易錯(cuò)點(diǎn)08三角函數(shù)與解三角易錯(cuò)點(diǎn)1：解三角函數(shù)的定義此類題主要考查三角函數(shù)的符號(hào)，二倍角公式，特殊角的三角函數(shù)值等知識(shí)，意在考查學(xué)生的轉(zhuǎn)化能力和計(jì)算求解能力.所以要求考生要熟記公式，并懂得靈活應(yīng)用。易錯(cuò)點(diǎn)2：三角函數(shù)圖象變換函數(shù)圖象的平移變換解題策略：（1）對(duì)函數(shù)y=sinx，y=Asin(ωx＋φ)或y=Acos(ωx＋φ)的圖象，無(wú)論是先平移再伸縮，還是先伸縮再平移，只要平移|φ|個(gè)單位，都是相應(yīng)的解析式中的x變?yōu)閤±|φ|，而不是ωx變?yōu)棣豿±|φ|.（2）注意平移前后兩個(gè)函數(shù)的名稱是否一致，若不一致，應(yīng)用誘導(dǎo)公式化為同名函數(shù)再平移.易錯(cuò)點(diǎn)3：由三角函數(shù)圖像求解析式結(jié)合圖象及性質(zhì)求解析式y(tǒng)=Asin(ωx＋φ)＋B(A>0，ω>0)的方法（1）求A，B，已知函數(shù)的最大值M和最小值m，則SKIPIF1<0.（2）求ω，已知函數(shù)的周期T，則SKIPIF1<0.（3）求φ，常用方法有：=1\*GB3①代入法：把圖象上的一個(gè)已知點(diǎn)代入(此時(shí)，A，ω，B已知)．=2\*GB3②確定φ值時(shí)，往往以尋找“五點(diǎn)法”中的第一個(gè)零點(diǎn)SKIPIF1<0作為突破口，具體如下：“第一點(diǎn)”(即圖象上升時(shí)與x軸的交點(diǎn)中距原點(diǎn)最近的交點(diǎn))為ωx＋φ=0；“第二點(diǎn)”(即圖象的“峰點(diǎn)”)為ωx＋φ=SKIPIF1<0；“第三點(diǎn)”(即圖象下降時(shí)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=SKIPIF1<0；“第五點(diǎn)”為ωx＋φ=2π.易錯(cuò)點(diǎn)4：給值（式）求角（值）解三角函數(shù)的給值求值問題的基本步驟（1）先化簡(jiǎn)所求式子或所給條件；（2）觀察已知條件與所求式子之間的聯(lián)系；（3）將已知條件代入所求式子，化簡(jiǎn)求值．易錯(cuò)點(diǎn)5：三角形中邊角關(guān)系此類題考查解三角形的相關(guān)知識(shí)，涉及到正弦定理角化邊的應(yīng)用、余弦定理的應(yīng)用、三角形周長(zhǎng)最大值的求解問題；求解周長(zhǎng)最大值的關(guān)鍵是能夠在余弦定理構(gòu)造的等式中，結(jié)合基本不等式構(gòu)造不等關(guān)系求得最值.題組一、三角函數(shù)的定義1.（2020?全國(guó)2卷）若α為第四象限角，則（）A.cos2α＞0B.cos2α＜0C.sin2α＞0D.sin2α＜0【答案】D【解析】方法一：由α為第四象限角，可得SKIPIF1<0，所以SKIPIF1<0此時(shí)SKIPIF1<0的終邊落在第三、四象限及SKIPIF1<0軸的非正半軸上，所以SKIPIF1<0故選：D.方法二：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，選項(xiàng)B錯(cuò)誤；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，選項(xiàng)A錯(cuò)誤；由SKIPIF1<0在第四象限可得：SKIPIF1<0，則SKIPIF1<0，選項(xiàng)C錯(cuò)誤，選項(xiàng)D正確；故選：D.2．（2014新課標(biāo)Ⅰ）若SKIPIF1<0，則（）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】C【解析】SKIPIF1<0知SKIPIF1<0的終邊在第一象限或第三象限，此時(shí)SKIPIF1<0與SKIPIF1<0同號(hào)，故SKIPIF1<0，選C．3．（2011新課標(biāo)）已知角SKIPIF1<0的頂點(diǎn)與原點(diǎn)重合，始邊與SKIPIF1<0軸的正半軸重合，終邊在直線SKIPIF1<0上，則SKIPIF1<0=（）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】B【解析】由角SKIPIF1<0的終邊在直線SKIPIF1<0上可得，SKIPIF1<0，SKIPIF1<0．故選B題組二、三角函數(shù)的圖像與變換4．(2021年高考全國(guó)乙卷理科)把函數(shù)SKIPIF1<0圖像上所有點(diǎn)的橫坐標(biāo)縮短到原來(lái)的SKIPIF1<0倍，縱坐標(biāo)不變，再把所得曲線向右平移SKIPIF1<0個(gè)單位長(zhǎng)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【解析】SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0.5．（2017新課標(biāo)Ⅰ）已知曲線SKIPIF1<0：SKIPIF1<0，SKIPIF1<0：SKIPIF1<0，則下面結(jié)論正確的是()A．把SKIPIF1<0上各點(diǎn)的橫坐標(biāo)伸長(zhǎng)到原來(lái)的2倍，縱坐標(biāo)不變，再把得到的曲線向右平移SKIPIF1<0個(gè)單位長(zhǎng)度，得到曲線SKIPIF1<0B．把SKIPIF1<0上各點(diǎn)的橫坐標(biāo)伸長(zhǎng)到原來(lái)的2倍，縱坐標(biāo)不變，再把得到的曲線向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，得到曲線SKIPIF1<0C．把SKIPIF1<0上各點(diǎn)的橫坐標(biāo)縮短到原來(lái)的SKIPIF1<0倍，縱坐標(biāo)不變，再把得到的曲線向右平移SKIPIF1<0QUOTE個(gè)單位長(zhǎng)度，得到曲線SKIPIF1<0D．把SKIPIF1<0上各點(diǎn)的橫坐標(biāo)縮短到原來(lái)的QUOTESKIPIF1<0倍，縱坐標(biāo)不變，再把得到的曲線向左平移SKIPIF1<0QUOTE個(gè)單位長(zhǎng)度，得到曲線SKIPIF1<0【答案】D【解析】把SKIPIF1<0的解析式運(yùn)用誘導(dǎo)公式變?yōu)橛嘞?，SKIPIF1<0：SKIPIF1<0則由SKIPIF1<0圖象橫坐標(biāo)縮短為原來(lái)的SKIPIF1<0，再把得到的曲線向左平移SKIPIF1<0QUOTE個(gè)單位長(zhǎng)度，得到曲線SKIPIF1<0．選D6．（2016全國(guó)II）若將函數(shù)的圖像向左平移個(gè)單位長(zhǎng)度，則平移后圖象的對(duì)稱軸為A．B．C．D．【答案】B【解析】函數(shù)SKIPIF1<0的圖像向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，得到的圖像對(duì)應(yīng)的函數(shù)表達(dá)式為SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，所以所求對(duì)稱軸的方程為SKIPIF1<0，故選B．7．（2016年全國(guó)III）函數(shù)的圖像可由函數(shù)的圖像至少向右平移_____________個(gè)單位長(zhǎng)度得到．【答案】SKIPIF1<0【解析】函數(shù)SKIPIF1<0的圖像可由函數(shù)SKIPIF1<0SKIPIF1<0的圖像至少向右平移SKIPIF1<0個(gè)單位長(zhǎng)度得到．題組三、由三角函數(shù)圖像求解析式8．(2020年高考數(shù)學(xué)課標(biāo)Ⅰ卷理科)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<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】由圖可得：函數(shù)圖象過(guò)點(diǎn)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(fù)半軸的第一個(gè)交點(diǎn)，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：C9.（2020?新全國(guó)1山東）（多選題）下圖是函數(shù)y＝sin(ωx＋φ)的部分圖像，則sin(ωx＋φ)＝（）A.SKIPIF1<0B.SKIPIF1<0C.SKIPIF1<0D.SKIPIF1<0【答案】C【解析】由函數(shù)圖像可知：SKIPIF1<0，則SKIPIF1<0，所以不選A,當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0，解得：SKIPIF1<0，即函數(shù)的解析式為：SKIPIF1<0.而SKIPIF1<0,故選：BC.10．（2015新課標(biāo)Ⅱ）函數(shù)SKIPIF1<0的部分圖像如圖所示，則SKIPIF1<0的單調(diào)遞減區(qū)間為（）．SKIPIF1<0，SKIPIF1<0B．SKIPIF1<0，SKIPIF1<0C．SKIPIF1<0，SKIPIF1<0D．SKIPIF1<0，SKIPIF1<0【答案】D【解析】由圖象可知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間為，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0．題組四、給值（式）求值（角）11．(2021年高考全國(guó)甲卷理科)若SKIPIF1<0，則SKIPIF1<0 ()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．故選：A．12．(2018全國(guó)卷Ⅲ)若SKIPIF1<0，則SKIPIF1<0（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】SKIPIF1<0．故選B．13.（2019全國(guó)Ⅱ理10）已知α∈(0，)，2sin2α=cos2α+1，則sinα=A．B．C． D．【答案】B【解析】由，得.因?yàn)?，所?由，得.故選B.14．（2016年全國(guó)II）若，則（）A．B．C．D．【答案】D【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0,所以SKIPIF1<0,故選D．題組五、三角形中的邊角關(guān)系15.（2020?全國(guó)3卷）在△ABC中，cosC＝SKIPIF1<0，AC＝4，BC＝3，則cosB＝（）A.SKIPIF1<0B.SKIPIF1<0C.SKIPIF1<0D.SKIPIF1<0【答案】A【解析】SKIPIF1<0在SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0根據(jù)余弦定理：SKIPIF1<0,SKIPIF1<0,可得SKIPIF1<0，即SKIPIF1<0,由SKIPIF1<0SKIPIF1<0,故SKIPIF1<0.故選：A.16．(2018年高考數(shù)學(xué)課標(biāo)Ⅲ卷（理）)SKIPIF1<0的內(nèi)角SKIPIF1<0的對(duì)邊分別為SKIPIF1<0，若SKIPIF1<0的面積為SKIPIF1<0，則SKIPIF1<0 ()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】由余弦定理可得SKIPIF1<0，所以由SKIPIF1<0所以SKIPIF1<0，而SKIPIF1<0，所以SKIPIF1<0，故選C．17.（2021年上海卷第18題）在SKIPIF1<0中，已知SKIPIF1<0（1）若SKIPIF1<0，求SKIPIF1<0的面積；（2）若SKIPIF1<0，求SKIPIF1<0的周長(zhǎng).【答案】（1）SKIPIF1<0；（2）SKIPIF1<0【解析】（1）由已知得，SKIPIF1<0SKIPIF1<0（2）SKIPIF1<0因?yàn)镾KIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0所以SKIPIF1<0因?yàn)镾KIPIF1<0，所以SKIPIF1<0所以SKIPIF1<018.（2021年天津卷）在SKIPIF1<0，角SKIPIF1<0所對(duì)的邊分別為SKIPIF1<0，已知SKIPIF1<0，SKIPIF1<0．（1）求a的值；（2）求SKIPIF1<0的值；（3）求SKIPIF1<0的值．【答案】（1）SKIPIF1<0；（2）SKIPIF1<0；（3）SKIPIF1<0【解析】（1）因?yàn)镾KIPIF1<0，由正弦定理可得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0；（2）由余弦定理可得SKIPIF1<0；（3）SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0.1．設(shè)函數(shù)SKIPIF1<0，則（）A．SKIPIF1<0在SKIPIF1<0單調(diào)遞增，其圖象關(guān)于直線SKIPIF1<0對(duì)稱B．SKIPIF1<0在SKIPIF1<0單調(diào)遞增，其圖象關(guān)于直線SKIPIF1<0對(duì)稱C．SKIPIF1<0在SKIPIF1<0單調(diào)遞減，其圖象關(guān)于直線SKIPIF1<0對(duì)稱D．SKIPIF1<0在SKIPIF1<0單調(diào)遞減，其圖象關(guān)于直線SKIPIF1<0對(duì)稱【答案】D【解析】∵SKIPIF1<0=SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0單調(diào)遞減，對(duì)稱軸為SKIPIF1<0，即SKIPIF1<0．2．已知SKIPIF1<0>0，SKIPIF1<0，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(duì)稱軸，則SKIPIF1<0=（）A．eq\f(π,4)B．eq\f(π,3)C．eq\f(π,2)D．eq\f(3π,4)【答案】A【解析】由題設(shè)知，SKIPIF1<0=SKIPIF1<0，∴SKIPIF1<0=1，∴SKIPIF1<0=SKIPIF1<0（SKIPIF1<0），∴SKIPIF1<0=SKIPIF1<0（SKIPIF1<0），∵SKIPIF1<0，∴SKIPIF1<0=SKIPIF1<0，故選A.3．設(shè)函數(shù)SKIPIF1<0，則下列結(jié)論錯(cuò)誤的是（）A．SKIPIF1<0的一個(gè)周期為SKIPIF1<0B．SKIPIF1<0的圖像關(guān)于直線SKIPIF1<0對(duì)稱C．SKIPIF1<0的一個(gè)零點(diǎn)為SKIPIF1<0D．SKIPIF1<0在SKIPIF1<0單調(diào)遞減【答案】D【解析】∵SKIPIF1<0的周期為SKIPIF1<0，SKIPIF1<0，所以A正確；∵SKIPIF1<0，所以B正確；設(shè)SKIPIF1<0，而SKIPIF1<0，C正確；選D．4．已知SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0單調(diào)遞減，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】函數(shù)SKIPIF1<0的圖像可看作是由函數(shù)SKIPIF1<0的圖像先向左平移QUOTESKIPIF1<0個(gè)單位得SKIPIF1<0的圖像，再將圖像上所有點(diǎn)的橫坐標(biāo)縮小到原來(lái)的SKIPIF1<0QUOTE1ω倍，縱坐標(biāo)不變得到的，而函數(shù)SKIPIF1<0的減區(qū)間是SKIPIF1<0，所以要使函數(shù)SKIPIF1<0在SKIPIF1<0上是減函數(shù)，需滿足SKIPIF1<0，解得SKIPIF1<0．故選A．5．已知函數(shù)SKIPIF1<0為SKIPIF1<0的零點(diǎn)，SKIPIF1<0為SKIPIF1<0圖像的對(duì)稱軸，且SKIPIF1<0在SKIPIF1<0單調(diào)，則SKIPIF1<0的最大值為（）A．11

B．9

C．7

D．5【答案】B【解析】因?yàn)镾KIPIF1<0為函數(shù)SKIPIF1<0的零點(diǎn)，SKIPIF1<0為SKIPIF1<0圖像的對(duì)稱軸，所以SKIPIF1<0SKIPIF1<0（SKIPIF1<0，SKIPIF1<0為周期），得SKIPIF1<0（SKIPIF1<0）．又SKIPIF1<0在SKIPIF1<0單調(diào)，所以SKIPIF1<0，又當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0不單調(diào)；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0單調(diào)，滿足題意，故SKIPIF1<0，即SKIPIF1<0的最大值為9．6.已知銳角的內(nèi)角的對(duì)邊分別為，SKIPIF1<0SKIPIF1<0，，，則（）A． B． C． D．【答案】D【解析】SKIPIF1<0，SKIPIF1<0，由余弦定理解得SKIPIF1<0．7.的內(nèi)角的對(duì)邊分別為，若，，，則．【答案】SKIPIF1<0【解析】∵SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，由正弦定理得：SKIPIF1<0解得SKIPIF1<0．8.若函數(shù)SKIPIF1<0的最大值為2，則常數(shù)SKIPIF1<0的一個(gè)取值為________．【答案】SKIPIF1<0【解析】根據(jù)兩角和的正弦公式以及輔助角公式即可求得SKIPIF1<0，可得SKIPIF1<0，即可解出.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，故可取SKIPIF1<0.故答