余弦函數(shù)經(jīng)典題目及答案

一、單項(xiàng)選擇題(每題2分,共20分)1.函數(shù)\(y=\cosx\)的周期是(）A.\(\pi\)B.\(2\pi\)C.\(3\pi\)D.\(4\pi\)2.\(\cos\frac{\pi}{3}\)的值是()A.\(\frac{1}{2}\)B.\(\frac{\sqrt{3}}{2}\)C.\(-\frac{1}{2}\)D.\(-\frac{\sqrt{3}}{2}\)3.函數(shù)\(y=\cos(2x)\)的最小正周期是()A.\(\pi\)B.\(2\pi\)C.\(\frac{\pi}{2}\)D.\(4\pi\)4.已知\(\cos\alpha=\frac{1}{3}\),\(\alpha\in(0,\frac{\pi}{2})\),則\(\sin\alpha\)的值為()A.\(\frac{2\sqrt{2}}{3}\)B.\(\frac{\sqrt{2}}{3}\)C.\(\frac{2}{3}\)D.\(\frac{1}{3}\)5.函數(shù)\(y=\cosx\)在區(qū)間\([0,\pi]\)上的單調(diào)性是()A.單調(diào)遞增B.單調(diào)遞減C.先增后減D.先減后增6.若\(\cos\theta=-\frac{4}{5}\),\(\theta\in(\frac{\pi}{2},\pi)\),則\(\tan\theta\)的值為()A.\(\frac{3}{4}\)B.\(-\frac{3}{4}\)C.\(\frac{4}{3}\)D.\(-\frac{4}{3}\)7.函數(shù)\(y=\cos(x+\frac{\pi}{4})\)的圖象是由\(y=\cosx\)的圖象()A.向左平移\(\frac{\pi}{4}\)個(gè)單位得到B.向右平移\(\frac{\pi}{4}\)個(gè)單位得到C.向上平移\(\frac{\pi}{4}\)個(gè)單位得到D.向下平移\(\frac{\pi}{4}\)個(gè)單位得到8.已知\(\cos(A-B)=\frac{1}{3}\),\(\cos(A+B)=-\frac{1}{3}\),則\(\cosA\cosB\)的值為()A.\(\frac{1}{3}\)B.\(0\)C.\(-\frac{1}{3}\)D.\(\frac{2}{3}\)9.函數(shù)\(y=\cos^2x-\sin^2x\)的最小正周期是()A.\(\pi\)B.\(2\pi\)C.\(\frac{\pi}{2}\)D.\(4\pi\)10.若\(\cosx=\frac{1}{2}\),\(x\in[0,2\pi]\),則\(x\)的值為()A.\(\frac{\pi}{3}\)B.\(\frac{5\pi}{3}\)C.\(\frac{\pi}{3}\)或\(\frac{5\pi}{3}\)D.\(\frac{\pi}{6}\)或\(\frac{11\pi}{6}\)二、多項(xiàng)選擇題(每題2分,共20分)1.以下關(guān)于余弦函數(shù)\(y=\cosx\)的性質(zhì)正確的是(）A.是偶函數(shù)B.值域是\([-1,1]\)C.周期是\(2\pi\)D.在\([0,\pi]\)上單調(diào)遞減2.已知\(\cos\alpha=\frac{\sqrt{3}}{2}\),則\(\alpha\)可能的值為()A.\(\frac{\pi}{6}\)B.\(\frac{11\pi}{6}\)C.\(-\frac{\pi}{6}\)D.\(\frac{5\pi}{6}\)3.函數(shù)\(y=\cos(3x)\)的性質(zhì)有()A.周期是\(\frac{2\pi}{3}\)B.是偶函數(shù)C.值域是\([-1,1]\)D.在\([0,\frac{\pi}{6}]\)上單調(diào)遞減4.下列等式成立的是()A.\(\cos(A+B)=\cosA\cosB-\sinA\sinB\)B.\(\cos(A-B)=\cosA\cosB+\sinA\sinB\)C.\(\cos^2x=\frac{1+\cos2x}{2}\)D.\(\sin^2x=\frac{1-\cos2x}{2}\)5.函數(shù)\(y=\cos(x-\frac{\pi}{3})\)的圖象可以由\(y=\cosx\)的圖象()A.向右平移\(\frac{\pi}{3}\)個(gè)單位得到B.向左平移\(\frac{\pi}{3}\)個(gè)單位得到C.關(guān)于\(x\)軸對稱變換得到D.關(guān)于\(y\)軸對稱變換得到6.若\(\cos\alpha=\frac{1}{2}\),\(\cos\beta=\frac{1}{2}\),\(\alpha,\beta\in[0,2\pi]\),則\(\alpha-\beta\)的值可能為()A.\(0\)B.\(\frac{2\pi}{3}\)C.\(\frac{4\pi}{3}\)D.\(2\pi\)7.函數(shù)\(y=\cosx\)在區(qū)間\([-\pi,\pi]\)上的對稱軸有()A.\(x=0\)B.\(x=\frac{\pi}{2}\)C.\(x=-\frac{\pi}{2}\)D.\(x=\pi\)8.已知\(\cos\alpha=m\),\(\alpha\in(\frac{\pi}{2},\pi)\),則\(\sin\alpha\)的值為()A.\(\sqrt{1-m^2}\)B.\(-\sqrt{1-m^2}\)C.\(\pm\sqrt{1-m^2}\)D.根據(jù)\(m\)的正負(fù)確定9.函數(shù)\(y=\cos^2x\)的性質(zhì)有()A.周期是\(\pi\)B.值域是\([0,1]\)C.是偶函數(shù)D.在\([0,\frac{\pi}{2}]\)上單調(diào)遞減10.若\(\cosx=a\),\(x\in[0,2\pi]\),則\(x\)的解的個(gè)數(shù)可能為()A.\(0\)B.\(1\)C.\(2\)D.\(3\)三、判斷題(每題2分,共20分)1.函數(shù)\(y=\cosx\)的圖象關(guān)于\(y\)軸對稱。()2.\(\cos(\frac{\pi}{2}+x)=\sinx\)。()3.函數(shù)\(y=\cos(2x+\frac{\pi}{3})\)的周期是\(\pi\)。()4.若\(\cos\alpha=\cos\beta\),則\(\alpha=\beta\)。（)5.函數(shù)\(y=\cosx\)在區(qū)間\((\frac{\pi}{2},\frac{3\pi}{2})\)上單調(diào)遞增。()6.\(\cos^2x-\sin^2x=\cos2x\)。()7.函數(shù)\(y=\cos(x+\pi)\)的圖象與\(y=\cosx\)的圖象關(guān)于\(x\)軸對稱。()8.已知\(\cos\alpha=\frac{1}{2}\),則\(\alpha=\frac{\pi}{3}+2k\pi\),\(k\inZ\)。()9.函數(shù)\(y=\cosx\)的最大值是\(1\),最小值是\(-1\)。(）10.若\(\cos(A+B)=0\),則\(A+B=\frac{\pi}{2}+k\pi\),\(k\inZ\)。（）四、簡答題(每題5分,共20分)1.求函數(shù)\(y=\cos(2x-\frac{\pi}{3})\)的單調(diào)遞增區(qū)間。2.已知\(\cos\alpha=\frac{3}{5}\),\(\alpha\in(0,\frac{\pi}{2})\),求\(\sin2\alpha\)的值。3.化簡\(\cos^2x+\sin^2x+\cos2x\)。4.求函數(shù)\(y=\cosx+1\)的值域。五、討論題(每題5分,共20分)1.討論余弦函數(shù)\(y=\cosx\)與正弦函數(shù)\(y=\sinx\)的關(guān)系。2.探討函數(shù)\(y=A\cos(\omegax+\varphi)\)(\(A\neq0\),\(\omega\gt0\))的圖象變換規(guī)律。3.分析余弦函數(shù)在實(shí)際生活中的應(yīng)用,如簡諧運(yùn)動(dòng)等。4.討論如何利用余弦函數(shù)的性質(zhì)解決三角函數(shù)方程的問題。答案：一、單項(xiàng)選擇題1.B2.A3.A4.A5.B6.B7.A8.B9.A10.C二、多項(xiàng)選擇題1.ABCD2.ABC3.ACD4.ABCD5.A6.ABCD7.AD8.B9.ABCD10.ABC三、判斷題1.√2.×3.√4.×5.×6.√7.√8.×9.√10.√四、簡答題1.令\(2k\pi-\pi\leqslant2x-\frac{\pi}{3}\leqslant2k\pi\),\(k\inZ\),解得\(k\pi-\frac{\pi}{3}\leqslantx\leqslantk\pi+\frac{\pi}{6}\),\(k\inZ\),所以單調(diào)遞增區(qū)間是\([k\pi-\frac{\pi}{3},k\pi+\frac{\pi}{6}]\),\(k\inZ\)。2.因?yàn)閈(\cos\alpha=\frac{3}{5}\),\(\alpha\in(0,\frac{\pi}{2})\),所以\(\sin\alpha=\frac{4}{5}\),則\(\sin2\alpha=2\sin\alpha\cos\alpha=2\times\frac{4}{5}\times\frac{3}{5}=\frac{24}{25}\)。3.\(\cos^2x+\sin^2x+\cos2x=1+\cos2x\)。4.因?yàn)閈(-1\leqslant\cosx\leqslant1\),所以\(0\leqslant\cosx+1\leqslant2\),值域是\([0,2]\)。五、討論題1.關(guān)系:\(\