新高考數(shù)學一輪復習考點精講練+易錯題型第25講 三角函數(shù)的圖像與性質（原卷版）

第25講三角函數(shù)的圖像與性質【基礎知識網(wǎng)絡圖】應用應用三角函數(shù)的圖象與性質正弦函數(shù)的圖象與性質余弦函數(shù)的圖象與性質正切函數(shù)的圖象與性質【基礎知識全通關】一、正弦函數(shù)SKIPIF1<0,余弦函數(shù)SKIPIF1<0,正切函數(shù)SKIPIF1<0的圖象與性質函數(shù)SKIPIF1<0SKIPIF1<0SKIPIF1<0圖象定義域SKIPIF1<0SKIPIF1<0SKIPIF1<0值域SKIPIF1<0SKIPIF1<0SKIPIF1<0最值當SKIPIF1<0時，SKIPIF1<0；當SKIPIF1<0時，SKIPIF1<0．當SKIPIF1<0時，SKIPIF1<0；當SKIPIF1<0時，SKIPIF1<0．既無最大值，也無最小值周期性最小正周期為SKIPIF1<0最小正周期為SKIPIF1<0最小正周期為SKIPIF1<0奇偶性SKIPIF1<0,奇函數(shù)SKIPIF1<0，偶函數(shù)SKIPIF1<0，奇函數(shù)單調性在SKIPIF1<0上是增函數(shù)；在SKIPIF1<0上是減函數(shù)．在SKIPIF1<0上是增函數(shù)；在SKIPIF1<0上是減函數(shù)．在SKIPIF1<0上是增函數(shù)．對稱性對稱中心SKIPIF1<0；對稱軸SKIPIF1<0，既是中心對稱圖形又是軸對稱圖形.對稱中心SKIPIF1<0；對稱軸SKIPIF1<0，既是中心對稱圖形又是軸對稱圖形.對稱中心SKIPIF1<0；無對稱軸，是中心對稱圖形但不是軸對稱圖形.二、函數(shù)SKIPIF1<0的圖象與性質1．函數(shù)SKIPIF1<0的圖象的畫法（1）變換作圖法由函數(shù)SKIPIF1<0的圖象通過變換得到SKIPIF1<0（A>0,ω>0）的圖象，有兩種主要途徑：“先平移后伸縮”與“先伸縮后平移”.如下圖.（2）五點作圖法找五個關鍵點，分別為使y取得最小值、最大值的點和曲線與x軸的交點.其步驟為：①先確定最小正周期T=SKIPIF1<0,在一個周期內作出圖象；②令SKIPIF1<0,令X分別取0，SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,求出對應的x值，列表如下：由此可得五個關鍵點；③描點畫圖，再利用函數(shù)的周期性把所得簡圖向左右分別擴展，從而得到SKIPIF1<0的簡圖.2．函數(shù)SKIPIF1<0（A>0,ω>0）的性質（1）奇偶性：SKIPIF1<0時，函數(shù)SKIPIF1<0為奇函數(shù)；SKIPIF1<0時，函數(shù)SKIPIF1<0為偶函數(shù).（2）周期性：SKIPIF1<0存在周期性，其最小正周期為T=SKIPIF1<0.（3）單調性：根據(jù)y=sint和t=SKIPIF1<0的單調性來研究，由SKIPIF1<0得單調增區(qū)間；由SKIPIF1<0得單調減區(qū)間.（4）對稱性：利用y=sinx的對稱中心為SKIPIF1<0求解，令SKIPIF1<0，求得x.利用y=sinx的對稱軸為SKIPIF1<0求解，令SKIPIF1<0，得其對稱軸.3．函數(shù)SKIPIF1<0（A>0,ω>0）的物理意義當函數(shù)SKIPIF1<0（A>0,ω>0,SKIPIF1<0）表示一個簡諧振動量時，則A叫做振幅，T=SKIPIF1<0叫做周期，f=SKIPIF1<0叫做頻率,SKIPIF1<0叫做相位，x=0時的相位SKIPIF1<0叫做初相.三、三角函數(shù)的綜合應用（1）函數(shù)SKIPIF1<0，SKIPIF1<0的定義域均為SKIPIF1<0；函數(shù)SKIPIF1<0的定義域均為SKIPIF1<0.（2）函數(shù)SKIPIF1<0，SKIPIF1<0的最大值為SKIPIF1<0，最小值為SKIPIF1<0；函數(shù)SKIPIF1<0的值域為SKIPIF1<0.（3）函數(shù)SKIPIF1<0，SKIPIF1<0的最小正周期為SKIPIF1<0；函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0．（4）對于SKIPIF1<0，當且僅當SKIPIF1<0時為奇函數(shù)，當且僅當SKIPIF1<0時為偶函數(shù)；對于SKIPIF1<0，當且僅當SKIPIF1<0時為奇函數(shù)，當且僅當SKIPIF1<0時為偶函數(shù)；對于SKIPIF1<0，當且僅當SKIPIF1<0時為奇函數(shù)．（5）函數(shù)SKIPIF1<0的單調遞增區(qū)間由不等式SKIPIF1<0SKIPIF1<0來確定，單調遞減區(qū)間由不等式SKIPIF1<0來確定；函數(shù)SKIPIF1<0的單調遞增區(qū)間由不等式SKIPIF1<0來確定，單調遞減區(qū)間由不等式SKIPIF1<0來確定；函數(shù)SKIPIF1<0的單調遞增區(qū)間由不等式SKIPIF1<0來確定．【注】函數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0（SKIPIF1<0有可能為負數(shù)）的單調區(qū)間：先利用誘導公式把SKIPIF1<0化為正數(shù)后再求解．（6）函數(shù)SKIPIF1<0圖象的對稱軸為SKIPIF1<0，對稱中心為SKIPIF1<0；函數(shù)SKIPIF1<0圖象的對稱軸為SKIPIF1<0，對稱中心為SKIPIF1<0；函數(shù)SKIPIF1<0圖象的對稱中心為SKIPIF1<0.【注】函數(shù)SKIPIF1<0，SKIPIF1<0的圖象與SKIPIF1<0軸的交點都為對稱中心，過最高點或最低點且垂直于SKIPIF1<0軸的直線都為對稱軸.函數(shù)SKIPIF1<0的圖象與SKIPIF1<0軸的交點和漸近線與SKIPIF1<0軸的交點都為對稱中心，無對稱軸.【考點研習一點通】1、定義域和值域（1）SKIPIF1<0（2）SKIPIF1<0（3）SKIPIF1<0（4）SKIPIF1<0SKIPIF1<0【變式1-2】已知函數(shù)SKIPIF1<0的定義域是SKIPIF1<0，值域是SKIPIF1<0，求常數(shù)SKIPIF1<0．考點02奇偶性、周期性、單調性2、已知函數(shù)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的大小關系為.【變式2-1】已知函數(shù)SKIPIF1<0若SKIPIF1<0是偶函數(shù)，則SKIPIF1<0.3、求函數(shù)SKIPIF1<0的單調區(qū)間?！咀兪?-1】求函數(shù)SKIPIF1<0的單調區(qū)間.4、已知函數(shù)f(x)=4tanxsin(SKIPIF1<0)cos(SKIPIF1<0)-SKIPIF1<0.(Ⅰ)求f(x)的定義域與最小正周期；（Ⅱ）討論f(x)在區(qū)間[SKIPIF1<0]上的單調性.【變式4-1】已知函數(shù)SKIPIF1<0，（1）求SKIPIF1<0的定義域及最小正周期；（2）求SKIPIF1<0的單調遞增區(qū)間.【變式4-2】設函數(shù)f(x)＝Asin(ωx＋φ)(其中A＞0，ω＞0，－π＜φ≤π)在x＝SKIPIF1<0處取得最大值2，其圖象與x軸的相鄰兩個交點的距離為SKIPIF1<0．（1）求f(x)的解析式；（2）求函數(shù)g(x)＝SKIPIF1<0的值域．【考點易錯】1、已知函數(shù)SKIPIF1<0.（Ⅰ）若SKIPIF1<0，求SKIPIF1<0的值；（II）設SKIPIF1<0，求函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值和最小值.【變式1-1】已知函數(shù)SKIPIF1<0（SKIPIF1<0）的最大值為SKIPIF1<0，最小值為SKIPIF1<0，求函數(shù)SKIPIF1<0的最大值和最小值.【變式1-2】已知函數(shù)SKIPIF1<0的定義域是SKIPIF1<0，值域是SKIPIF1<0，求常數(shù)SKIPIF1<0．2、已知函數(shù)f(x)=Asin(ωx+φ)A>0,ω>0,（1）求函數(shù)f(x)的解析式.（2）求函數(shù)f(x)在區(qū)間0,5π123、已知函數(shù)SKIPIF1<0.（1）求SKIPIF1<0的最小正周期；（2）若SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值與最小值的和為2，求SKIPIF1<0的值.【鞏固提升】1、函數(shù)SKIPIF1<0的最小正周期為A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02.已知函數(shù)SKIPIF1<0（SKIPIF1<0，SKIPIF1<0），若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0內沒有零點，則SKIPIF1<0的取值范圍是A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03.設函數(shù)SKIPIF1<0，SKIPIF1<0則下列判斷正確的是A．函數(shù)的一條對稱軸為SKIPIF1<0B．函數(shù)在區(qū)間SKIPIF1<0內單調遞增C．SKIPIF1<0，使SKIPIF1<0D．SKIPIF1<0，使得函數(shù)SKIPIF1<0在其定義域內為偶函數(shù)4、若SKIPIF1<0在SKIPIF1<0是減函數(shù)，則SKIPIF1<0的最大值是A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0