2025年三角函數(shù)歷屆真題及答案

一、單項(xiàng)選擇題1.已知\(\sin\alpha=\frac{3}{5}\)，且\(\alpha\)是第二象限角，則\(\cos\alpha\)的值為（）A.\(\frac{4}{5}\)B.\(-\frac{4}{5}\)C.\(\frac{3}{4}\)D.\(-\frac{3}{4}\)答案：B2.函數(shù)\(y=\sinx\)的最小正周期是（）A.\(\pi\)B.\(2\pi\)C.\(3\pi\)D.\(4\pi\)答案：B3.若\(\tan\alpha=2\)，則\(\frac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\)的值為（）A.\(3\)B.\(-3\)C.\(\frac{1}{3}\)D.\(-\frac{1}{3}\)答案：A4.函數(shù)\(y=\cos(2x+\frac{\pi}{3})\)的圖象的一條對(duì)稱軸方程是（）A.\(x=\frac{\pi}{6}\)B.\(x=\frac{\pi}{3}\)C.\(x=\frac{5\pi}{12}\)D.\(x=\frac{\pi}{12}\)答案：C5.已知\(\sin(\alpha+\frac{\pi}{6})=\frac{1}{3}\)，則\(\cos(\frac{\pi}{3}-\alpha)\)的值為（）A.\(\frac{1}{3}\)B.\(-\frac{1}{3}\)C.\(\frac{2\sqrt{2}}{3}\)D.\(-\frac{2\sqrt{2}}{3}\)答案：A6.函數(shù)\(y=A\sin(\omegax+\varphi)(A>0,\omega>0)\)的圖象上一個(gè)最高點(diǎn)的坐標(biāo)為\((\frac{\pi}{12},2)\)，與之相鄰的一個(gè)最低點(diǎn)的坐標(biāo)為\((\frac{7\pi}{12},-2)\)，則\(\omega\)的值為（）A.\(2\)B.\(4\)C.\(\frac{1}{2}\)D.\(\frac{1}{4}\)答案：A7.若\(\alpha\)是第四象限角，\(\sin\alpha=-\frac{5}{13}\)，則\(\tan\alpha\)等于（）A.\(-\frac{12}{5}\)B.\(\frac{12}{5}\)C.\(-\frac{5}{12}\)D.\(\frac{5}{12}\)答案：C8.函數(shù)\(y=\sin(2x-\frac{\pi}{4})\)的單調(diào)遞增區(qū)間是（）A.\([k\pi-\frac{\pi}{8},k\pi+\frac{3\pi}{8}](k\inZ)\)B.\([k\pi+\frac{3\pi}{8},k\pi+\frac{7\pi}{8}](k\inZ)\)C.\([k\pi-\frac{3\pi}{8},k\pi+\frac{\pi}{8}](k\inZ)\)D.\([k\pi+\frac{\pi}{8},k\pi+\frac{5\pi}{8}](k\inZ)\)答案：A9.已知\(\sin\alpha+\cos\alpha=\frac{1}{5}\)，\(\alpha\in(0,\pi)\)，則\(\sin\alpha-\cos\alpha\)的值為（）A.\(\frac{7}{5}\)B.\(-\frac{7}{5}\)C.\(\frac{49}{25}\)D.\(\frac{12}{25}\)答案：A10.函數(shù)\(y=\tan(2x+\frac{\pi}{4})\)的定義域?yàn)椋ǎ〢.\(\{x|x\neqk\pi+\frac{\pi}{2},k\inZ\}\)B.\(\{x|x\neq\frac{k\pi}{2}+\frac{\pi}{8},k\inZ\}\)C.\(\{x|x\neq\frac{k\pi}{2}+\frac{\pi}{4},k\inZ\}\)D.\(\{x|x\neq\frac{k\pi}{2}+\frac{3\pi}{8},k\inZ\}\)答案：D二、多項(xiàng)選擇題1.以下關(guān)于三角函數(shù)的說(shuō)法正確的是（）A.\(\sin^2\alpha+\cos^2\alpha=1\)B.\(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\)（\(\cos\alpha\neq0\)）C.\(\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta\)D.\(\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta\)答案：ABCD2.下列函數(shù)中，最小正周期為\(\pi\)的是（）A.\(y=\sin(2x+\frac{\pi}{3})\)B.\(y=\cos(2x-\frac{\pi}{6})\)C.\(y=\tan(2x+\frac{\pi}{4})\)D.\(y=|\sinx|\)答案：ABD3.已知\(\sin\alpha=\frac{1}{2}\)，則\(\alpha\)可能的值為（）A.\(\frac{\pi}{6}\)B.\(\frac{5\pi}{6}\)C.\(\frac{13\pi}{6}\)D.\(\frac{17\pi}{6}\)答案：ABCD4.函數(shù)\(y=A\sin(\omegax+\varphi)\)（\(A>0,\omega>0\)）的圖象的性質(zhì)正確的有（）A.振幅是\(A\)B.周期\(T=\frac{2\pi}{\omega}\)C.初相是\(\varphi\)D.當(dāng)\(\omegax+\varphi=2k\pi+\frac{\pi}{2}(k\inZ)\)時(shí)，\(y\)取得最大值\(A\)答案：ABCD5.若\(\alpha\)是第三象限角，則下列說(shuō)法正確的是（）A.\(\sin\alpha<0\)B.\(\cos\alpha<0\)C.\(\tan\alpha>0\)D.\(\sec\alpha<0\)（\(\sec\alpha=\frac{1}{\cos\alpha}\)）答案：ABCD6.對(duì)于函數(shù)\(y=\cosx\)，以下說(shuō)法正確的是（）A.它是偶函數(shù)B.它的圖象關(guān)于\(y\)軸對(duì)稱C.它在\([0,\pi]\)上單調(diào)遞減D.它的最大值為\(1\)答案：ABCD7.已知\(\tan\alpha=3\)，則下列式子的值為（）A.\(\frac{\sin\alpha}{\cos\alpha}=3\)B.\(\sin\alpha=3\cos\alpha\)C.\(\frac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}=2\)D.\(\sin^2\alpha=\frac{9}{10}\)答案：ABD8.函數(shù)\(y=\sin(2x-\frac{\pi}{3})\)的圖象可以由\(y=\sinx\)的圖象經(jīng)過(guò)怎樣的變換得到（）A.先將\(y=\sinx\)的圖象上所有點(diǎn)的橫坐標(biāo)縮短到原來(lái)的\(\frac{1}{2}\)（縱坐標(biāo)不變）B.再將所得圖象向右平移\(\frac{\pi}{6}\)個(gè)單位長(zhǎng)度C.先將\(y=\sinx\)的圖象向右平移\(\frac{\pi}{3}\)個(gè)單位長(zhǎng)度D.再將所得圖象上所有點(diǎn)的橫坐標(biāo)縮短到原來(lái)的\(\frac{1}{2}\)（縱坐標(biāo)不變）答案：AB9.以下三角函數(shù)值相等的是（）A.\(\sin\frac{\pi}{4}\)與\(\cos\frac{\pi}{4}\)B.\(\sin\frac{5\pi}{6}\)與\(\sin\frac{\pi}{6}\)C.\(\cos\frac{7\pi}{6}\)與\(\cos\frac{5\pi}{6}\)D.\(\tan\frac{\pi}{4}\)與\(\tan\frac{5\pi}{4}\)答案：ABCD10.已知\(\alpha\)為銳角，且\(\cos(\alpha+\frac{\pi}{6})=\frac{1}{3}\)，則（）A.\(\sin(\alpha+\frac{\pi}{6})=\frac{2\sqrt{2}}{3}\)B.\(\sin\alpha=\frac{2\sqrt{2}}{3}\times\frac{\sqrt{3}}{2}-\frac{1}{3}\times\frac{1}{2}\)C.\(\cos\alpha=\frac{1}{3}\times\frac{\sqrt{3}}{2}+\frac{2\sqrt{2}}{3}\times\frac{1}{2}\)D.\(\tan\alpha=\frac{2\sqrt{6}-1}{\sqrt{3}+2\sqrt{2}}\)答案：ABCD三、判斷題1.函數(shù)\(y=\sinx\)在\([0,2\pi]\)上是單調(diào)遞增的。（×）2.\(\cos(\alpha+\beta)\cos\beta+\sin(\alpha+\beta)\sin\beta=\cos\alpha\)。（√）3.若\(\sin\alpha=\sin\beta\)，則\(\alpha=\beta+2k\pi\)，\(k\inZ\)。（×）4.函數(shù)\(y=\tanx\)的圖象的對(duì)稱中心是\((\frac{k\pi}{2},0)\)，\(k\inZ\)。（√）5.函數(shù)\(y=A\sin(\omegax+\varphi)\)的圖象向左平移\(\frac{\varphi}{\omega}\)個(gè)單位長(zhǎng)度就得到\(y=A\sin\omegax\)的圖象。（×）6.\(\sin^2\frac{\alpha}{2}=\frac{1-\cos\alpha}{2}\)。（√）7.函數(shù)\(y=\cos(2x-\frac{\pi}{3})\)的最小正周期是\(\pi\)。（√）8.若\(\alpha\)是第二象限角，則\(\cos\alpha<0\)且\(\tan\alpha<0\)。（√）9.函數(shù)\(y=\sinx\)與\(y=\cosx\)的圖象的對(duì)稱軸完全相同。（×）10.\(\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\)對(duì)任意\(\alpha,\beta\)都成立。（×）四、簡(jiǎn)答題1.簡(jiǎn)述三角函數(shù)的誘導(dǎo)公式及其記憶方法。誘導(dǎo)公式是指三角函數(shù)中，利用周期性將角度比較大的三角函數(shù)，轉(zhuǎn)換為角度比較小的三角函數(shù)的公式。記憶方法可概括為“奇變偶不變，符號(hào)看象限”?！捌孀兣疾蛔儭敝傅氖强碶(\frac{\pi}{2}\)的奇數(shù)倍還是偶數(shù)倍，奇數(shù)倍則函數(shù)名改變，正弦變余弦，余弦變正弦等；偶數(shù)倍函數(shù)名不變?！胺?hào)看象限”是把\(\alpha\)看成銳角，看原函數(shù)在相應(yīng)象限的符號(hào)來(lái)確定誘導(dǎo)公式的符號(hào)。2.求函數(shù)\(y=\sin(2x+\frac{\pi}{4})\)的單調(diào)區(qū)間。令\(2k\pi-\frac{\pi}{2}\leq2x+\frac{\pi}{4}\leq2k\pi+\frac{\pi}{2}\)，\(k\inZ\)，解這個(gè)不等式得\(k\pi-\frac{3\pi}{8}\leqx\leqk\pi+\frac{\pi}{8}\)，\(k\inZ\)，此為函數(shù)的單調(diào)遞增區(qū)間。令\(2k\pi+\frac{\pi}{2}\leq2x+\frac{\pi}{4}\leq2k\pi+\frac{3\pi}{2}\)，\(k\inZ\)，解這個(gè)不等式得\(k\pi+\frac{\pi}{8}\leqx\leqk\pi+\frac{5\pi}{8}\)，\(k\inZ\)，此為函數(shù)的單調(diào)遞減區(qū)間。3.已知\(\sin\alpha=\frac{4}{5}\)，\(\alpha\)是第二象限角，求\(\sin2\alpha\)，\(\cos2\alpha\)，\(\tan2\alpha\)的值。因?yàn)閈(\sin\alpha=\frac{4}{5}\)，\(\alpha\)是第二象限角，所以\(\cos\alpha=-\sqrt{1-\sin^{2}\alpha}=-\frac{3}{5}\)。\(\sin2\alpha=2\sin\alpha\cos\alpha=2\times\frac{4}{5}\times(-\frac{3}{5})=-\frac{24}{25}\)；\(\cos2\alpha=1-2\sin^{2}\alpha=1-2\times(\frac{4}{5})^{2}=-\frac{7}{25}\)；\(\tan2\alpha=\frac{\sin2\alpha}{\cos2\alpha}=\frac{-\frac{24}{25}}{-\frac{7}{25}}=\frac{24}{7}\)。4.如何由\(y=\sinx\)的圖象得到\(y=3\sin(2x-\frac{\pi}{3})\)的圖象？先將\(y=\sinx\)的圖象上所有點(diǎn)的橫坐標(biāo)縮短到原來(lái)的\(\frac{1}{2}\)（縱坐標(biāo)不變），得到\(y=\sin2x\)的圖象；再將\(y=\sin2x\)的圖象向右平移\(\frac{\pi}{6}\)個(gè)單位長(zhǎng)度，得到\(y=\sin(2x-\frac{\pi}{3})\)的圖象；最后將\(y=\sin(2x-\frac{\pi}{3})\)的圖象上所有點(diǎn)的縱坐標(biāo)伸長(zhǎng)到原來(lái)的\(3\)倍（橫坐標(biāo)不變），就得到\(y=3\sin(2x-\frac{\pi}{3})\)的圖象。五