三角函數(shù)的圖像與性質(zhì)題型歸納總結(jié)

1、1三角函數(shù)的圖像與性質(zhì)題型歸納總結(jié)題型歸納及思路提示題型歸納及思路提示題型1已知函數(shù)解析式確定函數(shù)性質(zhì)【思路提示】一般所給函數(shù)為y＝Asin(ωx＋φ)或y＝Acos(ωx＋φ)，A0ω0要根據(jù)y＝sinx，y＝cosx的整體性質(zhì)求解。1、函數(shù)的奇偶性例1f(x)＝sin（0≤）是R上的偶函數(shù)，則上的偶函數(shù)，則等于（等于（）()x?????A.0BCD4?2??【評(píng)注】由是奇函數(shù)，是偶函數(shù)可拓展得到關(guān)于三角函數(shù)奇偶性的重sinyx?co

2、syx?要結(jié)論：sin()()yAxkkZ???????(1)若是奇函數(shù)，則sin()()2yAxkkZ????????(2)若是偶函數(shù)，則cos()()2yAxkkZ?????????(3)若是奇函數(shù)，則cos()()yAxkkZ???????(4)若是偶函數(shù)，則tan()().2kyAxkZ???????(5)若是奇函數(shù)，則.()sin||aRfxxaa???變式1已知，函數(shù)為奇函數(shù)，則等于（）A.0B1CD1?1?2.0()cos

3、()()RfxxxR????????變式設(shè)，則“”是“為偶函數(shù)”的（）A充分不必要條件B必要不充分條C充要條件D無(wú)關(guān)條件3.()sin()0()fxxfx??????變式設(shè)，其中，則是偶函數(shù)的充要條件是（）A.BCD(0)1f?(0)0f?(0)1f?(0)0f?2.()sin(2)()()2fxxxRfx????例設(shè)，則是()A.B?最小正周期為的奇函數(shù)?最小正周期為的偶函數(shù)CD2?最小正周期為的奇函數(shù)2?最小正周期為的偶函數(shù)2()s

4、in1()()fxxxRfx???變式1.若，則是()A.B?最小正周期為的奇函數(shù)?最小正周期為的偶函數(shù)CD?最小正周期為2的奇函數(shù)?最小正周期為2的偶函數(shù)331.sin()[()44yxfxfx?????變式函數(shù)在]內(nèi)單調(diào)遞增，則可以是()A.BCD1cosxsinxcosx?()sin()(0)(42fxx?????????變式2.若在)上單調(diào)遞增，則的取值范圍是（）A.BCD15[]2413[]241(0]2(02]3.()3si

5、ncos()cos()(0)33(1)()(2)()[0]()22fxxxxfxfxxfx???????????????變式已知函數(shù)求的值域；若的最小正周期為，的單調(diào)遞減區(qū)間.4、函數(shù)的對(duì)稱性（對(duì)稱軸、對(duì)稱中心）.sin(2)3yx???例5函數(shù)圖象的對(duì)稱軸方程可能是()A.BCD6x???12x???6x??12x??【評(píng)注】關(guān)于三角函數(shù)對(duì)稱性的幾個(gè)重要結(jié)論：sin()(0)()2cos()(0)()2tan(0)()22sin()(

6、)=()2:yxxkkZkkZyxxkkZkkZkyxkZkyAxbxkkZxkZxk????????????????????????????????????????????(1)函數(shù)的對(duì)稱軸為對(duì)稱中心(2)函數(shù)的對(duì)稱軸為對(duì)稱中心(3)函數(shù)無(wú)對(duì)稱軸，對(duì)稱中心(4)函數(shù)的對(duì)稱軸的求法：令得對(duì)稱中心的求法令()=()()()cos()()=()22:()=()()()2kkkZxkZbkZkyAxbxkkZxkZkkxkkZxkZbkZ??