新高考數(shù)學(xué)二輪復(fù)習(xí)培優(yōu)講義08 三角函數(shù)圖像與性質(zhì)（含解析）

講義09講：三角函數(shù)圖像與性質(zhì)【考點講義】1．正弦、余弦、正切函數(shù)的圖象與性質(zhì)(下表中k∈Z)函數(shù)y＝sinxy＝cosxy＝tanx圖象定義域RReq\b\lc\{\rc\}(\a\vs4\al\co1(x\b\lc\|\rc\(\a\vs4\al\co1(x≠kπ＋\f(π,2)))))值域[－1,1][－1,1]R周期性2π2ππ奇偶性奇函數(shù)偶函數(shù)奇函數(shù)遞增區(qū)間eq\b\lc\[\rc\](\a\vs4\al\co1(2kπ－\f(π,2)，2kπ＋\f(π,2)))[2kπ－π，2kπ]SKIPIF1<0遞減區(qū)間eq\b\lc\[\rc\](\a\vs4\al\co1(2kπ＋\f(π,2)，2kπ＋\f(3π,2)))[2kπ，2kπ＋π]對稱中心(kπ，0)eq\b\lc\(\rc\)(\a\vs4\al\co1(kπ＋\f(π,2)，0))eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(kπ,2)，0))對稱軸方程x＝kπ＋eq\f(π,2)x＝kπ2．簡諧運動的有關(guān)概念y＝Asin(ωx＋φ)(A>0，ω>0)，x≥0振幅周期頻率相位初相AT＝eq\f(2π,ω)f＝eq\f(1,T)＝eq\f(ω,2π)ωx＋φφ3．用“五點法”畫y＝Asin(ωx＋φ)(A>0，ω>0)一個周期內(nèi)的簡圖時，要找五個特征點xeq\f(0－φ,ω)eq\f(\f(π,2)－φ,ω)eq\f(π－φ,ω)eq\f(\f(3π,2)－φ,ω)eq\f(2π－φ,ω)ωx＋φ0eq\f(π,2)πeq\f(3π,2)2πy＝Asin(ωx＋φ)0A0－A04．函數(shù)y＝sinx的圖象經(jīng)變換得到y(tǒng)＝Asin(ωx＋φ)(A>0，ω>0)的圖象的兩種途徑【方法技巧】1．求解三角函數(shù)的值域(最值)常見到以下幾種類型(1)形如y＝asinx＋bcosx＋c的三角函數(shù)化為y＝Asin(ωx＋φ)＋c的形式，再求值域(最值)．求三角函數(shù)取最值時相應(yīng)自變量x的集合時，要注意考慮三角函數(shù)的周期性．(2)形如y＝asin2x＋bsinx＋c(或y＝acos2x＋bcosx＋c)，x∈D的函數(shù)的值域或最值時，通過換元，令t＝sinx(或cosx)，將原函數(shù)轉(zhuǎn)化為關(guān)于t的二次函數(shù)，利用配方法求值域或最值即可．求解過程中要注意t＝sinx(或cosx)的有界性．(3)形如y＝asinxcosx＋b(sinx±cosx)＋c的三角函數(shù)，可先設(shè)t＝sinx±cosx，化為關(guān)于t的二次函數(shù)求值域(最值)．2．求三角函數(shù)周期的方法(1)定義法：即利用周期函數(shù)的定義求解．(2)公式法：對形如y＝Asin(ωx＋φ)或y＝Acos(ωx＋φ)(A，ω，φ是常數(shù)，A≠0，ω≠0)的函數(shù)，T＝eq\f(2π,|ω|)；對形如y＝Atan(ωx＋φ)(A，ω，φ是常數(shù)，A≠0，ω≠0)的函數(shù)，SKIPIF1<0．形如y＝|Asinωx|(或y＝|Acosωx|)的函數(shù)的周期T＝eq\f(π,|ω|).(3)觀察法：即通過觀察函數(shù)圖象求其周期．3．三角函數(shù)周期性與奇偶性、對稱性的解題策略(1)探求三角函數(shù)的周期，常用方法是公式法，即將函數(shù)化為y＝Asin(ωx＋φ)或y＝Acos(ωx＋φ)的形式，再利用公式求解．(2)判斷函數(shù)y＝Asin(ωx＋φ)或y＝Acos(ωx＋φ)是否具備奇偶性，關(guān)鍵是看它能否通過誘導(dǎo)公式轉(zhuǎn)化為y＝Asinωx(Aω≠0)或y＝Acosωx(Aω≠0)其中的一個．(3)對于可化為f(x)＝Asin(ωx＋φ)(或f(x)＝Acos(ωx＋φ))形式的函數(shù)，如果求f(x)的對稱軸，只需令ωx＋φ＝eq\f(π,2)＋kπ(k∈Z)(或令ωx＋φ＝kπ(k∈Z))，求x即可；如果求f(x)的對稱中心的橫坐標(biāo)，只需令ωx＋φ＝kπ(k∈Z)(或令ωx＋φ＝eq\f(π,2)＋kπ(k∈Z))，求x即可．(4)對于可化為f(x)＝Atan(ωx＋φ)形式的函數(shù)，如果求f(x)的對稱中心的橫坐標(biāo)，只需令ωx＋φ＝eq\f(kπ,2)(k∈Z)，求x即可．4．求函數(shù)y＝tan(ωx＋φ)的單調(diào)區(qū)間的方法y＝tan(ωx＋φ)(ω>0)的單調(diào)區(qū)間的求法是把ωx＋φ看成一個整體，解－eq\f(π,2)＋kπ<ωx＋φ<eq\f(π,2)＋kπ，k∈Z即可．當(dāng)ω<0時，先用誘導(dǎo)公式把ω化為正值再求單調(diào)區(qū)間．5．(1)由函數(shù)y＝sinx的圖象通過變換得到y(tǒng)＝Asin(ωx＋φ)的圖象有兩條途徑：“先平移后伸縮”與“先伸縮后平移”．(2)當(dāng)x的系數(shù)不為1時，特別注意先提取系數(shù)，再加減．(3)橫向伸縮變換，只變ω，而φ不發(fā)生變化.6．若設(shè)所求解析式為y＝Asin(ωx＋φ)，則在觀察函數(shù)圖象的基礎(chǔ)上，可按以下規(guī)律來確定A，ω，φ.(1)由函數(shù)圖象上的最大值、最小值來確定|A|.(2)由函數(shù)圖象與x軸的交點確定T，由T＝eq\f(2π,ω)，確定ω.(3)y＝Asin(ωx＋φ)中φ的確定方法①代入法：把圖象上的一個已知點代入(此時A，ω已知)或代入圖象與x軸的交點求解(此時要注意交點在上升區(qū)間上還是在下降區(qū)間上)，或把圖象的最高點或最低點代入．②五點對應(yīng)法：確定φ值時，往往以尋找“五點法”中的特殊點作為突破口.“五點”的ωx＋φ的值具體如下：“第一點”(即圖象上升時與x軸的交點)為ωx＋φ＝0；“第二點”(即圖象的“峰點”)為ωx＋φ＝eq\f(π,2)；“第三點”(即圖象下降時與x軸的交點)為ωx＋φ＝π；“第四點”(即圖象的“谷點”)為ωx＋φ＝eq\f(3π,2)；“第五點”為ωx＋φ＝2π.【核心題型】題型一：整體代入法求三角函數(shù)的單調(diào)區(qū)間、對稱軸和對稱中心1．（2023春·河北·高二統(tǒng)考學(xué)業(yè)考試）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】B【分析】根據(jù)輔助角公式,化簡三角函數(shù)式,結(jié)合正弦函數(shù)的圖像與性質(zhì),即可求得其單調(diào)遞增區(qū)間.【詳解】由輔助角公式,化簡三角函數(shù)式SKIPIF1<0可得SKIPIF1<0SKIPIF1<0由正弦函數(shù)的圖像與性質(zhì)可知其單調(diào)遞增區(qū)間滿足SKIPIF1<0解得SKIPIF1<0即單調(diào)遞增區(qū)間為SKIPIF1<0,SKIPIF1<0故選：B2．（2022秋·安徽·高三校聯(lián)考開學(xué)考試）函數(shù)SKIPIF1<0的圖象的一個對稱中心為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】根據(jù)正切型函數(shù)的對稱中心為SKIPIF1<0SKIPIF1<0，求解即可.【詳解】由SKIPIF1<0，可得SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0為SKIPIF1<0圖象的一個對稱中心，故選：D3．（2022秋·廣西欽州·高三?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0且SKIPIF1<0，則函數(shù)SKIPIF1<0的圖象的一條對稱軸是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】首先利用兩角差的余弦公式及同角三角函數(shù)的基本關(guān)系求出SKIPIF1<0的取值，再根據(jù)正弦函數(shù)的性質(zhì)計算可得.【詳解】解：因為SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，所以函數(shù)的對稱軸為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故函數(shù)的一條對稱軸為SKIPIF1<0.故選：A題型二：代入檢驗法判斷三角函數(shù)的單調(diào)區(qū)間、對稱軸和對稱中心4．（2023春·河南·高三商丘市回民中學(xué)校聯(lián)考開學(xué)考試）將函數(shù)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，下列說法正確的是（

）．A．SKIPIF1<0為奇函數(shù) B．SKIPIF1<0在SKIPIF1<0上單調(diào)遞減C．SKIPIF1<0在SKIPIF1<0上的值域為SKIPIF1<0 D．點SKIPIF1<0是SKIPIF1<0圖象的一個對稱中心【答案】D【分析】由題意利用函數(shù)SKIPIF1<0的圖象變換規(guī)律，正弦函數(shù)的圖象和性質(zhì)，即可求解.【詳解】由題知，SKIPIF1<0，所以A錯誤；因為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上先增后減，所以B錯誤；因為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以C錯誤；因為SKIPIF1<0，所以點SKIPIF1<0是SKIPIF1<0圖象的一個對稱中心，所以D正確．故選：D.5．（2022秋·天津河西·高三天津市海河中學(xué)?？计谀┮阎瘮?shù)SKIPIF1<0，給出以下四個命題：①SKIPIF1<0的最小正周期為SKIPIF1<0；②SKIPIF1<0在SKIPIF1<0上的值域為SKIPIF1<0；③SKIPIF1<0的圖像關(guān)于點SKIPIF1<0中心對稱；④SKIPIF1<0的圖像關(guān)于直線SKIPIF1<0對稱．其中正確命題的個數(shù)是（

）A．1 B．2 C．3 D．4【答案】B【分析】由題知SKIPIF1<0，進而結(jié)合三角函數(shù)性質(zhì)依次討論各選項即可.【詳解】解：SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0的最小正周期為SKIPIF1<0，①正確；當(dāng)SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，故②錯誤；當(dāng)SKIPIF1<0時，SKIPIF1<0，故SKIPIF1<0的圖像關(guān)于SKIPIF1<0對稱，故③錯誤；當(dāng)SKIPIF1<0時，SKIPIF1<0，故SKIPIF1<0的圖像關(guān)于SKIPIF1<0對稱，故④正確.故正確命題的個數(shù)是2個.故選：B6．（多選）（2022秋·山西晉中·高三校聯(lián)考階段練習(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上單調(diào)遞減B．函數(shù)SKIPIF1<0的圖像關(guān)于SKIPIF1<0中心對稱C．函數(shù)SKIPIF1<0的對稱軸方程為SKIPIF1<0，SKIPIF1<0D．將SKIPIF1<0的圖像向右平移SKIPIF1<0個單位長度后，可以得到SKIPIF1<0的圖像【答案】ACD【分析】根據(jù)函數(shù)的解析式分別應(yīng)用對稱軸,對稱中心,單調(diào)性及平移逐個判斷選項即可.【詳解】對于A:SKIPIF1<0,SKIPIF1<0,所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減,故A正確;對于B:令SKIPIF1<0,則SKIPIF1<0,故函數(shù)SKIPIF1<0的對稱中心為SKIPIF1<0,故B錯誤;對于C:令SKIPIF1<0,則SKIPIF1<0,故函數(shù)SKIPIF1<0的對稱軸為SKIPIF1<0,故C正確;對于D:將SKIPIF1<0的圖像向右平移SKIPIF1<0個單位長度可得SKIPIF1<0,故D正確.故選:ACD.題型三：圖像法求三角函數(shù)最值或值域7．（2021春·上海普陀·高一曹楊二中?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0，則SKIPIF1<0的最小值是_________.【答案】SKIPIF1<0【分析】先將函數(shù)SKIPIF1<0轉(zhuǎn)化成正弦函數(shù)的形式，然后結(jié)合正弦函數(shù)的圖象判斷出函數(shù)SKIPIF1<0的最大值和最小值，從而得出結(jié)果．【詳解】解：由題意可得SKIPIF1<0，其中SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0．因為SKIPIF1<0,所以SKIPIF1<0.所以SKIPIF1<0的最小值是SKIPIF1<0.故答案為：SKIPIF1<08．（2022秋·北京·高三北京市八一中學(xué)?？茧A段練習(xí)）定義運算SKIPIF1<0例如，SKIPIF1<0，則函數(shù)SKIPIF1<0的值域為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】先閱讀理解題意，可得SKIPIF1<0，再作出函數(shù)SKIPIF1<0在一個周期內(nèi)的圖象，再由圖像觀察值域即可.【詳解】根據(jù)題設(shè)中的新定義，得SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，當(dāng)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0，SKIPIF1<0時，SKIPIF1<0，所以函數(shù)SKIPIF1<0為周期函數(shù)，周期為SKIPIF1<0，作出函數(shù)SKIPIF1<0在一個周期內(nèi)的圖象（實線部分），觀察圖象，可知函數(shù)SKIPIF1<0的值域為SKIPIF1<0，故選:D.9．（2022·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0，求SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值和最小值.【答案】最大值為SKIPIF1<0+1，最小值為0.【分析】利用三角函數(shù)恒等變換轉(zhuǎn)化為正弦型三角函數(shù)，根據(jù)自變量取值范圍，利用正弦函數(shù)圖象與性質(zhì)求最值即可得解.【詳解】因為SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0.由正弦函數(shù)SKIPIF1<0在SKIPIF1<0上的圖象與性質(zhì)知，當(dāng)SKIPIF1<0，即SKIPIF1<0時，SKIPIF1<0取最大值SKIPIF1<0；當(dāng)SKIPIF1<0，即SKIPIF1<0時，SKIPIF1<0取最小值0.綜上，SKIPIF1<0在SKIPIF1<0上的最大值為SKIPIF1<0，最小值為0.題型四：換元法求三角函數(shù)最值或值域10．（2023·全國·高三專題練習(xí)）函數(shù)SKIPIF1<0的最大值為（

）A．1 B．SKIPIF1<0 C．SKIPIF1<0 D．3【答案】C【分析】利用換元法，令SKIPIF1<0，則原函數(shù)可化為SKIPIF1<0，再根據(jù)二次函數(shù)的性質(zhì)可求得其最大值【詳解】SKIPIF1<0，令SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，所以原函數(shù)可化為SKIPIF1<0，SKIPIF1<0，對稱軸為SKIPIF1<0，所以當(dāng)SKIPIF1<0時，SKIPIF1<0取得最大值，所以函數(shù)的最大值為SKIPIF1<0，即SKIPIF1<0的最大值為SKIPIF1<0，故選：C11．（2022·全國·高三專題練習(xí)）函數(shù)SKIPIF1<0的值域為________．【答案】SKIPIF1<0【分析】令SKIPIF1<0，函數(shù)化為SKIPIF1<0，利用二次函數(shù)的性質(zhì)即可求出.【詳解】由于SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，于是函數(shù)化為SKIPIF1<0，而SKIPIF1<0SKIPIF1<0，所以當(dāng)SKIPIF1<0時，函數(shù)取最大值1，當(dāng)SKIPIF1<0時，函數(shù)取最小值SKIPIF1<0，故值域為SKIPIF1<0.故答案為：SKIPIF1<0.12．（2022·全國·高三專題練習(xí)）函數(shù)y＝cos2x－sinx的值域是__________________【答案】SKIPIF1<0【分析】將原函數(shù)轉(zhuǎn)換成同名三角函數(shù)即可.【詳解】SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時取最大值SKIPIF1<0，當(dāng)SKIPIF1<0時，取最小值SKIPIF1<0；故答案為：SKIPIF1<0.題型五：利用三角函數(shù)單調(diào)性、奇偶性、周期性和對稱性求參數(shù)的值13．（2023秋·廣西南寧·高三南寧二中校考期末）已知函數(shù)SKIPIF1<0的兩個相鄰的對稱中心的間距為SKIPIF1<0，現(xiàn)SKIPIF1<0的圖象向左平移SKIPIF1<0個單位后得到一個奇函數(shù)，則SKIPIF1<0的一個可能取值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．0 D．SKIPIF1<0【答案】D【分析】根據(jù)給定條件，求出函數(shù)的周期，進而求出SKIPIF1<0，再利用給定變換及奇函數(shù)求出SKIPIF1<0作答.【詳解】由于函數(shù)SKIPIF1<0的兩條相鄰的對稱軸的間距為SKIPIF1<0，該函數(shù)的最小正周期為π，即有SKIPIF1<0，則SKIPIF1<0，將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個單位后，得到函數(shù)SKIPIF1<0，而函數(shù)SKIPIF1<0為奇函數(shù)，則SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，D正確，不存在整數(shù)k使得選項A，B，C成立.故選：D14．（2023·全國·校聯(lián)考模擬預(yù)測）已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】由題知SKIPIF1<0，進而根據(jù)題意得SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，進而得SKIPIF1<0或SKIPIF1<0，再解不等式即可得答案.【詳解】解：SKIPIF1<0，因為SKIPIF1<0，所以SKIPIF1<0因為函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，即SKIPIF1<0.因為SKIPIF1<0，所以，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增等價于SKIPIF1<0或SKIPIF1<0，所以，解不等式得SKIPIF1<0或SKIPIF1<0，所以，SKIPIF1<0的取值范圍是SKIPIF1<0.故選：D15．（多選）（2023秋·黑龍江哈爾濱·高三哈爾濱市第六中學(xué)校?？计谀┮阎瘮?shù)SKIPIF1<0SKIPIF1<0的最小正周期為SKIPIF1<0，函數(shù)SKIPIF1<0圖象關(guān)于直線SKIPIF1<0對稱，且滿足函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】CD【分析】由周期為SKIPIF1<0，可得SKIPIF1<0.根據(jù)對稱軸SKIPIF1<0以及正弦函數(shù)的對稱性可得SKIPIF1<0或SKIPIF1<0.分別將SKIPIF1<0或SKIPIF1<0代入SKIPIF1<0，得出范圍，根據(jù)正弦函數(shù)的單調(diào)性即可得出SKIPIF1<0的值.【詳解】由已知函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0，可得SKIPIF1<0.又函數(shù)SKIPIF1<0圖象關(guān)于直線SKIPIF1<0對稱，所以有SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0.當(dāng)SKIPIF1<0時，SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，因為函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，滿足題意；當(dāng)SKIPIF1<0時，SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，因為函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，不滿足題意.所以，SKIPIF1<0.故選：CD.題型六：五點法求三角函數(shù)解析式16．（2020·全國·統(tǒng)考高考真題）設(shè)函數(shù)SKIPIF1<0在SKIPIF1<0的圖像大致如下圖，則f(x)的最小正周期為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】由圖可得：函數(shù)圖象過點SKIPIF1<0，即可得到SKIPIF1<0，結(jié)合SKIPIF1<0是函數(shù)SKIPIF1<0圖象與SKIPIF1<0軸負半軸的第一個交點即可得到SKIPIF1<0，即可求得SKIPIF1<0，再利用三角函數(shù)周期公式即可得解.【詳解】由圖可得：函數(shù)圖象過點SKIPIF1<0，將它代入函數(shù)SKIPIF1<0可得：SKIPIF1<0又SKIPIF1<0是函數(shù)SKIPIF1<0圖象與SKIPIF1<0軸負半軸的第一個交點，所以SKIPIF1<0，解得：SKIPIF1<0所以函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0故選：C【點睛】本題主要考查了三角函數(shù)的性質(zhì)及轉(zhuǎn)化能力，還考查了三角函數(shù)周期公式，屬于中檔題.17．（2022·山西運城·校聯(lián)考模擬預(yù)測）設(shè)函數(shù)SKIPIF1<0（SKIPIF1<0，SKIPIF1<0）的部分圖象如圖所示.若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】由圖像可求出函數(shù)的解析式SKIPIF1<0，由已知結(jié)合誘導(dǎo)公式知SKIPIF1<0，再利用二倍角公式可求解.【詳解】由圖可知，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0又SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0故選：A18．（多選）（2022·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0的部分圖象如圖所示，則（

）A．SKIPIF1<0的最小正周期為SKIPIF1<0B．SKIPIF1<0為偶函數(shù)C．SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)的最小值為1D．SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對稱【答案】AC【分析】由圖知，SKIPIF1<0的最小正周期為SKIPIF1<0，結(jié)論A正確；求出SKIPIF1<0，從而SKIPIF1<0不是偶函數(shù)，結(jié)論B錯誤；因為SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)的最小值為1，結(jié)論C正確；因為SKIPIF1<0為SKIPIF1<0的零點，不是最值點，結(jié)論D錯誤.【詳解】解：由圖知，SKIPIF1<0的最小正周期為SKIPIF1<0，結(jié)論A正確；因為SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0．因為SKIPIF1<0為SKIPIF1<0在SKIPIF1<0內(nèi)的最小零點，則SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，從而SKIPIF1<0不是偶函數(shù)，結(jié)論B錯誤；因為SKIPIF1<0，SKIPIF1<0，結(jié)合圖像可得SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)的最小值為1，結(jié)論C正確；因為SKIPIF1<0，則SKIPIF1<0為SKIPIF1<0的零點，不是最值點，結(jié)論D錯誤.故選：AC．題型七：三角函數(shù)圖像的伸縮變換問題19．（2023·全國·高三專題練習(xí)）為了得到函數(shù)SKIPIF1<0的圖象，只要把函數(shù)SKIPIF1<0圖象上所有的點（

）A．向左平移SKIPIF1<0個單位長度 B．向右平移SKIPIF1<0個單位長度C．向左平移SKIPIF1<0個單位長度 D．向右平移SKIPIF1<0個單位長度【答案】D【分析】根據(jù)三角函數(shù)圖象的變換法則即可求出．【詳解】因為SKIPIF1<0，所以把函數(shù)SKIPIF1<0圖象上的所有點向右平移SKIPIF1<0個單位長度即可得到函數(shù)SKIPIF1<0的圖象．故選：D.

20．（2021·全國·統(tǒng)考高考真題）把函數(shù)SKIPIF1<0圖像上所有點的橫坐標(biāo)縮短到原來的SKIPIF1<0倍，縱坐標(biāo)不變，再把所得曲線向右平移SKIPIF1<0個單位長度，得到函數(shù)SKIPIF1<0的圖像，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】解法一：從函數(shù)SKIPIF1<0的圖象出發(fā)，按照已知的變換順序，逐次變換，得到SKIPIF1<0，即得SKIPIF1<0，再利用換元思想求得SKIPIF1<0的解析表達式；解法二：從函數(shù)SKIPIF1<0出發(fā)，逆向?qū)嵤└鞑阶儞Q，利用平移伸縮變換法則得到SKIPIF1<0的解析表達式.【詳解】解法一：函數(shù)SKIPIF1<0圖象上所有點的橫坐標(biāo)縮短到原來的SKIPIF1<0倍，縱坐標(biāo)不變，得到SKIPIF1<0的圖象，再把所得曲線向右平移SKIPIF1<0個單位長度，應(yīng)當(dāng)?shù)玫絊KIPIF1<0的圖象，根據(jù)已知得到了函數(shù)SKIPIF1<0的圖象，所以SKIPIF1<0，令SKIPIF1<0,則SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0；解法二：由已知的函數(shù)SKIPIF1<0逆向變換，第一步：向左平移SKIPIF1<0個單位長度，得到SKIPIF1<0的圖象，第二步：圖象上所有點的橫坐標(biāo)伸長到原來的2倍，縱坐標(biāo)不變，得到SKIPIF1<0的圖象，即為SKIPIF1<0的圖象，所以SKIPIF1<0.故選：B.21．（2022·全國·高三專題練習(xí)）若函數(shù)SKIPIF1<0在SKIPIF1<0處有最小值，為了得到SKIPIF1<0的圖象，則只要將SKIPIF1<0的圖象（

）A．向右平移SKIPIF1<0個單位長度 B．向左平移SKIPIF1<0個單位長度C．向左平移SKIPIF1<0個單位長度 D．向右平移SKIPIF1<0個單位長度【答案】C【分析】由題意可得SKIPIF1<0，結(jié)合SKIPIF1<0可得SKIPIF1<0的值，進而可得SKIPIF1<0的解析式，再由圖象的平移變換即可求解.【詳解】因為函數(shù)SKIPIF1<0在SKIPIF1<0處有最小值，所以SKIPIF1<0，可得：SKIPIF1<0，因為SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，將SKIPIF1<0的圖象向左平移SKIPIF1<0個單位長度可得SKIPIF1<0，故選：C.【高考必刷】一、單選題1．（2023春·安徽安慶·高一安徽省宿松中學(xué)校考開學(xué)考試）設(shè)函數(shù)SKIPIF1<0，則下列結(jié)論正確的是（

）A．SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對稱B．SKIPIF1<0的圖象關(guān)于點SKIPIF1<0對稱C．SKIPIF1<0是偶函數(shù)D．SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增【答案】C【分析】對于A，求出函數(shù)的對稱軸，可知不存在SKIPIF1<0使得對稱軸為直線SKIPIF1<0，A錯誤；對于B，求出函數(shù)的對稱中心，可知不存在SKIPIF1<0使其一個對稱中心為SKIPIF1<0，B錯誤；對于C，由SKIPIF1<0求出SKIPIF1<0，利用誘導(dǎo)公式，結(jié)合偶函數(shù)的定義，可得C正確；對于D，當(dāng)SKIPIF1<0時，求出整體SKIPIF1<0的范圍，驗證SKIPIF1<0不是單調(diào)遞增，D錯誤.【詳解】由SKIPIF1<0解得SKIPIF1<0，所以函數(shù)SKIPIF1<0的對稱軸為SKIPIF1<0，由SKIPIF1<0解得SKIPIF1<0，故A錯誤；由SKIPIF1<0解得SKIPIF1<0，所以函數(shù)SKIPIF1<0的對稱中心為SKIPIF1<0，由SKIPIF1<0解得SKIPIF1<0，故B錯誤；SKIPIF1<0，而SKIPIF1<0，所以SKIPIF1<0是偶函數(shù)，C正確；令SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0即SKIPIF1<0，此時SKIPIF1<0在SKIPIF1<0不是單調(diào)遞增函數(shù)，故D錯誤.故選：C.2．（2022·高一課時練習(xí)）函數(shù)SKIPIF1<0的單調(diào)增區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】首先利用誘導(dǎo)公式將函數(shù)化簡為SKIPIF1<0，再根據(jù)正弦函數(shù)的性質(zhì)計算可得；【詳解】解：因為SKIPIF1<0，所以SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，故函數(shù)的單調(diào)遞增區(qū)間為SKIPIF1<0故選：D.3．（2021秋·云南昆明·高三昆明市第三中學(xué)?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0的圖像關(guān)于SKIPIF1<0對稱，則函數(shù)SKIPIF1<0的圖像的一條對稱軸是（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】先由函數(shù)SKIPIF1<0的圖像關(guān)于SKIPIF1<0對稱，求出SKIPIF1<0，再對SKIPIF1<0化簡即可求出.【詳解】函數(shù)SKIPIF1<0變?yōu)镾KIPIF1<0，（令SKIPIF1<0）.因為函數(shù)SKIPIF1<0的圖像關(guān)于SKIPIF1<0對稱，所以SKIPIF1<0,解得：SKIPIF1<0.所以SKIPIF1<0.所以函數(shù)SKIPIF1<0，其中SKIPIF1<0,其對稱軸方程SKIPIF1<0,所以SKIPIF1<0.因為SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.當(dāng)SKIPIF1<0時，SKIPIF1<0符合題意.對照四個選項，D正確.故選：D.4．（2022秋·河南鄭州·高三統(tǒng)考期末）若將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個單位長度，再將圖象上所有點的橫坐標(biāo)縮短到原來的SKIPIF1<0，得到函數(shù)SKIPIF1<0的圖象，則函數(shù)SKIPIF1<0圖象的對稱軸可能是（

）A．直線SKIPIF1<0 B．直線SKIPIF1<0C．直線SKIPIF1<0 D．直線SKIPIF1<0【答案】C【分析】利用輔助角公式將函數(shù)SKIPIF1<0化簡，再根據(jù)平移變換和周期變換的特征求出函數(shù)SKIPIF1<0的解析式，再根據(jù)正弦函數(shù)的對稱性即可得出答案.【詳解】解：由題得SKIPIF1<0，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，再將圖象上所有點的橫坐標(biāo)縮短到原來的SKIPIF1<0，得到函數(shù)SKIPIF1<0的圖象，令SKIPIF1<0，得SKIPIF1<0，當(dāng)SKIPIF1<0時，得函數(shù)SKIPIF1<0圖象的一條對稱軸為直線SKIPIF1<0，而SKIPIF1<0，所以SKIPIF1<0都不是函數(shù)的對稱軸.故選：C.5．（2022秋·河南洛陽·高三校聯(lián)考階段練習(xí)）函數(shù)SKIPIF1<0的最大值為2，且對任意的SKIPIF1<0，SKIPIF1<0恒成立，SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0的值為（

）A．1 B．SKIPIF1<0 C．SKIPIF1<0 D．2【答案】B【分析】根據(jù)函數(shù)的最值可得A＝2，根據(jù)SKIPIF1<0恒成立可得SKIPIF1<0，由函數(shù)的單調(diào)性可得SKIPIF1<0，進而求得SKIPIF1<0，求出函數(shù)解析式，即可求解.【詳解】因為SKIPIF1<0的最大值為2，所以A＝2，因為SKIPIF1<0恒成立，所以當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0取得最大值，則SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0．當(dāng)SKIPIF1<0時，SKIPIF1<0，因為SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0．所以SKIPIF1<0，故選：B．6．（2023秋·江蘇泰州·高三統(tǒng)考期末）已知函數(shù)SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的最小正周期SKIPIF1<0，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】由已知可得SKIPIF1<0，解不等式求出SKIPIF1<0，再由周期公式求出SKIPIF1<0，最后由SKIPIF1<0可得答案.【詳解】SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，解得SKIPIF1<0，因為SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，又SKIPIF1<0∴SKIPIF1<0.故選：D.7．（2023·甘肅·模擬預(yù)測）設(shè)函數(shù)SKIPIF1<0的部分圖象如圖所示，若SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】根據(jù)圖像求出SKIPIF1<0，由SKIPIF1<0得到SKIPIF1<0，代入即可求解.【詳解】根據(jù)函數(shù)SKIPIF1<0的部分圖象，可得：A=1;因為SKIPIF1<0，SKIPIF1<0，結(jié)合五點法作圖可得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．如果SKIPIF1<0，且SKIPIF1<0，結(jié)合SKIPIF1<0，可得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故選：C．8．（2022秋·江蘇南通·高三校考期中）函數(shù)SKIPIF1<0（SKIPIF1<0，SKIPIF1<0，SKIPIF1<0）的部分圖象如圖所示，將SKIPIF1<0的圖象上所有點的橫坐標(biāo)擴大到原來的4倍（縱坐標(biāo)不變），再把所得的圖象沿SKIPIF1<0軸向左平移SKIPIF1<0個單位長度，得到函數(shù)SKIPIF1<0的圖象，則函數(shù)SKIPIF1<0的一個單調(diào)遞增區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】由函數(shù)的圖象的頂點坐標(biāo)求出A的值，由周期求出SKIPIF1<0的值，由五點法作圖求出SKIPIF1<0的值，可得函數(shù)SKIPIF1<0的解析式，結(jié)合圖象的變換規(guī)則，可得出SKIPIF1<0的解析式，再利用正弦函數(shù)的單調(diào)性即可求解.【詳解】根據(jù)函數(shù)SKIPIF1<0（SKIPIF1<0，SKIPIF1<0，SKIPIF1<0）的部分圖象，可得SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0．結(jié)合五點法作圖可得SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0．將SKIPIF1<0的圖象上所有點的橫坐標(biāo)擴大到原來的4倍（縱坐標(biāo)不變），可得SKIPIF1<0的圖象．再把所得的圖象沿SKIPIF1<0軸向左平移SKIPIF1<0個單位長度，得到函數(shù)SKIPIF1<0的圖象．令SKIPIF1<0，求得SKIPIF1<0，可得函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，可得一個增區(qū)間為SKIPIF1<0.故選：A.9．（2020秋·北京·高三北京八中?？计谥校┮阎猄KIPIF1<0,SKIPIF1<0，直線SKIPIF1<0＝SKIPIF1<0和SKIPIF1<0＝SKIPIF1<0是函數(shù)SKIPIF1<0圖象的兩條相鄰的對稱軸，則SKIPIF1<0＝（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】由于直線SKIPIF1<0＝SKIPIF1<0和SKIPIF1<0＝SKIPIF1<0是函數(shù)SKIPIF1<0圖像的兩條相鄰的對稱軸，所以可得SKIPIF1<0，從而可求出SKIPIF1<0，又由直線SKIPIF1<0＝SKIPIF1<0為函數(shù)SKIPIF1<0圖象的對稱軸，可得SKIPIF1<0，從而可求出SKIPIF1<0的值【詳解】解：因為直線SKIPIF1<0＝SKIPIF1<0和SKIPIF1<0＝SKIPIF1<0是函數(shù)SKIPIF1<0圖像的兩條相鄰的對稱軸，所以SKIPIF1<0，即SKIPIF1<0所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，因為直線SKIPIF1<0＝SKIPIF1<0為函數(shù)SKIPIF1<0圖象的對稱軸，所以SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，因為SKIPIF1<0，所以SKIPIF1<0故選：A【點睛】此題考查正弦函數(shù)的圖像和性質(zhì)的應(yīng)用，屬于基礎(chǔ)題.10．（2022·四川內(nèi)江·四川省內(nèi)江市第六中學(xué)?？寄M預(yù)測）設(shè)函數(shù)SKIPIF1<0，其中SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上的單調(diào)減區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】根據(jù)SKIPIF1<0的對稱中心、零點求得SKIPIF1<0，進而求得SKIPIF1<0，結(jié)合三角函數(shù)單調(diào)區(qū)間的求法求得正確答案.【詳解】據(jù)題意可以得出直線SKIPIF1<0和點SKIPIF1<0分別是的圖象的一條對稱軸和一個對稱中心，所以SKIPIF1<0，即SKIPIF1<0（SKIPIF1<0），所以SKIPIF1<0；又由SKIPIF1<0得SKIPIF1<0，即SKIPIF1<0（SKIPIF1<0），SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0；由SKIPIF1<0得SKIPIF1<0的單調(diào)減區(qū)間為SKIPIF1<0（SKIPIF1<0），所以SKIPIF1<0在SKIPIF1<0上的單調(diào)減區(qū)間是SKIPIF1<0.故選：C11．（2022·全國·高三專題練習(xí)）如圖，點SKIPIF1<0和點SKIPIF1<0分別是函數(shù)SKIPIF1<0(SKIPIF1<0，SKIPIF1<0，SKIPIF1<0)圖像上的最低點和最高點，若SKIPIF1<0、SKIPIF1<0兩點間的距離為SKIPIF1<0，則關(guān)于函數(shù)SKIPIF1<0的說法正確的是（

）A．在區(qū)間SKIPIF1<0上單調(diào)遞增 B．在區(qū)間SKIPIF1<0上單調(diào)遞減C．在區(qū)間SKIPIF1<0上單調(diào)遞減 D．在區(qū)間SKIPIF1<0上單調(diào)遞增【答案】C【分析】首先利用二倍角公式將SKIPIF1<0化簡為SKIPIF1<0，再由SKIPIF1<0，SKIPIF1<0分別為SKIPIF1<0的圖像上的最低點和最高點得到SKIPIF1<0，再由SKIPIF1<0，SKIPIF1<0兩點之間距離為SKIPIF1<0