新高考數(shù)學(xué)二輪復(fù)習(xí)解答題培優(yōu)練習(xí)專題02 三角函數(shù)的圖象與性質(zhì)（五點(diǎn)法作圖）(典型題型歸類訓(xùn)練)（解析版）

專題02三角函數(shù)的圖象與性質(zhì)（五點(diǎn)法作圖）(典型題型歸類訓(xùn)練)目錄TOC\o"1-2"\h\u一、必備秘籍 1二、典型題型 1題型一：用五點(diǎn)法畫出一個周期內(nèi)的圖象，不限制具體范圍 1題型二：用五點(diǎn)法畫出具體某個范圍內(nèi)的圖象 4三、專項(xiàng)訓(xùn)練 6一、必備秘籍必備方法：SKIPIF1<0五點(diǎn)法步驟③SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0①SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0②SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0對于復(fù)合函數(shù)SKIPIF1<0，第一步：將SKIPIF1<0看做一個整體，用五點(diǎn)法作圖列表時，分別令SKIPIF1<0等于SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，對應(yīng)的SKIPIF1<0則取SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0。，（如上表中，先列出序號①②兩行）第二步：逆向解出SKIPIF1<0（如上表中，序號③行。）第三步：得到五個關(guān)鍵點(diǎn)為：SKIPIF1<0，SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0二、典型題型題型一：用五點(diǎn)法畫出一個周期內(nèi)的圖象，不限制具體范圍1．（2023·高一課時練習(xí)）已知函數(shù)SKIPIF1<0.(1)試用“五點(diǎn)法”畫出它的圖象；列表：SKIPIF1<0xy作圖：(2)求它的振幅、周期和初相.【答案】(1)答案見解析(2)振幅為SKIPIF1<0，周期SKIPIF1<0，初相為SKIPIF1<0【詳解】（1）列表如下：SKIPIF1<00SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0020SKIPIF1<00描點(diǎn)連線并向左右兩邊分別擴(kuò)展，得到如圖所示的函數(shù)圖象：（2）由SKIPIF1<0可知，振幅SKIPIF1<0，初相為SKIPIF1<0，最小正周期SKIPIF1<0.2．（2023春·云南昆明·高一校考階段練習(xí)）（1）利用“五點(diǎn)法”畫出函數(shù)SKIPIF1<0在長度為一個周期的閉區(qū)間的簡圖.列表:

SKIPIF1<0xy作圖:

（2）并說明該函數(shù)圖象可由SKIPIF1<0的圖象經(jīng)過怎么變換得到的.【答案】（1）答案見解析；（2）答案見解析.【詳解】（1）先列表，后描點(diǎn)并畫圖.SKIPIF1<00SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0xSKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0y010SKIPIF1<00

（2）把SKIPIF1<0的圖象上所有的點(diǎn)向左平移SKIPIF1<0個單位，得到SKIPIF1<0的圖象，再把所得圖象的點(diǎn)的橫坐標(biāo)伸長到原來的2倍（縱坐標(biāo)不變），得到SKIPIF1<0的圖象.題型二：用五點(diǎn)法畫出具體某個范圍內(nèi)的圖象1．（2023秋·江蘇連云港·高一統(tǒng)考期末）已知函數(shù)SKIPIF1<0．(1)用“五點(diǎn)法”畫出函數(shù)一個周期的簡圖；SKIPIF1<0xy(2)寫出函數(shù)在區(qū)間SKIPIF1<0上的單調(diào)遞增區(qū)間．【答案】(1)答案見解析(2)SKIPIF1<0，SKIPIF1<0【詳解】（1）用“五點(diǎn)法”畫出函數(shù)一個周期的簡圖，列表如下：SKIPIF1<00SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0xSKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0y030SKIPIF1<00函數(shù)一個周期的簡圖，如圖，（2）由SKIPIF1<0，解得SKIPIF1<0，當(dāng)SKIPIF1<0時，得SKIPIF1<0或SKIPIF1<0，所以函數(shù)在區(qū)間SKIPIF1<0上的單調(diào)遞增區(qū)間為SKIPIF1<0，SKIPIF1<0.2．（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，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：

三、專項(xiàng)訓(xùn)練1．（2023春·江西南昌·高一?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0(1)用“五點(diǎn)法”畫出函數(shù)SKIPIF1<0在一個周期內(nèi)的圖象；(2)直接寫出函數(shù)SKIPIF1<0的值域和最小正周期．列表：SKIPIF1<0SKIPIF1<0SKIPIF1<0作圖：

【答案】(1)答案見解析(2)值域SKIPIF1<0，最小正周期為SKIPIF1<0【詳解】（1）解：列表:SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<00SKIPIF1<0SKIPIF1<0圖象如圖所示：

（2）解：因?yàn)镾KIPIF1<0，則SKIPIF1<0，故函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，最小正周期為SKIPIF1<0.2．（2023春·廣西河池·高一校聯(lián)考階段練習(xí)）已知函數(shù)SKIPIF1<0，SKIPIF1<0．(1)在用“五點(diǎn)法”作函數(shù)SKIPIF1<0的圖象時，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；

(2)求函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間；(3)求函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最值．【答案】(1)表格見解析，圖象見解析(2)SKIPIF1<0，SKIPIF1<0(3)最大值為SKIPIF1<0，最小值為SKIPIF1<0【詳解】（1）SKIPIF1<0SKIPIF1<00SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0x0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0020－2SKIPIF1<0函數(shù)圖象如圖所示，

（2）令SKIPIF1<0，SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，所以函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，SKIPIF1<0（3）因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0．當(dāng)SKIPIF1<0，即SKIPIF1<0時，SKIPIF1<0；當(dāng)SKIPIF1<0，即SKIPIF1<0時，SKIPIF1<0．∴SKIPIF1<0的最大值為SKIPIF1<0，最小值為SKIPIF1<0．3．（2023春·四川資陽·高一四川省樂至中學(xué)校考階段練習(xí)）已知函數(shù)SKIPIF1<0．(1)請用“五點(diǎn)法”畫出函數(shù)SKIPIF1<0在一個周期上的圖象；(2)寫出SKIPIF1<0的單調(diào)遞減區(qū)間．【答案】(1)作圖見解析(2)減區(qū)間為SKIPIF1<0【詳解】（1）列表如下，xSKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<00SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0010SKIPIF1<00描點(diǎn)作圖即可（2）由SKIPIF1<0，SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0，（SKIPIF1<0）,或?qū)懗砷_區(qū)間.4．（2023秋·福建福州·高一福建省福州第一中學(xué)校考期末）已知函數(shù)SKIPIF1<0（其中SKIPIF1<0，SKIPIF1<0，SKIPIF1<0）的圖象過點(diǎn)SKIPIF1<0，且圖象上與點(diǎn)SKIPIF1<0最近的一個最低點(diǎn)的坐標(biāo)為SKIPIF1<0．(1)求函數(shù)SKIPIF1<0的解析式并用“五點(diǎn)法”作出函數(shù)在一個周期內(nèi)的圖象簡圖；(2)將函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個單位長度得到的函數(shù)SKIPIF1<0是偶函數(shù)，求SKIPIF1<0的最小值．【答案】(1)SKIPIF1<0，圖象見解析；(2)SKIPIF1<0【詳解】（1）由題意可得，SKIPIF1<0，且周期SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0又SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0（2）SKIPIF1<0，函數(shù)SKIPIF1<0是偶函數(shù)，則SKIPIF1<0，解得SKIPIF1<0又SKIPIF1<0，則當(dāng)SKIPIF1<0時，SKIPIF1<0的最小值為SKIPIF1<0．5．（2023秋·福建廈門·高一統(tǒng)考期末）某同學(xué)用“五點(diǎn)法”畫函數(shù)SKIPIF1<0在某一個周期內(nèi)的圖像時，列表并填入了部分?jǐn)?shù)據(jù)，如下表：SKIPIF1<00SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0xSKIPIF1<0SKIPIF1<0SKIPIF1<0020SKIPIF1<00(1)請將上表數(shù)據(jù)補(bǔ)充完整，并根據(jù)表格數(shù)據(jù)做出函數(shù)SKIPIF1<0在一個周期內(nèi)的圖像；(2)將SKIPIF1<0的圖形向右平移SKIPIF1<0個單位長度，得到SKIPIF1<0的圖像，若SKIPIF1<0的圖像關(guān)于y軸對稱，求SKIPIF1<0的最小值.【答案】(1)答案見解析(2)SKIPIF1<0【詳解】（1）SKIPIF1<00SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0xSKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0020SKIPIF1<00由表中數(shù)據(jù)可得，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0（2）由題意可得，SKIPIF1<0，因?yàn)镾KIPIF1<0的圖像關(guān)于y軸對稱，則SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0且SKIPIF1<0，所以當(dāng)SKIPIF1<0時，SKIPIF1<06．（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：7．（2023·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0．

用五點(diǎn)法畫出函數(shù)SKIPIF1<0在SKIPIF1<0上的大致圖像【答案】作圖見解析【詳解】由SKIPIF1<0，列表如下：SKIPIF1<0SKIPIF1<0SKIPIF1<00SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0020SKIPIF1<0函數(shù)圖像如圖：

8．（2023·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0．完成下面表格，并用“五點(diǎn)法”作函數(shù)SKIPIF1<0在SKIPIF1<0上的簡圖：

x0SKIPIF1<0πSKIPIF1<02πSKIPIF1<0【答案】填表見解析；作圖見解析【詳解】補(bǔ)充完整的表格如下：x0SKIPIF1<0πSKIPIF1<02πSKIPIF1<013531描點(diǎn)、連線得函數(shù)SKIPIF1<0的圖象如圖所示，

9．（2023·全國·高三專題練習(xí)）要得到函數(shù)SKIPIF1<0的圖象，可以從正弦函數(shù)或余弦函數(shù)圖象出發(fā)，通過圖象變換得到，也可以用“五點(diǎn)法”列表、描點(diǎn)、連線得到．(1)由SKIPIF1<0圖象變換得到函數(shù)SKIPIF1<0的圖象，寫出變換的步驟和函數(shù)；(2)用“五點(diǎn)法”畫出函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的簡圖．

【答案】(1)答案見解析(2)作圖見解析【詳解】（1）步驟1：把SKIPIF1<0圖象上所有點(diǎn)向左平移SKIPIF1<0個單位長度，得到函數(shù)SKIPIF1<0的圖象；步驟2：把SKIPIF1<0圖象上所有點(diǎn)的橫坐標(biāo)變?yōu)樵瓉淼腟KIPIF1<0倍（縱坐標(biāo)不變），得到函數(shù)SKIPIF1<0的圖象；步驟3：最后把函數(shù)SKIPIF1<0的圖象的縱坐標(biāo)變?yōu)樵瓉淼?倍（橫坐標(biāo)不變），得到函數(shù)SKIPIF1<0的圖象．或者步驟1：步驟1：把SKIPIF1<0圖象上所有點(diǎn)的橫坐標(biāo)變?yōu)樵瓉淼腟KIPIF1<0倍（縱坐標(biāo)不變），得到函數(shù)SKIPIF1<0的圖象；步驟2：把SKIPIF1<0圖象上所有點(diǎn)向左平移SKIPIF1<0個單位長度，得到函數(shù)SKIPIF1<0的圖象；步驟3：最后把函數(shù)SKIPIF1<0的圖象的縱坐標(biāo)變?yōu)樵瓉淼?倍（橫坐標(biāo)不變），得到函數(shù)SKIPIF1<0的圖象．（2）因?yàn)镾KIPIF1<0列表：SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0

10．（2023春·江西·高一統(tǒng)考期中）已知變換SKIPIF1<0：先縱坐標(biāo)不變，橫坐標(biāo)伸長為原來的2倍，再向左平移SKIPIF1<0個單位長度；變換SKIPIF1<0：先向左平移SKIPIF1<0個單位長度，再縱坐標(biāo)不變，橫坐標(biāo)伸長為原來的2倍.請從SKIPIF1<0，SKIPIF1<0兩種變換中選擇一種變換，將函數(shù)SKIPIF1<0的圖象變換得到函數(shù)SKIPIF1<0的圖象，并求解下列問題.(1)求SKIPIF1<0的解析式，并用五點(diǎn)法畫出函數(shù)SKIPIF1<0在一個周期內(nèi)的閉區(qū)間SKIPIF1<0上的圖象；(2)求函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間，并求SKIPIF1<0的最大值以及對應(yīng)SKIPIF1<0的取值集合.【答案】(1)SKIPIF1<0，圖象見解析(2)SKIPIF1<0，SKIPIF1<0；最大值為SKIPIF1<0，SKIPIF1<0【詳解】（1）選擇SKIPIF1<0，SKIPIF1<0兩種變換均得SKIPIF1<0，列表如下：SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0圖象如圖所示：（2）令SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，所以函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0，SKIPIF1<0.當(dāng)SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0時，SKIPIF1<0取得最大值SKIPIF1<0，此時對應(yīng)的SKIPIF1<0的取值集合為SKIPIF1<0.11．（2023春·遼寧本溪·高一?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0.(1)用五點(diǎn)法畫出函數(shù)SKIPIF1<0在SKIPIF1<0上的大致圖像，并寫出SKIPIF1<0的最小正周期；(2)解不等式SKIPIF1<0.【答案】(1)作圖見解析，SKIPIF1<0(2)SKIPIF1<0，SKIPIF1<0.【詳解】（1）由SKIPIF1<0，列表如下：SKIPIF1<0SKIPIF1<0SKIPIF1<00SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0020SKIPIF1<0函數(shù)圖像如圖：函數(shù)SKIPIF1<0的最小正周期SKIPIF1<