2）因?yàn)镾KIPIF1<0的最小正周期為π，且函數(shù)SKIPIF1<0的圖象是將函數(shù)SKIPIF1<0的圖象在x軸下方的部分對折到SKIPIF1<0軸上方，并且保留在SKIPIF1<0軸上方圖象而得到的．由此可知所求函數(shù)的最小正周期為SKIPIF1<0.【變式1】（2023·全國·高三專題練習(xí)）下列函數(shù)中，最小正周期為π的函數(shù)是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】對于A，函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0，故A不符合題意；對于B，作出函數(shù)SKIPIF1<0的圖象，

由圖可知，函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0，故B符合題意；對于C，函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0，故C不符合題意；對于D，函數(shù)SKIPIF1<0，其圖象如圖，

由圖可知，函數(shù)SKIPIF1<0不是周期函數(shù)，故D不符合題意.故選：B.【變式2】（多選）（2023·全國·高一假期作業(yè)）下列函數(shù)中，是周期函數(shù)的是（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】ABC【詳解】對于A，SKIPIF1<0，SKIPIF1<0的最小正周期為SKIPIF1<0；對于B，SKIPIF1<0，SKIPIF1<0的最小正周期為SKIPIF1<0；對于C，SKIPIF1<0，SKIPIF1<0的最小正周期為SKIPIF1<0；對于D，∵SKIPIF1<0，∴函數(shù)圖象關(guān)于SKIPIF1<0軸對稱，不具有奇偶性，故錯(cuò)誤.故選：ABC【變式3】（2023·全國·高一課堂例題）求下列函數(shù)的最小正周期．(1)SKIPIF1<0；(2)SKIPIF1<0．【答案】(1)最小正周期為SKIPIF1<0．(2)最小正周期為SKIPIF1<0．【詳解】（1）∵SKIPIF1<0，∴SKIPIF1<0．又最小正周期SKIPIF1<0，∴函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0．（2）畫出函數(shù)SKIPIF1<0的圖象，如圖所示，

由圖象可知，函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0．題型02三角函數(shù)的奇偶性及其應(yīng)用【典例1】（2023春·新疆塔城·高一塔城地區(qū)第一高級中學(xué)校考階段練習(xí)）已知函數(shù)SKIPIF1<0，則SKIPIF1<0是SKIPIF1<0為奇函數(shù)的（

）A．充要條件 B．充分不必要條件C．必要不充分條件 D．既不充分也不必要條件【答案】B【詳解】SKIPIF1<0時(shí)，可得SKIPIF1<0，定義域?yàn)镽，此時(shí)SKIPIF1<0，故SKIPIF1<0為奇函數(shù)，故充分性成立，而當(dāng)SKIPIF1<0為奇函數(shù)時(shí)，得SKIPIF1<0，故SKIPIF1<0不一定為SKIPIF1<0，故必要性不成立，SKIPIF1<0是SKIPIF1<0為奇函數(shù)的充分不必要條件.故選：B【典例2】（多選）（2023秋·重慶沙坪壩·高三重慶一中?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0為偶函數(shù)，則SKIPIF1<0的取值可以為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】ACD【詳解】SKIPIF1<0為偶函數(shù)，因此SKIPIF1<0或SKIPIF1<0．所以SKIPIF1<0，故SKIPIF1<0正確，故選：SKIPIF1<0.【典例3】（2023秋·河北張家口·高三統(tǒng)考開學(xué)考試）已知函數(shù)SKIPIF1<0，SKIPIF1<0是奇函數(shù)，則SKIPIF1<0的值為．【答案】SKIPIF1<0【詳解】∵SKIPIF1<0為偶函數(shù)，所以SKIPIF1<0，SKIPIF1<0為奇函數(shù)，∴SKIPIF1<0，SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0．故答案為：SKIPIF1<0【典例4】（2023·貴州·校聯(lián)考模擬預(yù)測）若函數(shù)SKIPIF1<0為偶函數(shù)，則SKIPIF1<0的最小正值為.【答案】SKIPIF1<0/SKIPIF1<0【詳解】函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0