新高考數(shù)學(xué)一輪復(fù)習(xí)講義 第21講 函數(shù)y＝Asin(ωx＋φ)的圖象性質(zhì)及其應(yīng)用（含解析）

第21講函數(shù)y＝Asin(ωx＋φ)的圖象性質(zhì)及其應(yīng)用（精講）題型目錄一覽①函數(shù)y＝Asin(ωx＋φ)的單調(diào)性②函數(shù)y＝Asin(ωx＋φ)的奇偶性、對稱性③函數(shù)y＝Asin(ωx＋φ)的圖像變換④根據(jù)圖像求函數(shù)y＝Asin(ωx＋φ)的解析式⑤三角函數(shù)圖像與性質(zhì)的綜合應(yīng)用一、知識點梳理一、知識點梳理一、SKIPIF1<0的圖像與性質(zhì)（1）最小正周期：SKIPIF1<0.（2）定義域與值域：SKIPIF1<0的定義域為R，值域為[-A,A].（3）最值（以下SKIPIF1<0）SKIPIF1<0（4）單調(diào)性SKIPIF1<0（5）對稱軸與對稱中心.SKIPIF1<0正弦曲線的對稱軸是相應(yīng)函數(shù)取最大（小）值的位置，對稱中心是與SKIPIF1<0軸交點的位置.（6）平移與伸縮函數(shù)y＝sinx的圖象變換得到y(tǒng)＝Asin(ωx＋φ)(ω>0)的圖象的步驟注：每一個變換總是對變量SKIPIF1<0而言的，即圖像變換要看“變量SKIPIF1<0”發(fā)生多大變化，而不是“角SKIPIF1<0”變化多少.【常用結(jié)論】1.根據(jù)圖像求解析式一般步驟①根據(jù)最高最低點求出A②根據(jù)周期算出SKIPIF1<0,題目一般會提供周期的一部分③通過帶最高或最低點算出φ2.對稱與周期(1)y＝Asin(ωx＋φ)相鄰兩條對稱軸之間的距離是SKIPIF1<0；(2)y＝Asin(ωx＋φ)相鄰兩個對稱中心的距離是SKIPIF1<0；(3)y＝Asin(ωx＋φ)相鄰兩條對稱軸與對稱中心距離SKIPIF1<0；3.函數(shù)具有奇、偶性的充要條件(1)函數(shù)y＝Asin(ωx＋φ)(x∈R)是奇函數(shù)?φ＝kπ(k∈Z)；(2)函數(shù)y＝Asin(ωx＋φ)(x∈R)是偶函數(shù)?φ＝kπ＋eq\f(π,2)(k∈Z)；(3)函數(shù)y＝Acos(ωx＋φ)(x∈R)是奇函數(shù)?φ＝kπ＋eq\f(π,2)(k∈Z)；(4)函數(shù)y＝Acos(ωx＋φ)(x∈R)是偶函數(shù)?φ＝kπ(k∈Z)．二、題型分類精講二、題型分類精講題型一函數(shù)y＝Asin(ωx＋φ)的單調(diào)性【典例1】函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】化簡可得SKIPIF1<0，整體法求出函數(shù)的單調(diào)遞增區(qū)間，結(jié)合已知范圍，即可得出答案.【詳解】因為SKIPIF1<0.由SKIPIF1<0可得，SKIPIF1<0.當(dāng)SKIPIF1<0時，SKIPIF1<0，且SKIPIF1<0；當(dāng)SKIPIF1<0時，所以SKIPIF1<0，SKIPIF1<0.所以，函數(shù)在SKIPIF1<0上的單調(diào)遞增區(qū)間是SKIPIF1<0.故選：A.【題型訓(xùn)練】一、單選題1．（2023春·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0，則（

）A．SKIPIF1<0在SKIPIF1<0上單調(diào)遞減 B．SKIPIF1<0在SKIPIF1<0上單調(diào)遞增C．SKIPIF1<0在SKIPIF1<0上單調(diào)遞減 D．SKIPIF1<0在SKIPIF1<0上單調(diào)遞增【答案】C【分析】利用余弦函數(shù)的二倍角公式化簡得出SKIPIF1<0，利用余弦型函數(shù)的單調(diào)性逐項判斷可得出合適的選項.【詳解】因為SKIPIF1<0.對于A選項，當(dāng)SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，A錯；對于B選項，當(dāng)SKIPIF1<0時，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，故B錯；對于C選項，當(dāng)SKIPIF1<0時，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，C對；對于D選項，當(dāng)SKIPIF1<0時，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故D錯.故選：C.2．（2023·吉林通化·梅河口市第五中學(xué)校考模擬預(yù)測）下列區(qū)間中，函數(shù)SKIPIF1<0單調(diào)遞減的區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】化簡為SKIPIF1<0，再結(jié)合余弦函數(shù)的單調(diào)區(qū)間即可判斷各項.【詳解】SKIPIF1<0SKIPIF1<0對于A，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，A錯誤；對于B，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0沒有單調(diào)性，B錯誤；對于C，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0單調(diào)遞減，C正確；對于D，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0沒有單調(diào)性，D錯誤.故選：C3．（2023春·高三課時練習(xí)）函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】根據(jù)正弦函數(shù)的性質(zhì)、復(fù)合函數(shù)的單調(diào)性以及整體代換技巧進(jìn)行求解.【詳解】因為SKIPIF1<0，由SKIPIF1<0有：SKIPIF1<0，故B，C，D錯誤.故選：A.4．（2023春·四川綿陽·高三四川省綿陽江油中學(xué)?？计谥校┫铝胁坏仁匠闪⒌氖牵?/p>

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】分別判斷正弦、余弦、正切、的單調(diào)性，判斷選項A，B，C，再結(jié)和正弦余弦正切單調(diào)性及誘導(dǎo)公式找中間值比較即可判斷D選項.【詳解】A選項：因為SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，所以有SKIPIF1<0，故A錯誤；B選項：因為SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，所以有SKIPIF1<0，故B錯誤；C選項：SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，且SKIPIF1<0，所以有SKIPIF1<0，故C錯誤.D選項：因為SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，所以SKIPIF1<0，由SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，故選：D.二、多選題5．（2023·湖南邵陽·統(tǒng)考三模）已知函數(shù)SKIPIF1<0，則（

）A．SKIPIF1<0的最小正周期為SKIPIF1<0 B．SKIPIF1<0在SKIPIF1<0上單調(diào)遞增C．SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對稱 D．若SKIPIF1<0，則SKIPIF1<0的最小值為SKIPIF1<0【答案】BC【分析】利用整體思想，結(jié)合余弦函數(shù)的周期性、對稱性、單調(diào)性，可得答案.【詳解】對于A，由函數(shù)SKIPIF1<0，則SKIPIF1<0，故A錯誤；對于B，由SKIPIF1<0，則SKIPIF1<0，因為函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故B正確；對于C，由SKIPIF1<0，則SKIPIF1<0，因為函數(shù)SKIPIF1<0的對稱軸為直線SKIPIF1<0SKIPIF1<0，故C正確；對于D，由SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，因為函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，故SKIPIF1<0，故D錯誤.故選：BC.6．（2023·全國·高三專題練習(xí)）關(guān)于函數(shù)SKIPIF1<0，下列說法正確的是（

）A．函數(shù)SKIPIF1<0在SKIPIF1<0上最大值為SKIPIF1<0 B．函數(shù)SKIPIF1<0的圖象關(guān)于點SKIPIF1<0對稱C．函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增 D．函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0【答案】BD【分析】根據(jù)給定條件，利用正弦函數(shù)的圖象性質(zhì)，逐項分析判斷作答.【詳解】對于A，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0最大值為2，A錯誤；對于B，因為SKIPIF1<0，則函數(shù)SKIPIF1<0的圖象關(guān)于點SKIPIF1<0對稱，B正確；對于C，當(dāng)SKIPIF1<0時，SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上不單調(diào)，則SKIPIF1<0在SKIPIF1<0上不單調(diào)，C錯誤；對于D，函數(shù)SKIPIF1<0的最小正周期SKIPIF1<0，D正確.故選：BD.7．（2023·全國·高三專題練習(xí)）設(shè)函數(shù)SKIPIF1<0，則（

）A．SKIPIF1<0是偶函數(shù) B．SKIPIF1<0在SKIPIF1<0上單調(diào)遞減C．SKIPIF1<0的最大值為2 D．SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對稱【答案】ABD【分析】先利用輔助角公式化簡為SKIPIF1<0，再根據(jù)余弦函數(shù)的性質(zhì)即可一一判斷各選項.【詳解】SKIPIF1<0SKIPIF1<0，對于A，SKIPIF1<0，所以SKIPIF1<0是偶函數(shù)，故A正確；對于B，令SKIPIF1<0，SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故B正確；對于C，當(dāng)SKIPIF1<0，即SKIPIF1<0時，SKIPIF1<0有最大值SKIPIF1<0，故C錯誤；對于D，SKIPIF1<0，是最小值，故D正確．故選：ABD．三、填空題8．（2023春·遼寧鐵嶺·高三昌圖縣第一高級中學(xué)?？茧A段練習(xí)）函數(shù)SKIPIF1<0的遞增區(qū)間為___________.【答案】SKIPIF1<0【分析】根據(jù)余弦函數(shù)的單調(diào)性和單調(diào)區(qū)間的求法求解.【詳解】因為SKIPIF1<0，令SKIPIF1<0,解得SKIPIF1<0，所以遞增區(qū)間為SKIPIF1<0，故答案為:SKIPIF1<0.9．（2023春·廣東深圳·高三深圳市高級中學(xué)?？计谥校┖瘮?shù)SKIPIF1<0的單調(diào)遞減區(qū)間為______.【答案】SKIPIF1<0，SKIPIF1<0【分析】利用誘導(dǎo)公式及輔助角公式化簡，再結(jié)合正弦函數(shù)的性質(zhì)計算可得.【詳解】因為SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，所以函數(shù)的單調(diào)遞減區(qū)間為SKIPIF1<0，SKIPIF1<0.故答案為：SKIPIF1<0，SKIPIF1<010．（2023·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0，SKIPIF1<0.則SKIPIF1<0的最大值為___________.【答案】1【分析】利用整體法，結(jié)合余弦函數(shù)的單調(diào)性即可求出函數(shù)的最值.【詳解】因為SKIPIF1<0，所以SKIPIF1<0，又函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0，所以SKIPIF1<0的最大值為1．故答案為：1.11．（2023·全國·高三專題練習(xí)）函數(shù)SKIPIF1<0的單調(diào)增區(qū)間是______．【答案】SKIPIF1<0【分析】先根據(jù)二倍角的余弦將函數(shù)化簡，然后再利用余弦函數(shù)單調(diào)增區(qū)間即可求解.【詳解】函數(shù)SKIPIF1<0，令SKIPIF1<0，解得：SKIPIF1<0，SKIPIF1<0，所以函數(shù)的單調(diào)增區(qū)間為SKIPIF1<0，故答案為：SKIPIF1<0.12．（2023·湖北·統(tǒng)考模擬預(yù)測）請寫出一個滿足下列3個條件的函數(shù)SKIPIF1<0的表達(dá)式__________.①SKIPIF1<0；②在SKIPIF1<0上單調(diào)遞減；③SKIPIF1<0.【答案】SKIPIF1<0（答案不唯一）【分析】由①知SKIPIF1<0為偶函數(shù)，由③知SKIPIF1<0的周期為2，再結(jié)合SKIPIF1<0的單調(diào)區(qū)間即可求解.【詳解】由SKIPIF1<0得：SKIPIF1<0，又SKIPIF1<0SKIPIF1<0，SKIPIF1<0的一個正周期為2.故函數(shù)應(yīng)該是最小正周期為2的偶函數(shù)為：SKIPIF1<0（答案不唯一）.故答案為:SKIPIF1<0（答案不唯一）.四、解答題13．（2023春·高三單元測試）已知函數(shù)SKIPIF1<0，再從①SKIPIF1<0的最大值與最小值之和為0，②SKIPIF1<0這兩個條件中選擇一個作為已知條件.(1)求m的值；(2)求函數(shù)SKIPIF1<0在SKIPIF1<0上的單調(diào)遞增區(qū)間.注：如果選擇條件①和條件②分別解答，按第一個解答計分.【答案】(1)選①，SKIPIF1<0；選②，SKIPIF1<0(2)SKIPIF1<0【分析】（1）選①，利用三角恒等變換化簡SKIPIF1<0，求出SKIPIF1<0的最值，可求得SKIPIF1<0的值；選②，利用三角恒等變換化簡SKIPIF1<0，由SKIPIF1<0列式求解；（2）利用正弦函數(shù)的單調(diào)性求解.【詳解】（1）選①：SKIPIF1<0SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，由條件①可得SKIPIF1<0，解得SKIPIF1<0，此時SKIPIF1<0.選②：SKIPIF1<0SKIPIF1<0，由條件②，得SKIPIF1<0，解得SKIPIF1<0，此時SKIPIF1<0.（2）由（1）知，SKIPIF1<0.當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0，即SKIPIF1<0時，函數(shù)SKIPIF1<0單調(diào)遞增，故函數(shù)SKIPIF1<0在SKIPIF1<0上的單調(diào)遞增區(qū)間為SKIPIF1<0.14．（2023春·浙江·高三期中）已知函數(shù)SKIPIF1<0．(1)求函數(shù)SKIPIF1<0的周期及在SKIPIF1<0上的單調(diào)遞增區(qū)間：(2)若關(guān)于SKIPIF1<0的方程SKIPIF1<0在SKIPIF1<0上有兩個不同的實數(shù)根．求實數(shù)SKIPIF1<0的取值范圍．【答案】(1)SKIPIF1<0；遞增區(qū)間是SKIPIF1<0(2)SKIPIF1<0【分析】（1）根據(jù)二倍角公式和輔助角公式化簡函數(shù)后應(yīng)用周期公式和單調(diào)遞增區(qū)間求解即得；（2）根據(jù)方程有兩個不同的實數(shù)根求解值域即可.【詳解】（1）SKIPIF1<0SKIPIF1<0周期為SKIPIF1<0SKIPIF1<0所以遞增區(qū)間是SKIPIF1<0；（2）SKIPIF1<0因為方程SKIPIF1<0在SKIPIF1<0上有兩個不同的實數(shù)根,SKIPIF1<0SKIPIF1<0.題型二函數(shù)y＝Asin(ωx＋φ)的奇偶性、對稱性【典例1】使函數(shù)SKIPIF1<0為偶函數(shù)的最小正數(shù)φ＝（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】由函數(shù)SKIPIF1<0為偶函數(shù)，得SKIPIF1<0，由此能求出使函數(shù)SKIPIF1<0為偶函數(shù)的最小正數(shù)φ的值．【詳解】∵函數(shù)SKIPIF1<0為偶函數(shù)，∴SKIPIF1<0，∴使函數(shù)SKIPIF1<0為偶函數(shù)的最小正數(shù)SKIPIF1<0．故選：B【典例2】已知函數(shù)SKIPIF1<0，則下列說法正確的是（

）A．SKIPIF1<0的最小正周期是SKIPIF1<0 B．SKIPIF1<0的最大值是SKIPIF1<0C．SKIPIF1<0的圖象的一條對稱軸是直線SKIPIF1<0 D．SKIPIF1<0的圖象的一個對稱中心是SKIPIF1<0【答案】D【分析】先利用二倍角公式和輔助角公式化簡SKIPIF1<0，再根據(jù)正弦函數(shù)的圖象和性質(zhì)依次判斷即可.【詳解】因為SKIPIF1<0SKIPIF1<0，對于A，函數(shù)的最小正周期SKIPIF1<0，故A錯誤；對于B，因為SKIPIF1<0，所以函數(shù)的最大值為SKIPIF1<0，故B錯誤；對于C，令SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0的圖象的對稱軸為直線SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，故C錯誤；對于D，當(dāng)SKIPIF1<0時，SKIPIF1<0，將SKIPIF1<0代入函數(shù)解析式得SKIPIF1<0，故SKIPIF1<0為函數(shù)圖象的一個對稱中心，故D正確，故選：D【題型訓(xùn)練】一、單選題1．（2023·貴州貴陽·校聯(lián)考模擬預(yù)測）使函數(shù)SKIPIF1<0為偶函數(shù)，則SKIPIF1<0的一個值可以是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】根據(jù)題意，化簡SKIPIF1<0，由SKIPIF1<0為偶函數(shù)，求得SKIPIF1<0，結(jié)合選項，即可求解.【詳解】由SKIPIF1<0，因為SKIPIF1<0為偶函數(shù)，可得SKIPIF1<0，所以SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0.故選：A.2．（2023春·重慶沙坪壩·高三重慶八中?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0，若SKIPIF1<0是函數(shù)SKIPIF1<0圖象的一條對稱軸，則其圖象的一個對稱中心為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】根據(jù)余弦函數(shù)的對稱軸公式求解SKIPIF1<0，再由對稱中心公式求得結(jié)果.【詳解】因為SKIPIF1<0是函數(shù)SKIPIF1<0圖象的對稱軸，所以SKIPIF1<0，則SKIPIF1<0，又因為SKIPIF1<0，所以SKIPIF1<0.令SKIPIF1<0，得SKIPIF1<0，所以函數(shù)SKIPIF1<0圖象的一個對稱中心為SKIPIF1<0.故選：A.3．（2023·貴州遵義·統(tǒng)考三模）已知曲線SKIPIF1<0的一條對稱軸是SKIPIF1<0，則SKIPIF1<0的值可能為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】根據(jù)余弦函數(shù)的性質(zhì)先寫出其對稱軸的一般形式，然后檢查符合條件的選項.【詳解】由題意，SKIPIF1<0，即SKIPIF1<0，于是SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，經(jīng)檢驗，只有當(dāng)SKIPIF1<0時即SKIPIF1<0時符合.故選：C4．（2023·?？寄M預(yù)測）已知函數(shù)SKIPIF1<0的最小正周期為T，且SKIPIF1<0，若SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對稱，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】運用二倍角公式化簡SKIPIF1<0，結(jié)合SKIPIF1<0與SKIPIF1<0的對稱性求得SKIPIF1<0的值，進(jìn)而求得結(jié)果.【詳解】因為SKIPIF1<0，所以SKIPIF1<0．又因為SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，①又因為SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對稱，所以SKIPIF1<0，SKIPIF1<0．所以SKIPIF1<0，SKIPIF1<0，②所以由①②得SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0．故選：A.5．（2023·山東日照·三模）函數(shù)SKIPIF1<0的圖像向左平移SKIPIF1<0個單位得到函數(shù)SKIPIF1<0的圖像，若函數(shù)SKIPIF1<0是偶函數(shù)，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】根據(jù)圖像平移得函數(shù)SKIPIF1<0的解析式，由函數(shù)SKIPIF1<0是偶函數(shù)，解出SKIPIF1<0，可得SKIPIF1<0.【詳解】函數(shù)SKIPIF1<0的圖像向左平移SKIPIF1<0個單位，得SKIPIF1<0的圖像，又函數(shù)SKIPIF1<0是偶函數(shù)，則有SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0；所以SKIPIF1<0.故選：C．6．（2023·全國·統(tǒng)考高考真題）已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0單調(diào)遞增，直線SKIPIF1<0和SKIPIF1<0為函數(shù)SKIPIF1<0的圖像的兩條對稱軸，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】根據(jù)題意分別求出其周期，再根據(jù)其最小值求出初相，代入SKIPIF1<0即可得到答案.【詳解】因為SKIPIF1<0在區(qū)間SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0取得最小值，則SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，不妨取SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，故選：D.7．（2023·北京西城·統(tǒng)考二模）已知函數(shù)SKIPIF1<0．則“SKIPIF1<0”是“SKIPIF1<0為偶函數(shù)”的（

）A．充分而不必要條件 B．必要而不充分條件C．充分必要條件 D．既不充分也不必要條件【答案】C【分析】根據(jù)充分，必要條件的定義，結(jié)合三角函數(shù)變換，即可判斷選項.【詳解】當(dāng)SKIPIF1<0，即SKIPIF1<0則SKIPIF1<0，化簡為SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，為偶函數(shù)，當(dāng)SKIPIF1<0時，SKIPIF1<0，為偶函數(shù)，所以SKIPIF1<0，能推出函數(shù)SKIPIF1<0是偶函數(shù)反過來，若函數(shù)SKIPIF1<0是偶函數(shù)，則有SKIPIF1<0，所以“SKIPIF1<0”是“SKIPIF1<0為偶函數(shù)”的充分必要條件.故選：C8．（2023·山東·山東省實驗中學(xué)校考二模）將函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個單位長度后的函數(shù)圖象關(guān)于原點對稱，則實數(shù)SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】利用三角恒等變換化簡函數(shù)SKIPIF1<0的解析式，利用三角函數(shù)圖象變換求出平移后所得函數(shù)的解析式，利用正弦型函數(shù)的對稱性可求出SKIPIF1<0的表達(dá)式，即可求得SKIPIF1<0的最小值.【詳解】因為SKIPIF1<0，將函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個單位長度可得到函數(shù)SKIPIF1<0的圖象，由題意可知，函數(shù)SKIPIF1<0的圖象關(guān)于原點對稱，所以，SKIPIF1<0，所以，SKIPIF1<0，因為SKIPIF1<0，故當(dāng)SKIPIF1<0時，SKIPIF1<0取最小值SKIPIF1<0.故選：A.9．（2023·河南新鄉(xiāng)·統(tǒng)考三模）已知函數(shù)SKIPIF1<0圖象的一個對稱中心是SKIPIF1<0，點SKIPIF1<0在SKIPIF1<0的圖象上，下列說法錯誤的是（

）A．SKIPIF1<0 B．直線SKIPIF1<0是SKIPIF1<0圖象的一條對稱軸C．SKIPIF1<0在SKIPIF1<0上單調(diào)遞減 D．SKIPIF1<0是奇函數(shù)【答案】B【分析】由SKIPIF1<0可得SKIPIF1<0，由對稱中心SKIPIF1<0可求得SKIPIF1<0，從而知函數(shù)SKIPIF1<0的解析式，再根據(jù)余弦函數(shù)的圖象與性質(zhì)，逐一分析選項即可.【詳解】因為點SKIPIF1<0在SKIPIF1<0的圖象上，所以SKIPIF1<0.又SKIPIF1<0，所以SKIPIF1<0.因為SKIPIF1<0圖象的一個對稱中心是SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0.又SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，A正確.SKIPIF1<0，則直線SKIPIF1<0不是SKIPIF1<0圖象的一條對稱軸，B不正確.當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0單調(diào)遞減，C正確.SKIPIF1<0，是奇函數(shù)，D正確.故選：B.10．（2023·河南開封·校考模擬預(yù)測）已知函數(shù)SKIPIF1<0圖象的兩個相鄰零點的差的絕對值為SKIPIF1<0，則（

）A．SKIPIF1<0的最小正周期為SKIPIF1<0B．將SKIPIF1<0的圖象向左平移SKIPIF1<0個單位長度，得到函數(shù)SKIPIF1<0的圖象C．SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對稱D．SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0【答案】D【分析】根據(jù)函數(shù)SKIPIF1<0的性質(zhì)可得SKIPIF1<0的解析式，由正弦函數(shù)的圖象性質(zhì)逐項判斷即可.【詳解】因為函數(shù)SKIPIF1<0圖象的兩個相鄰零點的差的絕對值為SKIPIF1<0，所以最小正周期SKIPIF1<0，即SKIPIF1<0，故A不正確；則SKIPIF1<0，所以SKIPIF1<0的圖象向左平移SKIPIF1<0個單位長度，得到函數(shù)SKIPIF1<0，故B不正確；函數(shù)SKIPIF1<0的圖象的對稱軸方程滿足SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，故C不正確；函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間滿足SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，故D正確.故選：D.二、多選題11．（2023·全國·高三專題練習(xí)）設(shè)函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0為偶函數(shù)，則SKIPIF1<0的值可以是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】BC【分析】根據(jù)三角函數(shù)變換結(jié)合條件可得SKIPIF1<0，進(jìn)而SKIPIF1<0，即得.【詳解】因為SKIPIF1<0，所以SKIPIF1<0，又函數(shù)SKIPIF1<0為偶函數(shù)，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0的值可以是SKIPIF1<0，SKIPIF1<0.故選：BC.12．（2023·全國·高三專題練習(xí)）若函數(shù)SKIPIF1<0的圖象關(guān)于坐標(biāo)原點對稱，則SKIPIF1<0的可能取值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】AD【分析】化簡SKIPIF1<0，得SKIPIF1<0，由SKIPIF1<0，求出SKIPIF1<0，結(jié)合四個選項可得答案.【詳解】由已知，得SKIPIF1<0SKIPIF1<0．因為SKIPIF1<0的圖象關(guān)于坐標(biāo)原點對稱，所以SKIPIF1<0，解得SKIPIF1<0．結(jié)合選項知，A，D符合題意，B，C不符合題意．故選：AD．13．（2023·江蘇淮安·江蘇省鄭梁梅高級中學(xué)?？寄M預(yù)測）關(guān)于函數(shù)SKIPIF1<0，下列結(jié)論正確的是（

）A．函數(shù)SKIPIF1<0的周期為SKIPIF1<0 B．函數(shù)SKIPIF1<0圖象關(guān)于直線SKIPIF1<0對稱C．函數(shù)SKIPIF1<0在SKIPIF1<0上遞增 D．函數(shù)SKIPIF1<0的最大值為1【答案】BC【分析】由平方關(guān)系可得SKIPIF1<0，根據(jù)SKIPIF1<0、SKIPIF1<0是否成立判斷A、B；令SKIPIF1<0得SKIPIF1<0，并利用導(dǎo)數(shù)研究單調(diào)性、最值判斷C、D.【詳解】因為SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0的周期不是SKIPIF1<0，A錯誤；因為SKIPIF1<0，故SKIPIF1<0關(guān)于直線SKIPIF1<0對稱，B正確；當(dāng)SKIPIF1<0時，設(shè)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0、SKIPIF1<0上SKIPIF1<0，SKIPIF1<0上SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上遞增，而SKIPIF1<0，則SKIPIF1<0，而SKIPIF1<0在SKIPIF1<0上遞增，所以SKIPIF1<0在SKIPIF1<0上遞增，故C正確；由C分析知：SKIPIF1<0在SKIPIF1<0上遞減，在SKIPIF1<0上遞增，在SKIPIF1<0上遞減；且SKIPIF1<0，SKIPIF1<0，所以函數(shù)SKIPIF1<0的最大值為SKIPIF1<0，故D錯誤.故選：BC.14．（2023·遼寧大連·大連二十四中校考模擬預(yù)測）已知函數(shù)SKIPIF1<0的圖象關(guān)于點SKIPIF1<0中心對稱，則（

）A．SKIPIF1<0在區(qū)間SKIPIF1<0單調(diào)遞減 B．SKIPIF1<0在區(qū)間SKIPIF1<0有兩個極值點C．直線SKIPIF1<0是曲線SKIPIF1<0的對稱軸 D．直線SKIPIF1<0是曲線SKIPIF1<0的切線【答案】ACD【分析】根據(jù)函數(shù)SKIPIF1<0的圖象關(guān)于點SKIPIF1<0中心對稱，由SKIPIF1<0求得SKIPIF1<0后，再逐項求解判斷.【詳解】解：因為函數(shù)SKIPIF1<0的圖象關(guān)于點SKIPIF1<0中心對稱，所以SKIPIF1<0，則SKIPIF1<0，因為SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0在區(qū)間SKIPIF1<0單調(diào)遞減，故A正確；B.若SKIPIF1<0，則SKIPIF1<0，由函數(shù)的單調(diào)性知：SKIPIF1<0在區(qū)間SKIPIF1<0有一個極值點，故B錯誤；C.令SKIPIF1<0，得SKIPIF1<0，所以直線SKIPIF1<0是曲線SKIPIF1<0的對稱軸，故正確；D.由SKIPIF1<0，得SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以曲線SKIPIF1<0在SKIPIF1<0處的切線方程為SKIPIF1<0，故D正確；故選：ACD三、填空題15．（2023·全國·高三專題練習(xí)）設(shè)函數(shù)SKIPIF1<0的圖象關(guān)于點SKIPIF1<0成中心對稱，若SKIPIF1<0，則SKIPIF1<0______.【答案】SKIPIF1<0【分析】先根據(jù)對稱性列方程，再根據(jù)范圍確定結(jié)果【詳解】因為函數(shù)SKIPIF1<0的圖象關(guān)于點SKIPIF1<0成中心對稱，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0因為SKIPIF1<0，所以SKIPIF1<0時，SKIPIF1<0.故答案為：SKIPIF1<016．（2023·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0圖象關(guān)于直線SKIPIF1<0對稱，則函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上零點的個數(shù)為_______.【答案】3【分析】根據(jù)余弦函數(shù)的對稱軸方程，結(jié)合圖象關(guān)于直線SKIPIF1<0對稱可得SKIPIF1<0，再求解SKIPIF1<0零點的表達(dá)式，分析在區(qū)間SKIPIF1<0內(nèi)的解的個數(shù)即可【詳解】SKIPIF1<0函數(shù)SKIPIF1<0圖象關(guān)于直線SKIPIF1<0對稱，SKIPIF1<0，(SKIPIF1<0的對稱軸是SKIPIF1<0)SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0知，SKIPIF1<0時，SKIPIF1<0，故SKIPIF1<0，令SKIPIF1<0得SKIPIF1<0，SKIPIF1<0．因為SKIPIF1<0，所以SKIPIF1<0時，SKIPIF1<0滿足條件，故零點有三個．故答案為：317．（2023·吉林通化·梅河口市第五中學(xué)校考模擬預(yù)測）某函數(shù)SKIPIF1<0滿足以下三個條件：①SKIPIF1<0是偶函數(shù)；②SKIPIF1<0；③SKIPIF1<0的最大值為4．請寫出一個滿足上述條件的函數(shù)SKIPIF1<0的解析式______．【答案】SKIPIF1<0（答案不唯一）【分析】根據(jù)所給條件分析函數(shù)的性質(zhì)，結(jié)合所學(xué)函數(shù)可得.【詳解】因為SKIPIF1<0是偶函數(shù)，所以SKIPIF1<0的圖象關(guān)于y軸對稱，因為SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0所以SKIPIF1<0的圖象關(guān)于點SKIPIF1<0對稱，所以4為SKIPIF1<0的一個周期，又SKIPIF1<0的最大值為4，所以SKIPIF1<0滿足條件.故答案為：SKIPIF1<0（答案不唯一）18．（2023·陜西咸陽·統(tǒng)考模擬預(yù)測）已知函數(shù)SKIPIF1<0的最小正周期為π，對于下列說法:①SKIPIF1<0;

②SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，（SKIPIF1<0）;③將SKIPIF1<0的圖象向左平移SKIPIF1<0個單位長度后所得圖象關(guān)于y軸對稱;④SKIPIF1<0.其中正確的序號是__________.【答案】①③④【分析】先化簡為SKIPIF1<0，再根據(jù)正弦型函數(shù)的性質(zhì)對各項一一判斷即可.【詳解】SKIPIF1<0對于①：因為SKIPIF1<0，∴SKIPIF1<0，故①正確；對于②：SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0SKIPIF1<0，SKIPIF1<0，所以單調(diào)遞增區(qū)間為SKIPIF1<0，SKIPIF1<0，故②錯誤；對于③：將SKIPIF1<0圖像向左平移SKIPIF1<0個單位得到SKIPIF1<0SKIPIF1<0，關(guān)于y軸對稱，故③正確；對于④：SKIPIF1<0SKIPIF1<0，所以④正確；故答案為：①③④．四、解答題19．（2023春·湖南長沙·高三雅禮中學(xué)?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0，SKIPIF1<0．（1）設(shè)SKIPIF1<0是函數(shù)SKIPIF1<0圖象的一條對稱軸，求SKIPIF1<0的值．（2）求函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間．【答案】（1）SKIPIF1<0或SKIPIF1<0,（2）SKIPIF1<0（SKIPIF1<0）．【詳解】試題分析：（Ⅰ）先利用倍角公式把函數(shù)解析式化為SKIPIF1<0，再由對稱軸的計算方法得SKIPIF1<0SKIPIF1<0，即SKIPIF1<0（SKIPIF1<0）．所以SKIPIF1<0．最后分SKIPIF1<0為奇數(shù)或偶數(shù)兩種情況求出SKIPIF1<0的值為SKIPIF1<0或SKIPIF1<0．（Ⅱ）先求出SKIPIF1<0，再由SKIPIF1<0，得函數(shù)的單調(diào)遞增區(qū)間為SKIPIF1<0（SKIPIF1<0）試題解析：（I）由題設(shè)知SKIPIF1<0．因為SKIPIF1<0是函數(shù)SKIPIF1<0圖象的一條對稱軸，所以SKIPIF1<0SKIPIF1<0，即SKIPIF1<0（SKIPIF1<0）．所以SKIPIF1<0．當(dāng)SKIPIF1<0為偶數(shù)時，SKIPIF1<0，當(dāng)SKIPIF1<0為奇數(shù)時，．（II）SKIPIF1<0SKIPIF1<0SKIPIF1<0．當(dāng)SKIPIF1<0，即SKIPIF1<0（SKIPIF1<0）時，函數(shù)SKIPIF1<0是增函數(shù)，故函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間是SKIPIF1<0（SKIPIF1<0）．考點?：輔助角公式的應(yīng)用?對稱軸的求法?求三角函數(shù)單調(diào)性區(qū)間20．（2023·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0，SKIPIF1<0．(1)求SKIPIF1<0的對稱軸；(2)若SKIPIF1<0在SKIPIF1<0內(nèi)的最大值與最小值之和為SKIPIF1<0，求a．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【分析】（1）先化簡為SKIPIF1<0，令SKIPIF1<0求解即可；（2）求得SKIPIF1<0的一個單調(diào)遞增區(qū)間為SKIPIF1<0，一個單調(diào)遞減區(qū)間為SKIPIF1<0，從而分三種情況討論：SKIPIF1<0、SKIPIF1<0、SKIPIF1<0結(jié)合單調(diào)性可求得最值，從而可求解.【詳解】（1）SKIPIF1<0SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0的對稱軸為SKIPIF1<0．（2）由（1）得SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，則SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，令k＝0，得SKIPIF1<0的一個單調(diào)遞增區(qū)間為SKIPIF1<0．令SKIPIF1<0，解得SKIPIF1<0，則SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0，令k＝－1，得SKIPIF1<0的一個單調(diào)遞減區(qū)間為SKIPIF1<0，又SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0內(nèi)單調(diào)遞減，且SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0，因為SKIPIF1<0，即SKIPIF1<0，所以不滿足SKIPIF1<0，舍去；當(dāng)SKIPIF1<0，此時SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，且SKIPIF1<0在SKIPIF1<0內(nèi)單調(diào)遞增，所以SKIPIF1<0；當(dāng)SKIPIF1<0，此時SKIPIF1<0，SKIPIF1<0