2025年三角函数历届真题及答案

2025年三角函数历届真题及答案

一、单项选择题1.已知\(\sin\alpha=\frac{3}{5}\)，且\(\alpha\)是第二象限角，则\(\cos\alpha\)的值为（）A.\(\frac{4}{5}\)B.\(-\frac{4}{5}\)C.\(\frac{3}{4}\)D.\(-\frac{3}{4}\)答案：B2.函数\(y=\sinx\)的最小正周期是（）A.\(\pi\)B.\(2\pi\)C.\(3\pi\)D.\(4\pi\)答案：B3.若\(\tan\alpha=2\)，则\(\frac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\)的值为（）A.\(3\)B.\(-3\)C.\(\frac{1}{3}\)D.\(-\frac{1}{3}\)答案：A4.函数\(y=\cos(2x+\frac{\pi}{3})\)的图象的一条对称轴方程是（）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\(\sin(\alpha+\frac{\pi}{6})=\frac{1}{3}\)，则\(\cos(\frac{\pi}{3}-\alpha)\)的值为（）A.\(\frac{1}{3}\)B.\(-\frac{1}{3}\)C.\(\frac{2\sqrt{2}}{3}\)D.\(-\frac{2\sqrt{2}}{3}\)答案：A6.函数\(y=A\sin(\omegax+\varphi)(A>0,\omega>0)\)的图象上一个最高点的坐标为\((\frac{\pi}{12},2)\)，与之相邻的一个最低点的坐标为\((\frac{7\pi}{12},-2)\)，则\(\omega\)的值为（）A.\(2\)B.\(4\)C.\(\frac{1}{2}\)D.\(\frac{1}{4}\)答案：A7.若\(\alpha\)是第四象限角，\(\sin\alpha=-\frac{5}{13}\)，则\(\tan\alpha\)等于（）A.\(-\frac{12}{5}\)B.\(\frac{12}{5}\)C.\(-\frac{5}{12}\)D.\(\frac{5}{12}\)答案：C8.函数\(y=\sin(2x-\frac{\pi}{4})\)的单调递增区间是（）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\(\sin\alpha+\cos\alpha=\frac{1}{5}\)，\(\alpha\in(0,\pi)\)，则\(\sin\alpha-\cos\alpha\)的值为（）A.\(\frac{7}{5}\)B.\(-\frac{7}{5}\)C.\(\frac{49}{25}\)D.\(\frac{12}{25}\)答案：A10.函数\(y=\tan(2x+\frac{\pi}{4})\)的定义域为（）A.\(\{x|x\neqk\pi+\frac{\pi}{2},k\inZ\}\)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、多项选择题1.以下关于三角函数的说法正确的是（）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.下列函数中，最小正周期为\(\pi\)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\(\sin\alpha=\frac{1}{2}\)，则\(\alpha\)可能的值为（）A.\(\frac{\pi}{6}\)B.\(\frac{5\pi}{6}\)C.\(\frac{13\pi}{6}\)D.\(\frac{17\pi}{6}\)答案：ABCD4.函数\(y=A\sin(\omegax+\varphi)\)（\(A>0,\omega>0\)）的图象的性质正确的有（）A.振幅是\(A\)B.周期\(T=\frac{2\pi}{\omega}\)C.初相是\(\varphi\)D.当\(\omegax+\varphi=2k\pi+\frac{\pi}{2}(k\inZ)\)时，\(y\)取得最大值\(A\)答案：ABCD5.若\(\alpha\)是第三象限角，则下列说法正确的是（）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.对于函数\(y=\cosx\)，以下说法正确的是（）A.它是偶函数B.它的图象关于\(y\)轴对称C.它在\([0,\pi]\)上单调递减D.它的最大值为\(1\)答案：ABCD7.已知\(\tan\alpha=3\)，则下列式子的值为（）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.函数\(y=\sin(2x-\frac{\pi}{3})\)的图象可以由\(y=\sinx\)的图象经过怎样的变换得到（）A.先将\(y=\sinx\)的图象上所有点的横坐标缩短到原来的\(\frac{1}{2}\)（纵坐标不变）B.再将所得图象向右平移\(\frac{\pi}{6}\)个单位长度C.先将\(y=\sinx\)的图象向右平移\(\frac{\pi}{3}\)个单位长度D.再将所得图象上所有点的横坐标缩短到原来的\(\frac{1}{2}\)（纵坐标不变）答案：AB9.以下三角函数值相等的是（）A.\(\sin\frac{\pi}{4}\)与\(\cos\frac{\pi}{4}\)B.\(\sin\frac{5\pi}{6}\)与\(\sin\frac{\pi}{6}\)C.\(\cos\frac{7\pi}{6}\)与\(\cos\frac{5\pi}{6}\)D.\(\tan\frac{\pi}{4}\)与\(\tan\frac{5\pi}{4}\)答案：ABCD10.已知\(\alpha\)为锐角，且\(\cos(\alpha+\frac{\pi}{6})=\frac{1}{3}\)，则（）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、判断题1.函数\(y=\sinx\)在\([0,2\pi]\)上是单调递增的。（×）2.\(\cos(\alpha+\beta)\cos\beta+\sin(\alpha+\beta)\sin\beta=\cos\alpha\)。（√）3.若\(\sin\alpha=\sin\beta\)，则\(\alpha=\beta+2k\pi\)，\(k\inZ\)。（×）4.函数\(y=\tanx\)的图象的对称中心是\((\frac{k\pi}{2},0)\)，\(k\inZ\)。（√）5.函数\(y=A\sin(\omegax+\varphi)\)的图象向左平移\(\frac{\varphi}{\omega}\)个单位长度就得到\(y=A\sin\omegax\)的图象。（×）6.\(\sin^2\frac{\alpha}{2}=\frac{1-\cos\alpha}{2}\)。（√）7.函数\(y=\cos(2x-\frac{\pi}{3})\)的最小正周期是\(\pi\)。（√）8.若\(\alpha\)是第二象限角，则\(\cos\alpha<0\)且\(\tan\alpha<0\)。（√）9.函数\(y=\sinx\)与\(y=\cosx\)的图象的对称轴完全相同。（×）10.\(\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\)对任意\(\alpha,\beta\)都成立。（×）四、简答题1.简述三角函数的诱导公式及其记忆方法。诱导公式是指三角函数中，利用周期性将角度比较大的三角函数，转换为角度比较小的三角函数的公式。记忆方法可概括为“奇变偶不变，符号看象限”。“奇变偶不变”指的是看\(\frac{\pi}{2}\)的奇数倍还是偶数倍，奇数倍则函数名改变，正弦变余弦，余弦变正弦等；偶数倍函数名不变。“符号看象限”是把\(\alpha\)看成锐角，看原函数在相应象限的符号来确定诱导公式的符号。2.求函数\(y=\sin(2x+\frac{\pi}{4})\)的单调区间。令\(2k\pi-\frac{\pi}{2}\leq2x+\frac{\pi}{4}\leq2k\pi+\frac{\pi}{2}\)，\(k\inZ\)，解这个不等式得\(k\pi-\frac{3\pi}{8}\leqx\leqk\pi+\frac{\pi}{8}\)，\(k\inZ\)，此为函数的单调递增区间。令\(2k\pi+\frac{\pi}{2}\leq2x+\frac{\pi}{4}\leq2k\pi+\frac{3\pi}{2}\)，\(k\inZ\)，解这个不等式得\(k\pi+\frac{\pi}{8}\leqx\leqk\pi+\frac{5\pi}{8}\)，\(k\inZ\)，此为函数的单调递减区间。3.已知\(\sin\alpha=\frac{4}{5}\)，\(\alpha\)是第二象限角，求\(\sin2\alpha\)，\(\cos2\alpha\)，\(\tan2\alpha\)的值。因为\(\sin\alpha=\frac{4}{5}\)，\(\alpha\)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\(y=\sinx\)的图象得到\(y=3\sin(2x-\frac{\pi}{3})\)的图象？先将\(y=\sinx\)的图象上所有点的横坐标缩短到原来的\(\frac{1}{2}\)（纵坐标不变），得到\(y=\sin2x\)的图象；再将\(y=\sin2x\)的图象向右平移\(\frac{\pi}{6}\)个单位长度，得到\(y=\sin(2x-\frac{\pi}{3})\)的图象；最后将\(y=\sin(2x-\frac{\pi}{3})\)的图象上所有点的纵坐标伸长到原来的\(3\)倍（横坐标不变），就得到\(y=3\sin(2x-\frac{\pi}{3})\)的图象。五