新高考數(shù)學(xué)二輪復(fù)習(xí)易錯題專練易錯點(diǎn)05 三角變換及三角函數(shù)的性質(zhì)（含解析）

易錯點(diǎn)05三角變換及三角函數(shù)的性質(zhì)易錯題【01】忽略角的范圍限制利用SKIPIF1<0求值一般要涉及到開方運(yùn)算,此時要注意利用角的范圍判斷三角函數(shù)值的正負(fù)；根據(jù)三角函數(shù)值求角,一般是先求出該角的某一個三角函數(shù)值,再確定角的范圍,確定角的范圍時不僅要看已知條件中角的范圍,還要挖掘隱含條件,根據(jù)三角函數(shù)值的符號及大小縮小角的范圍易錯題【02】不能全面理解三角函數(shù)性質(zhì)致誤本易錯點(diǎn)主要包含以下幾個問題：(1)求三角函數(shù)值域忽略定義域的限制；(2)確定三角函數(shù)的最小正周期,忽略三角變換的等價性；(3)求復(fù)合函數(shù)的單調(diào)性忽略對內(nèi)函數(shù)單調(diào)性的判斷.易錯題【03】對平移理解不準(zhǔn)確致誤三角函數(shù)圖象的左右平移是自變量x發(fā)生變化,如ωx→ωx±φ(φ>0)這個變化的實(shí)質(zhì)是x→x±eq\f(φ,ω),所以平移的距離并不一定是φ.易錯題【04】忽視弦函數(shù)的有界性在求與SKIPIF1<0,SKIPIF1<0有關(guān)的值域與范圍是要注意SKIPIF1<0.易錯題【05】用零點(diǎn)確定SKIPIF1<0的SKIPIF1<0,忽略圖象的升降確定φ值時,往往以尋找“五點(diǎn)法”中的特殊點(diǎn)作為突破口．具體如下：“第一點(diǎn)”(即圖象上升時與x軸的交點(diǎn))為ωx＋φ＝0；“第二點(diǎn)”(即圖象的“峰點(diǎn)”)為ωx＋φ＝eq\f(π,2)；“第三點(diǎn)”(即圖象下降時與x軸的交點(diǎn))為ωx＋φ＝π；“第四點(diǎn)”(即圖象的“谷點(diǎn)”)為ωx＋φ＝eq\f(3π,2)；“第五點(diǎn)”為ωx＋φ＝2π. 01已知sinα＝eq\f(\r(5),5),sinβ＝eq\f(\r(10),10),且α,β為銳角,則α＋β＝________.【警示】本題出錯的主要原因是忽略α＋β縮小的范圍.【答案】eq\f(π,4)【問診】因?yàn)棣?β為銳角,所以cosα＝eq\r(1－sin2α)＝eq\f(2\r(5),5),cosβ＝eq\r(1－sin2β)＝eq\f(3\r(10),10).所以cos(α＋β)＝cosαcosβ－sinαsinβ＝eq\f(2\r(5),5)×eq\f(3\r(10),10)－eq\f(\r(5),5)×eq\f(\r(10),10)＝eq\f(\r(2),2),又因?yàn)?<α＋β<π,所以α＋β＝eq\f(π,4).【叮囑】根據(jù)三角函數(shù)值求角,要注意三角函數(shù)值的符號及大小對角的范圍的限制1.(2022屆福建省永安市高三10月月考)已知SKIPIF1<0的終邊在第四象限,若SKIPIF1<0,則SKIPIF1<0()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】SKIPIF1<0的終邊在第四象限,SKIPIF1<0,所以SKIPIF1<0,則SKIPIF1<0．故選A.2.已知SKIPIF1<0為銳角,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0的值為()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】∵SKIPIF1<0為銳角,SKIPIF1<0,SKIPIF1<0,∴SKIPIF1<0,SKIPIF1<0,∴SKIPIF1<0SKIPIF1<0,又SKIPIF1<0,∴SKIPIF1<0,故選B 02函數(shù)SKIPIF1<0得最小正周期為.【警示】根據(jù)SKIPIF1<0,得出SKIPIF1<0的最小正周期SKIPIF1<0是本題出錯的主要原因.【答案】SKIPIF1<0【問診】忽略了SKIPIF1<0與SKIPIF1<0定義域不相同,SKIPIF1<0對定義域內(nèi)的SKIPIF1<0不成立,所以SKIPIF1<0不是SKIPIF1<0的周期.【叮囑】研究三角函數(shù)的性質(zhì),若對三角函數(shù)進(jìn)行變換,一點(diǎn)要保證變換的等價性.1．求函數(shù)SKIPIF1<0的值域.【答案】因?yàn)镾KIPIF1<0=SKIPIF1<0=SKIPIF1<0,當(dāng)SKIPIF1<0時SKIPIF1<0取得最小值SKIPIF1<0,當(dāng)SKIPIF1<0時SKIPIF1<0取得最大值4,又因?yàn)镾KIPIF1<0,所以SKIPIF1<0的值域?yàn)镾KIPIF1<0.2.(2022屆山東省青島市高三上學(xué)期期中)若函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0,則函數(shù)SKIPIF1<0在SKIPIF1<0上的值域?yàn)開______．【答案】SKIPIF1<0.【解析】函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0,故SKIPIF1<0；則SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0.由SKIPIF1<0SKIPIF1<0,故可得SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0的值域?yàn)镾KIPIF1<0. 03(2022全國乙卷理T7)把函數(shù)SKIPIF1<0圖像上所有點(diǎn)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<0SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【警示】平移方向或平移長度易出錯【答案】B【問診】SKIPIF1<0把函數(shù)SKIPIF1<0圖像上所有點(diǎn)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,SKIPIF1<0把函數(shù)SKIPIF1<0的圖像,向左平移SKIPIF1<0個單位長度,得到SKIPIF1<0的圖像；再把圖像上所有點(diǎn)的橫坐標(biāo)變?yōu)樵瓉淼?倍,縱坐標(biāo)不變,可得SKIPIF1<0的圖像．故選SKIPIF1<0．【叮囑】對于左右平移要注意要確定平移方向,法則是甲左減右,然后再確定平移長度,注意平移長度由x的變化來確定.1．要得到y(tǒng)＝sin(－3x)的圖象,需將y＝eq\f(\r(2),2)(cos3x－sin3x)的圖象A．向右平移eq\f(π,4)個單位 B．向左平移eq\f(π,4)個單位C．向右平移eq\f(π,12)個單位 D．向左平移eq\f(π,12)個單位【答案】D【解析】y＝eq\f(\r(2),2)(cos3x－sin3x)＝sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,4)－3x))＝sineq\b\lc\[\rc\](\a\vs4\al\co1(－3\b\lc\(\rc\)(\a\vs4\al\co1(x－\f(π,12))))),要由y＝sineq\b\lc\[\rc\](\a\vs4\al\co1(－3\b\lc\(\rc\)(\a\vs4\al\co1(x－\f(π,12)))))到y(tǒng)＝sin(－3x)只需對x加上eq\f(π,12)即可,因而是對y＝eq\f(\r(2),2)(cos3x－sin3x)向左平移eq\f(π,12)個單位,故選D.2．(2022屆海南省華僑中學(xué)高三11月月考)將函數(shù)y＝f(x)的圖象向左平移SKIPIF1<0個單位長度,再把所得圖象上所有點(diǎn)的橫坐標(biāo)伸長到原來的2倍得到y(tǒng)＝sinSKIPIF1<0的圖象,則f(x)＝()A．sinSKIPIF1<0 B．sinSKIPIF1<0 C．sinSKIPIF1<0 D．sinSKIPIF1<0【答案】B【解析】將SKIPIF1<0的圖象上各個點(diǎn)的橫坐標(biāo)變?yōu)樵瓉淼腟KIPIF1<0,可得函數(shù)SKIPIF1<0的圖象,再把函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個單位,即可得到SKIPIF1<0的圖象,所以SKIPIF1<0SKIPIF1<0,故選B. 04已知SKIPIF1<0,則SKIPIF1<0的取值范圍是.【警示】由SKIPIF1<0得SKIPIF1<0,因?yàn)镾KIPIF1<0,所以SKIPIF1<0.忽略了SKIPIF1<0.【答案】SKIPIF1<0【問診】由題意可得SKIPIF1<0,所以SKIPIF1<0,由SKIPIF1<0得SKIPIF1<0,因?yàn)镾KIPIF1<0,所以SKIPIF1<0.所以SKIPIF1<0的取值范圍是SKIPIF1<0.【叮囑】若一個式子中含有2個變量的弦函數(shù),要保證每個弦函數(shù)的范圍都在SKIPIF1<0上.1.(2022屆甘肅省部分名校年高三上學(xué)期月考)已知函數(shù)SKIPIF1<0．若關(guān)于x的方程SKIPIF1<0在SKIPIF1<0上有解,則實(shí)數(shù)m的取值范圍是()A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】SKIPIF1<0SKIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0,所以SKIPIF1<0,故SKIPIF1<0的值域?yàn)镾KIPIF1<0,因?yàn)镾KIPIF1<0在SKIPIF1<0上有解即SKIPIF1<0在SKIPIF1<0上有解,故SKIPIF1<0即SKIPIF1<0,故選C.2．(2022屆四川省綿陽高三上學(xué)期一診)已知SKIPIF1<0,則SKIPIF1<0的最大值為____________【答案】SKIPIF1<0【解析】SKIPIF1<0,SKIPIF1<0SKIPIF1<0,SKIPIF1<0,即SKIPIF1<0SKIPIF1<0又SKIPIF1<0,利用二次函數(shù)的性質(zhì)知,當(dāng)SKIPIF1<0時,SKIPIF1<0. 05函數(shù)f(x)＝sin(ωx＋φ)(x∈R,ω>0,|φ|<SKIPIF1<0)的部分圖象如圖所示,則SKIPIF1<0.【警示】本題出錯的主要原因是觀察圖象可知,A＝1,T＝π,∴ω＝2,f(x)＝sin(2x＋φ),把SKIPIF1<0代入得SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0.忽略了SKIPIF1<0在遞減區(qū)間內(nèi).【答案】sin(2x＋eq\f(π,3))【問診】觀察圖象可知,A＝1,T＝π,∴ω＝2,f(x)＝sin(2x＋φ),將(－eq\f(π,6),0)代入上式得sin(－eq\f(π,3)＋φ)＝0,由|φ|<SKIPIF1<0,得φ＝eq\f(π,3),則f(x)＝sin(2x＋eq\f(π,3))．【叮囑】利用函數(shù)的零點(diǎn)確定確定φ值時,要注意該零點(diǎn)是在遞增區(qū)間還是在遞減區(qū)間.1.已知函數(shù)SKIPIF1<0(SKIPIF1<0,SKIPIF1<0)的部分圖象如圖所示,將SKIPIF1<0圖象進(jìn)行怎樣的平移變換后得到的圖象對應(yīng)的函數(shù)SKIPIF1<0為奇函數(shù)()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【解析】由圖象可知SKIPIF1<0,所以SKIPIF1<0,設(shè)SKIPIF1<0的最小正周期為SKIPIF1<0,則SKIPIF1<0,所以SKIPIF1<0,則SKIPIF1<0,所以SKIPIF1<0,因?yàn)楹瘮?shù)SKIPIF1<0經(jīng)過點(diǎn)SKIPIF1<0,可得SKIPIF1<0,解得SKIPIF1<0,因?yàn)镾KIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,結(jié)合選項,當(dāng)SKIPIF1<0時,SKIPIF1<0為奇函數(shù),所以將SKIPIF1<0的圖象向右平移SKIPIF1<0個單位長度后得到的函數(shù)SKIPIF1<0為奇函數(shù).故選D.2.(2022屆重慶市八中高三上學(xué)期月考)(多選題)函數(shù)SKIPIF1<0的圖象如圖所示,則關(guān)于函數(shù)SKIPIF1<0下列結(jié)論中正確的是()A．SKIPIF1<0 B．SKIPIF1<0C．對稱軸為SKIPIF1<0 D．對稱中心為SKIPIF1<0【答案】ABD【解析】對于A：由SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,A正確；對于B：因?yàn)镾KIPIF1<0,SKIPIF1<0,解得SKIPIF1<0,故SKIPIF1<0,SKIPIF1<0,B正確；對于C：令SKIPIF1<0,解得SKIPIF1<0,即對稱軸為SKIPIF1<0,故C錯誤；對于D：令SKIPIF1<0,解得SKIPIF1<0,即對稱中心為SKIPIF1<0,故D正確.故選ABD.錯1．(2022屆河北省衡水市冀州區(qū)高三上學(xué)期期中)已知SKIPIF1<0且SKIPIF1<0,若SKIPIF1<0,則SKIPIF1<0()A．SKIPIF1<0或SKIPIF1<0 B．SKIPIF1<0或1 C．1 D．SKIPIF1<0【答案】D【解析】由SKIPIF1<0,得SKIPIF1<0,所以SKIPIF1<0,求得SKIPIF1<0(舍),SKIPIF1<0．又SKIPIF1<0,將SKIPIF1<0的值代入上式可得：SKIPIF1<0．故選D．2．若SKIPIF1<0,且SKIPIF1<0,則SKIPIF1<0()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】因?yàn)镾KIPIF1<0,所以SKIPIF1<0,又SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,故選A.3．設(shè)銳角SKIPIF1<0的內(nèi)角SKIPIF1<0,SKIPIF1<0,SKIPIF1<0的對邊分別為SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,若SKIPIF1<0,則SKIPIF1<0的取值范圍是()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由正弦定理得SKIPIF1<0SKIPIF1<0.因?yàn)镾KIPIF1<0為銳角三角形,所以SKIPIF1<0即SKIPIF1<0所以SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0的取值范圍是SKIPIF1<0.故選A.4．(2021屆安徽省合肥市高三下學(xué)期第二次教學(xué)質(zhì)量檢測)在SKIPIF1<0ABC中,已知SKIPIF1<0,SKIPIF1<0,則cosC=()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0或SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】在SKIPIF1<0ABC中,∵SKIPIF1<0,∴SKIPIF1<0,∴SKIPIF1<0.∵SKIPIF1<0,∴SKIPIF1<0或SKIPIF1<0(舍去),∴SKIPIF1<0,∴SKIPIF1<0,SKIPIF1<0.故選A.5．(2022屆北京師范大學(xué)附屬中學(xué)高三上學(xué)期期中)將SKIPIF1<0的圖象向右平移SKIPIF1<0個單位,則所得圖象的函數(shù)解析式為()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】將SKIPIF1<0的圖象向右平移SKIPIF1<0個單位,得到SKIPIF1<0,故選D6．(2021屆湘豫名校名校高三5月聯(lián)考)已知函數(shù)SKIPIF1<0的圖象沿SKIPIF1<0軸向右平移SKIPIF1<0個單位長度后得到函數(shù)SKIPIF1<0的圖象,則SKIPIF1<0的一個對稱中心為()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】SKIPIF1<0,因?yàn)楹瘮?shù)SKIPIF1<0的圖象沿SKIPIF1<0軸向右平移SKIPIF1<0個單位長度后得到函數(shù)SKIPIF1<0的圖象,所以函數(shù)SKIPIF1<0的圖象沿SKIPIF1<0軸向左平移SKIPIF1<0個單位長度后得到函數(shù)SKIPIF1<0的圖象,所以SKIPIF1<0,所以SKIPIF1<0,顯然SKIPIF1<0,因此SKIPIF1<0,因?yàn)镾KIPIF1<0,所以令SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,令SKIPIF1<0,令SKIPIF1<0,可得一個對稱中心為SKIPIF1<0,故選B7．(2022屆四川省成都高三上學(xué)期階段性檢)函數(shù)SKIPIF1<0,(其中SKIPIF1<0,SKIPIF1<0,SKIPIF1<0)其圖象如圖所示,為了得到SKIPIF1<0的圖象,可以將SKIPIF1<0的圖象()A．向右平移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．向左平移SKIPIF1<0個單位長度【答案】B【解析】由函數(shù)圖象可知：SKIPIF1<0,函數(shù)過SKIPIF1<0兩點(diǎn),設(shè)SKIPIF1<0的最小正周期為SKIPIF1<0,因?yàn)镾KIPIF1<0,所以有SKIPIF1<0,而SKIPIF1<0,因此SKIPIF1<0,即SKIPIF1<0,因?yàn)镾KIPIF1<0,所以SKIPIF1<0,因?yàn)镾KIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0,因此SKIPIF1<0,而SKIPIF1<0,而SKIPIF1<0,因此該函數(shù)向右平移SKIPIF1<0個單位長度得到函數(shù)SKIPIF1<0的圖象,故選B8．(2022屆江蘇省南通市高三上學(xué)期期中)已知函數(shù)SKIPIF1<0(SKIPIF1<0,SKIPIF1<0,SKIPIF1<0)的部分圖象如圖所示,將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個單位長度后得到SKIPIF1<0的圖象,則下列說法正確的是()A．SKIPIF1<0B．SKIPIF1<0C．函數(shù)SKIPIF1<0為奇函數(shù)D．函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減【答案】BCD【解析】SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,∴SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,∴SKIPIF1<0,A錯.SKIPIF1<0,SKIPIF