新高考数学二轮复习易错题专练易错点05 三角变换及三角函数的性质（含解析）

易错点05三角变换及三角函数的性质易错题【01】忽略角的范围限制利用SKIPIF1<0求值一般要涉及到开方运算,此时要注意利用角的范围判断三角函数值的正负；根据三角函数值求角,一般是先求出该角的某一个三角函数值,再确定角的范围,确定角的范围时不仅要看已知条件中角的范围,还要挖掘隐含条件,根据三角函数值的符号及大小缩小角的范围易错题【02】不能全面理解三角函数性质致误本易错点主要包含以下几个问题：(1)求三角函数值域忽略定义域的限制；(2)确定三角函数的最小正周期,忽略三角变换的等价性；(3)求复合函数的单调性忽略对内函数单调性的判断.易错题【03】对平移理解不准确致误三角函数图象的左右平移是自变量x发生变化,如ωx→ωx±φ(φ>0)这个变化的实质是x→x±eq\f(φ,ω),所以平移的距离并不一定是φ.易错题【04】忽视弦函数的有界性在求与SKIPIF1<0,SKIPIF1<0有关的值域与范围是要注意SKIPIF1<0.易错题【05】用零点确定SKIPIF1<0的SKIPIF1<0,忽略图象的升降确定φ值时,往往以寻找“五点法”中的特殊点作为突破口．具体如下：“第一点”(即图象上升时与x轴的交点)为ωx＋φ＝0；“第二点”(即图象的“峰点”)为ωx＋φ＝eq\f(π,2)；“第三点”(即图象下降时与x轴的交点)为ωx＋φ＝π；“第四点”(即图象的“谷点”)为ωx＋φ＝eq\f(3π,2)；“第五点”为ωx＋φ＝2π. 01已知sinα＝eq\f(\r(5),5),sinβ＝eq\f(\r(10),10),且α,β为锐角,则α＋β＝________.【警示】本题出错的主要原因是忽略α＋β缩小的范围.【答案】eq\f(π,4)【问诊】因为α,β为锐角,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),又因为0<α＋β<π,所以α＋β＝eq\f(π,4).【叮嘱】根据三角函数值求角,要注意三角函数值的符号及大小对角的范围的限制1.(2022届福建省永安市高三10月月考)已知SKIPIF1<0的终边在第四象限,若SKIPIF1<0,则SKIPIF1<0()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】SKIPIF1<0的终边在第四象限,SKIPIF1<0,所以SKIPIF1<0,则SKIPIF1<0．故选A.2.已知SKIPIF1<0为锐角,SKIPIF1<0,SKIPIF1<0,则SKIPIF1<0的值为()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】∵SKIPIF1<0为锐角,SKIPIF1<0,SKIPIF1<0,∴SKIPIF1<0,SKIPIF1<0,∴SKIPIF1<0SKIPIF1<0,又SKIPIF1<0,∴SKIPIF1<0,故选B 02函数SKIPIF1<0得最小正周期为.【警示】根据SKIPIF1<0,得出SKIPIF1<0的最小正周期SKIPIF1<0是本题出错的主要原因.【答案】SKIPIF1<0【问诊】忽略了SKIPIF1<0与SKIPIF1<0定义域不相同,SKIPIF1<0对定义域内的SKIPIF1<0不成立,所以SKIPIF1<0不是SKIPIF1<0的周期.【叮嘱】研究三角函数的性质,若对三角函数进行变换,一点要保证变换的等价性.1．求函数SKIPIF1<0的值域.【答案】因为SKIPIF1<0=SKIPIF1<0=SKIPIF1<0,当SKIPIF1<0时SKIPIF1<0取得最小值SKIPIF1<0,当SKIPIF1<0时SKIPIF1<0取得最大值4,又因为SKIPIF1<0,所以SKIPIF1<0的值域为SKIPIF1<0.2.(2022届山东省青岛市高三上学期期中)若函数SKIPIF1<0的最小正周期为SKIPIF1<0,则函数SKIPIF1<0在SKIPIF1<0上的值域为_______．【答案】SKIPIF1<0.【解析】函数SKIPIF1<0的最小正周期为SKIPIF1<0,故SKIPIF1<0；则SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0.由SKIPIF1<0SKIPIF1<0,故可得SKIPIF1<0,则SKIPIF1<0,SKIPIF1<0的值域为SKIPIF1<0. 03(2022全国乙卷理T7)把函数SKIPIF1<0图像上所有点的横坐标缩短到原来的SKIPIF1<0倍,纵坐标不变,再把所得曲线向右平移SKIPIF1<0个单位长度,得到函数SKIPIF1<0的图像,则SKIPIF1<0SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【警示】平移方向或平移长度易出错【答案】B【问诊】SKIPIF1<0把函数SKIPIF1<0图像上所有点的横坐标缩短到原来的SKIPIF1<0倍,纵坐标不变,再把所得曲线向右平移SKIPIF1<0个单位长度,得到函数SKIPIF1<0的图像,SKIPIF1<0把函数SKIPIF1<0的图像,向左平移SKIPIF1<0个单位长度,得到SKIPIF1<0的图像；再把图像上所有点的横坐标变为原来的2倍,纵坐标不变,可得SKIPIF1<0的图像．故选SKIPIF1<0．【叮嘱】对于左右平移要注意要确定平移方向,法则是甲左减右,然后再确定平移长度,注意平移长度由x的变化来确定.1．要得到y＝sin(－3x)的图象,需将y＝eq\f(\r(2),2)(cos3x－sin3x)的图象A．向右平移eq\f(π,4)个单位 B．向左平移eq\f(π,4)个单位C．向右平移eq\f(π,12)个单位 D．向左平移eq\f(π,12)个单位【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＝sin(－3x)只需对x加上eq\f(π,12)即可,因而是对y＝eq\f(\r(2),2)(cos3x－sin3x)向左平移eq\f(π,12)个单位,故选D.2．(2022届海南省华侨中学高三11月月考)将函数y＝f(x)的图象向左平移SKIPIF1<0个单位长度,再把所得图象上所有点的横坐标伸长到原来的2倍得到y＝sinSKIPIF1<0的图象,则f(x)＝()A．sinSKIPIF1<0 B．sinSKIPIF1<0 C．sinSKIPIF1<0 D．sinSKIPIF1<0【答案】B【解析】将SKIPIF1<0的图象上各个点的横坐标变为原来的SKIPIF1<0,可得函数SKIPIF1<0的图象,再把函数SKIPIF1<0的图象向右平移SKIPIF1<0个单位,即可得到SKIPIF1<0的图象,所以SKIPIF1<0SKIPIF1<0,故选B. 04已知SKIPIF1<0,则SKIPIF1<0的取值范围是.【警示】由SKIPIF1<0得SKIPIF1<0,因为SKIPIF1<0,所以SKIPIF1<0.忽略了SKIPIF1<0.【答案】SKIPIF1<0【问诊】由题意可得SKIPIF1<0,所以SKIPIF1<0,由SKIPIF1<0得SKIPIF1<0,因为SKIPIF1<0,所以SKIPIF1<0.所以SKIPIF1<0的取值范围是SKIPIF1<0.【叮嘱】若一个式子中含有2个变量的弦函数,要保证每个弦函数的范围都在SKIPIF1<0上.1.(2022届甘肃省部分名校年高三上学期月考)已知函数SKIPIF1<0．若关于x的方程SKIPIF1<0在SKIPIF1<0上有解,则实数m的取值范围是()A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】SKIPIF1<0SKIPIF1<0,当SKIPIF1<0时,SKIPIF1<0,所以SKIPIF1<0,故SKIPIF1<0的值域为SKIPIF1<0,因为SKIPIF1<0在SKIPIF1<0上有解即SKIPIF1<0在SKIPIF1<0上有解,故SKIPIF1<0即SKIPIF1<0,故选C.2．(2022届四川省绵阳高三上学期一诊)已知SKIPIF1<0,则SKIPIF1<0的最大值为____________【答案】SKIPIF1<0【解析】SKIPIF1<0,SKIPIF1<0SKIPIF1<0,SKIPIF1<0,即SKIPIF1<0SKIPIF1<0又SKIPIF1<0,利用二次函数的性质知,当SKIPIF1<0时,SKIPIF1<0. 05函数f(x)＝sin(ωx＋φ)(x∈R,ω>0,|φ|<SKIPIF1<0)的部分图象如图所示,则SKIPIF1<0.【警示】本题出错的主要原因是观察图象可知,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在递减区间内.【答案】sin(2x＋eq\f(π,3))【问诊】观察图象可知,A＝1,T＝π,∴ω＝2,f(x)＝sin(2x＋φ),将(－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),则f(x)＝sin(2x＋eq\f(π,3))．【叮嘱】利用函数的零点确定确定φ值时,要注意该零点是在递增区间还是在递减区间.1.已知函数SKIPIF1<0(SKIPIF1<0,SKIPIF1<0)的部分图象如图所示,将SKIPIF1<0图象进行怎样的平移变换后得到的图象对应的函数SKIPIF1<0为奇函数()A．向左平移SKIPIF1<0个单位长度 B．向右平移SKIPIF1<0个单位长度C．向左平移SKIPIF1<0个单位长度 D．向右平移SKIPIF1<0个单位长度【答案】D【解析】由图象可知SKIPIF1<0,所以SKIPIF1<0,设SKIPIF1<0的最小正周期为SKIPIF1<0,则SKIPIF1<0,所以SKIPIF1<0,则SKIPIF1<0,所以SKIPIF1<0,因为函数SKIPIF1<0经过点SKIPIF1<0,可得SKIPIF1<0,解得SKIPIF1<0,因为SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,结合选项,当SKIPIF1<0时,SKIPIF1<0为奇函数,所以将SKIPIF1<0的图象向右平移SKIPIF1<0个单位长度后得到的函数SKIPIF1<0为奇函数.故选D.2.(2022届重庆市八中高三上学期月考)(多选题)函数SKIPIF1<0的图象如图所示,则关于函数SKIPIF1<0下列结论中正确的是()A．SKIPIF1<0 B．SKIPIF1<0C．对称轴为SKIPIF1<0 D．对称中心为SKIPIF1<0【答案】ABD【解析】对于A：由SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,A正确；对于B：因为SKIPIF1<0,SKIPIF1<0,解得SKIPIF1<0,故SKIPIF1<0,SKIPIF1<0,B正确；对于C：令SKIPIF1<0,解得SKIPIF1<0,即对称轴为SKIPIF1<0,故C错误；对于D：令SKIPIF1<0,解得SKIPIF1<0,即对称中心为SKIPIF1<0,故D正确.故选ABD.错1．(2022届河北省衡水市冀州区高三上学期期中)已知SKIPIF1<0且SKIPIF1<0,若SKIPIF1<0,则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,将SKIPIF1<0的值代入上式可得：SKIPIF1<0．故选D．2．若SKIPIF1<0,且SKIPIF1<0,则SKIPIF1<0()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】因为SKIPIF1<0,所以SKIPIF1<0,又SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,故选A.3．设锐角SKIPIF1<0的内角SKIPIF1<0,SKIPIF1<0,SKIPIF1<0的对边分别为SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,若SKIPIF1<0,则SKIPIF1<0的取值范围是()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.因为SKIPIF1<0为锐角三角形,所以SKIPIF1<0即SKIPIF1<0所以SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0的取值范围是SKIPIF1<0.故选A.4．(2021届安徽省合肥市高三下学期第二次教学质量检测)在SKIPIF1<0ABC中,已知SKIPIF1<0,SKIPIF1<0,则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.故选A.5．(2022届北京师范大学附属中学高三上学期期中)将SKIPIF1<0的图象向右平移SKIPIF1<0个单位,则所得图象的函数解析式为()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】将SKIPIF1<0的图象向右平移SKIPIF1<0个单位,得到SKIPIF1<0,故选D6．(2021届湘豫名校名校高三5月联考)已知函数SKIPIF1<0的图象沿SKIPIF1<0轴向右平移SKIPIF1<0个单位长度后得到函数SKIPIF1<0的图象,则SKIPIF1<0的一个对称中心为()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】SKIPIF1<0,因为函数SKIPIF1<0的图象沿SKIPIF1<0轴向右平移SKIPIF1<0个单位长度后得到函数SKIPIF1<0的图象,所以函数SKIPIF1<0的图象沿SKIPIF1<0轴向左平移SKIPIF1<0个单位长度后得到函数SKIPIF1<0的图象,所以SKIPIF1<0,所以SKIPIF1<0,显然SKIPIF1<0,因此SKIPIF1<0,因为SKIPIF1<0,所以令SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,令SKIPIF1<0,令SKIPIF1<0,可得一个对称中心为SKIPIF1<0,故选B7．(2022届四川省成都高三上学期阶段性检)函数SKIPIF1<0,(其中SKIPIF1<0,SKIPIF1<0,SKIPIF1<0)其图象如图所示,为了得到SKIPIF1<0的图象,可以将SKIPIF1<0的图象()A．向右平移SKIPIF1<0个单位长度 B．向右平移SKIPIF1<0个单位长度C．向左平移SKIPIF1<0个单位长度 D．向左平移SKIPIF1<0个单位长度【答案】B【解析】由函数图象可知：SKIPIF1<0,函数过SKIPIF1<0两点,设SKIPIF1<0的最小正周期为SKIPIF1<0,因为SKIPIF1<0,所以有SKIPIF1<0,而SKIPIF1<0,因此SKIPIF1<0,即SKIPIF1<0,因为SKIPIF1<0,所以SKIPIF1<0,因为SKIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0,因此SKIPIF1<0,而SKIPIF1<0,而SKIPIF1<0,因此该函数向右平移SKIPIF1<0个单位长度得到函数SKIPIF1<0的图象,故选B8．(2022届江苏省南通市高三上学期期中)已知函数SKIPIF1<0(SKIPIF1<0,SKIPIF1<0,SKIPIF1<0)的部分图象如图所示,将函数SKIPIF1<0的图象向左平移SKIPIF1<0个单位长度后得到SKIPIF1<0的图象,则下列说法正确的是()A．SKIPIF1<0B．SKIPIF1<0C．函数SKIPIF1<0为奇函数D．函数SKIPIF1<0在区间SKIPIF1<0上单调递减【答案】BCD【解析】SKIPIF1<0,则SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,∴SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,∴SKIPIF1<0,A错.SKIPIF1<0,SKIPIF