2025年上學(xué)期高二數(shù)學(xué)誘導(dǎo)公式應(yīng)用測試題一、選擇題（本大題共12小題，每小題5分，共60分）下列各式中，與$\sin(-330^\circ)$的值相等的是（）A.$\sin30^\circ$B.$\cos30^\circ$C.$\sin60^\circ$D.$\cos60^\circ$已知$\cos(\pi+\alpha)=\frac{1}{3}$，且$\alpha$是第二象限角，則$\sin(2\pi-\alpha)$的值為（）A.$-\frac{2\sqrt{2}}{3}$B.$\frac{2\sqrt{2}}{3}$C.$-\frac{1}{3}$D.$\frac{1}{3}$若$\tan(\pi-\alpha)=2$，則$\frac{\sin(\alpha-2\pi)+\cos(\pi-\alpha)}{\sin(\frac{\pi}{2}-\alpha)-\sin(\pi+\alpha)}$的值為（）A.$\frac{1}{3}$B.$-\frac{1}{3}$C.3D.-3已知角$\alpha$終邊上一點$P(-3,4)$，則$\cos(\frac{\pi}{2}+\alpha)$的值為（）A.$-\frac{3}{5}$B.$\frac{3}{5}$C.$-\frac{4}{5}$D.$\frac{4}{5}$函數(shù)$f(x)=\sin(\frac{3\pi}{2}-x)\cos(\pi+x)$是（）A.奇函數(shù)，最小正周期為$\pi$B.偶函數(shù)，最小正周期為$\pi$C.奇函數(shù)，最小正周期為$2\pi$D.偶函數(shù)，最小正周期為$2\pi$若$\sin(\frac{\pi}{2}+\theta)=\frac{3}{5}$，則$\cos2\theta$的值為（）A.$-\frac{7}{25}$B.$\frac{7}{25}$C.$-\frac{16}{25}$D.$\frac{16}{25}$已知$\alpha$為銳角，且$\tan(\pi-\alpha)+3=0$，則$\sin\alpha$的值為（）A.$\frac{1}{3}$B.$\frac{3\sqrt{10}}{10}$C.$\frac{\sqrt{10}}{10}$D.$\frac{3}{4}$函數(shù)$y=\sin(\frac{\pi}{2}-x)\sin(\pi+x)$的單調(diào)遞增區(qū)間是（）A.$k\pi-\frac{\pi}{4},k\pi+\frac{\pi}{4}$B.$k\pi+\frac{\pi}{4},k\pi+\frac{3\pi}{4}$C.$2k\pi-\frac{\pi}{4},2k\pi+\frac{\pi}{4}$D.$2k\pi+\frac{\pi}{4},2k\pi+\frac{3\pi}{4}$已知$\sin(\alpha+\beta)=\frac{1}{2}$，$\sin(\alpha-\beta)=\frac{1}{3}$，則$\frac{\tan\alpha}{\tan\beta}$的值為（）A.5B.-5C.$\frac{1}{5}$D.$-\frac{1}{5}$若$\theta\in(\frac{\pi}{2},\pi)$，則$\sqrt{1-2\sin(\pi+\theta)\sin(\frac{3\pi}{2}-\theta)}$等于（）A.$\sin\theta-\cos\theta$B.$\cos\theta-\sin\theta$C.$\pm(\sin\theta-\cos\theta)$D.$\sin\theta+\cos\theta$已知函數(shù)$f(x)=\sin(\omegax+\varphi)(\omega>0,|\varphi|<\frac{\pi}{2})$的部分圖象如圖所示，則$f(\frac{\pi}{3})$的值為（）A.$-\frac{1}{2}$B.$\frac{1}{2}$C.$-\frac{\sqrt{3}}{2}$D.$\frac{\sqrt{3}}{2}$已知$\sin\alpha+\cos\alpha=\frac{1}{5}$，且$\alpha\in(0,\pi)$，則$\tan(\frac{\pi}{4}-\alpha)$的值為（）A.$-\frac{1}{7}$B.$\frac{1}{7}$C.-7D.7二、填空題（本大題共4小題，每小題5分，共20分）計算：$\sin(-1560^\circ)\cos(-930^\circ)+\cos(-1380^\circ)\sin(-1410^\circ)=$________.已知$\tan\alpha=2$，則$\sin^2(\frac{\pi}{2}-\alpha)+\sin(3\pi+\alpha)\cos(2\pi-\alpha)=$________.函數(shù)$f(x)=\sin(\frac{\pi}{2}x+\frac{\pi}{3})$的最小正周期為________，$f(1)+f(2)+\cdots+f(2025)=$________.已知$\alpha$、$\beta$均為銳角，且$\sin\alpha=\frac{\sqrt{5}}{5}$，$\cos\beta=\frac{\sqrt{10}}{10}$，則$\alpha+\beta=$________.三、解答題（本大題共6小題，共70分）（本小題滿分10分）已知角$\alpha$的終邊經(jīng)過點$P(4,-3)$，求下列各式的值：(1)$\sin(\pi+\alpha)+\cos(-\alpha)$；(2)$\frac{\sin(\frac{3\pi}{2}-\alpha)\tan(\alpha-2\pi)}{\cos(\pi+\alpha)}$.（本小題滿分12分）化簡下列各式：(1)$\frac{\sin(2\pi-\alpha)\cos(\pi+\alpha)\cos(\frac{\pi}{2}+\alpha)\cos(\frac{11\pi}{2}-\alpha)}{\cos(\pi-\alpha)\sin(3\pi-\alpha)\sin(-\alpha)\sin(\frac{9\pi}{2}+\alpha)}$；(2)$\sin(k\pi-\alpha)\cos(k\pi+\alpha)(k\in\mathbb{Z})$.（本小題滿分12分）已知$\sin(\frac{\pi}{4}-\alpha)=\frac{5}{13}$，且$\alpha\in(0,\frac{\pi}{4})$，求$\frac{\cos2\alpha}{\cos(\frac{\pi}{4}+\alpha)}$的值.（本小題滿分12分）已知函數(shù)$f(x)=\sin(\frac{\pi}{2}-x)\sinx-\sqrt{3}\cos^2x$.(1)求$f(x)$的最小正周期和最大值；(2)討論$f(x)$在$[\frac{\pi}{6},\frac{2\pi}{3}]$上的單調(diào)性.（本小題滿分12分）已知$\alpha$、$\beta$為銳角，$\tan\alpha=\frac{1}{7}$，$\sin\beta=\frac{\sqrt{10}}{10}$.(1)求$\sin(\alpha+2\beta)$的值；(2)若$\alpha+\beta+\gamma=\frac{\pi}{2}$，求$\tan\gamma$的值.（本小題滿分12分）已知函數(shù)$f(x)=\sin(\omegax+\varphi)(\omega>0,0<\varphi<\pi)$的圖象經(jīng)過點$(\frac{\pi}{6},\frac{1}{2})$和$(\frac{\pi}{2},-1)$，且在區(qū)間$(\frac{\pi}{6},\frac{\pi}{2})$上單調(diào)遞減.(1)求函數(shù)$f(x)$的解析式；(2)設(shè)$\alpha\in(0,\pi)$，且$f(\frac{\alpha}{2}+\frac{\pi}{12})=-\frac{3}{5}$，求$\cos\alpha$的值.參考答案與解析一、選擇題A解析：$\sin(-330^\circ)=\sin(-360^\circ+30^\circ)=\sin30^\circ=\frac{1}{2}$.B解析：由$\cos(\pi+\alpha)=-\cos\alpha=\frac{1}{3}$，得$\cos\alpha=-\frac{1}{3}$.又$\alpha$是第二象限角，所以$\sin\alpha=\sqrt{1-\cos^2\alpha}=\frac{2\sqrt{2}}{3}$，則$\sin(2\pi-\alpha)=-\sin\alpha=-\frac{2\sqrt{2}}{3}$.A解析：由$\tan(\pi-\alpha)=-\tan\alpha=2$，得$\tan\alpha=-2$.原式$=\frac{\sin\alpha-\cos\alpha}{\cos\alpha+\sin\alpha}=\frac{\tan\alpha-1}{1+\tan\alpha}=\frac{-2-1}{1-2}=\frac{-3}{-1}=3$.C解析：由三角函數(shù)定義知，$\sin\alpha=\frac{4}{5}$，所以$\cos(\frac{\pi}{2}+\alpha)=-\sin\alpha=-\frac{4}{5}$.B解析：$f(x)=\sin(\frac{3\pi}{2}-x)\cos(\pi+x)=(-\cosx)(-\cosx)=\cos^2x=\frac{1+\cos2x}{2}$，所以$f(x)$是偶函數(shù)，最小正周期為$\pi$.A解析：由$\sin(\frac{\pi}{2}+\theta)=\cos\theta=\frac{3}{5}$，得$\cos2\theta=2\cos^2\theta-1=2\times(\frac{3}{5})^2-1=-\frac{7}{25}$.B解析：由$\tan(\pi-\alpha)=-\tan\alpha=-3$，得$\tan\alpha=3$.設(shè)直角三角形中對邊為3k，鄰邊為k，則斜邊為$\sqrt{10}k$，所以$\sin\alpha=\frac{3k}{\sqrt{10}k}=\frac{3\sqrt{10}}{10}$.A解析：$y=\sin(\frac{\pi}{2}-x)\sin(\pi+x)=\cosx(-\sinx)=-\frac{1}{2}\sin2x$，其單調(diào)遞增區(qū)間為$k\pi-\frac{\pi}{4},k\pi+\frac{\pi}{4}$.A解析：由$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta=\frac{1}{2}$，$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta=\frac{1}{3}$，解得$\sin\alpha\cos\beta=\frac{5}{12}$，$\cos\alpha\sin\beta=\frac{1}{12}$，則$\frac{\tan\alpha}{\tan\beta}=\frac{\sin\alpha\cos\beta}{\cos\alpha\sin\beta}=5$.A解析：原式$=\sqrt{1-2\sin\theta\cos\theta}=\sqrt{(\sin\theta-\cos\theta)^2}=|\sin\theta-\cos\theta|$.因為$\theta\in(\frac{\pi}{2},\pi)$，所以$\sin\theta>\cos\theta$，故原式$=\sin\theta-\cos\theta$.D解析：由圖象知，函數(shù)周期$T=4\times(\frac{\pi}{3}-\frac{\pi}{12})=\pi$，所以$\omega=2$.又$f(\frac{\pi}{12})=1$，即$\sin(2\times\frac{\pi}{12}+\varphi)=1$，得$\varphi=\frac{\pi}{3}$，則$f(x)=\sin(2x+\frac{\pi}{3})$，所以$f(\frac{\pi}{3})=\sin(\frac{2\pi}{3}+\frac{\pi}{3})=\sin\pi=0$.C解析：由$\sin\alpha+\cos\alpha=\frac{1}{5}$，平方得$1+2\sin\alpha\cos\alpha=\frac{1}{25}$，所以$\sin2\alpha=-\frac{24}{25}$.因為$\alpha\in(0,\pi)$，所以$\sin\alpha>0$，$\cos\alpha<0$，則$\sin\alpha-\cos\alpha=\sqrt{1-2\sin\alpha\cos\alpha}=\frac{7}{5}$.解得$\sin\alpha=\frac{4}{5}$，$\cos\alpha=-\frac{3}{5}$，所以$\tan\alpha=-\frac{4}{3}$，則$\tan(\frac{\pi}{4}-\alpha)=\frac{1-\tan\alpha}{1+\tan\alpha}=\frac{1+\frac{4}{3}}{1-\frac{4}{3}}=-7$.二、填空題$\frac{1}{2}$解析：原式$=\sin(-4\times360^\circ-120^\circ)\cos(-5\times180^\circ-30^\circ)+\cos(-4\times360^\circ+60^\circ)\sin(-4\times360^\circ+30^\circ)$$=\sin(-120^\circ)\cos(-30^\circ)+\cos60^\circ\sin30^\circ$$=(-\sin120^\circ)\cos30^\circ+\cos60^\circ\sin30^\circ$$=(-\frac{\sqrt{3}}{2})\times\frac{\sqrt{3}}{2}+\frac{1}{2}\times\frac{1}{2}=-\frac{3}{4}+\frac{1}{4}=-\frac{1}{2}$.$\frac{2}{5}$解析：原式$=\cos^2\alpha+\sin(\pi+\alpha)\cos(-\alpha)=\cos^2\alpha-\sin\alpha\cos\alpha=\frac{\cos^2\alpha-\sin\alpha\cos\alpha}{\sin^2\alpha+\cos^2\alpha}=\frac{1-\tan\alpha}{\tan^2\alpha+1}=\frac{1-2}{4+1}=-\frac{1}{5}$.4；$\frac{1}{2}+\frac{\sqrt{3}}{2}$解析：最小正周期$T=\frac{2\pi}{\frac{\pi}{2}}=4$.因為函數(shù)周期為4，所以$f(1)+f(2)+f(3)+f(4)=\sin(\frac{\pi}{2}+\frac{\pi}{3})+\sin(\pi+\frac{\pi}{3})+\sin(\frac{3\pi}{2}+\frac{\pi}{3})+\sin(2\pi+\frac{\pi}{3})=\cos\frac{\pi}{3}-\sin\frac{\pi}{3}-\cos\frac{\pi}{3}+\sin\frac{\pi}{3}=0$.2025=4×506+1，所以$f(1)+f(2)+\cdots+f(2025)=f(1)=\sin(\frac{\pi}{2}+\frac{\pi}{3})=\cos\frac{\pi}{3}=\frac{1}{2}$.$\frac{\pi}{4}$解析：因為$\alpha$、$\beta$均為銳角，所以$\cos\alpha=\sqrt{1-\sin^2\alpha}=\frac{2\sqrt{5}}{5}$，$\cos\beta=\sqrt{1-\sin^2\beta}=\frac{3\sqrt{10}}{10}$.則$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\frac{2\sqrt{5}}{5}\times\frac{3\sqrt{10}}{10}-\frac{\sqrt{5}}{5}\times\frac{\sqrt{10}}{10}=\frac{\sqrt{2}}{2}$.又$\alpha+\beta\in(0,\pi)$，所以$\alpha+\beta=\frac{\pi}{4}$.三、解答題解：(1)由三角函數(shù)定義知，$\sin\alpha=-\frac{3}{5}$，$\cos\alpha=\frac{4}{5}$.$\sin(\pi+\alpha)+\cos(-\alpha)=-\sin\alpha+\cos\alpha=\frac{3}{5}+\frac{4}{5}=\frac{7}{5}$.(2)$\frac{\sin(\frac{3\pi}{2}-\alpha)\tan(\alpha-2\pi)}{\cos(\pi+\alpha)}=\frac{(-\cos\alpha)\tan\alpha}{(-\cos\alpha)}=\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=-\frac{3}{4}$.解：(1)原式$=\frac{(-\sin\alpha)(-\cos\alpha)(-\sin\alpha)(-\sin\alpha)}{(-\cos\alpha)\sin\alpha(-\sin\alpha)\cos\alpha}=\frac{\sin^4\alpha}{\cos^2\alpha\sin^2\alpha}=\tan^2\alpha$.(2)當(dāng)k為偶數(shù)時，設(shè)k=2n(n∈Z)，則原式$=\sin(2n\pi-\alpha)\cos(2n\pi+\alpha)=(-\sin\alpha)\cos\alpha=-\sin\alpha\cos\alpha$；當(dāng)k為奇數(shù)時，設(shè)k=2n+1(n∈Z)，則原式$=\sin((2n+1)\pi-\alpha)\cos((2n+1)\pi+\alpha)=\sin(\pi-\alpha)\cos(\pi+\alpha)=\sin\alpha(-\cos\alpha)=-\sin\alpha\cos\alpha$.綜上，原式$=-\sin\alpha\cos\alpha=-\frac{1}{2}\sin2\alpha$.解：因為$\sin(\frac{\pi}{4}-\alpha)=\frac{5}{13}$，且$\alpha\in(0,\frac{\pi}{4})$，所以$\frac{\pi}{4}-\alpha\in(0,\frac{\pi}{4})$，則$\cos(\frac{\pi}{4}-\alpha)=\sqrt{1-\sin^2(\frac{\pi}{4}-\alpha)}=\frac{12}{13}$.$\frac{\cos2\alpha}{\cos(\frac{\pi}{4}+\alpha)}=\frac{\sin(\frac{\pi}{2}-2\alpha)}{\cos(\frac{\pi}{4}+\alpha)}=\frac{2\sin(\frac{\pi}{4}-\alpha)\cos(\frac{\pi}{4}-\alpha)}{\sin(\frac{\pi}{4}-\alpha)}=2\cos(\frac{\pi}{4}-\alpha)=2\times\frac{12}{13}=\frac{24}{13}$.解：(1)$f(x)=\sin(\frac{\pi}{2}-x)\sinx-\sqrt{3}\cos^2x=\cosx\sinx-\frac{\sqrt{3}}{2}(1+\cos2x)=\frac{1}{2}\sin2x-\frac{\sqrt{3}}{2}\cos2x-\frac{\sqrt{3}}{2}=\sin(2x-\frac{\pi}{3})-\frac{\sqrt{3}}{2}$.所以$f(x)$的最小正周期$T=\frac{2\pi}{2}=\pi$，最大值為$1-\frac{\sqrt{3}}{2}$.(2)令$2k\pi-\frac{\pi}{2}\leq2x-\frac{\pi}{3}\leq2k\pi+\frac{\pi}{2}(k\in\mathbb{Z})$，得$k\pi-\frac{\pi}{12}\leqx\leqk\pi+\frac{5\pi}{12}(k\in\mathbb{Z})$.當(dāng)$k=0$時，增區(qū)間為$[-\frac{\pi}{12},\frac{5\pi}{12}]$；當(dāng)$k=1$時，增區(qū)間為$[\frac{11\pi}{12},\frac{17\pi}{12}]$.所以$f(x)$在$[\frac{\pi}{6},\frac{5\pi}{12}]$上單調(diào)遞增，在$[\frac{5\pi}{12},\frac{2\pi}{3}]$上單調(diào)遞減.解：(1)因為$\beta$為銳角，$\sin\beta=\frac{\sqrt{10}}{10}$，所以$\cos\beta=\frac{3\sqrt{10}}{10}$，$\tan\beta=\frac{1}{3}$.$\tan2\beta=\frac{2\tan\beta}{1-\tan^2\beta}=\frac{2\times\frac{1}{3}}{1-\frac{1}{9}}=\frac{3}{4}$，則$\sin2\beta=\frac{3}{5}$，$\cos2\beta=\frac{4}{5}$.又$\tan\alpha=\frac{1}{7}$，且$\alpha$為銳角，所以$\sin\alpha=\frac{1}{\sqrt{50}}=\frac{\sqrt{2}}{10}$，$\cos\alpha=\frac{7\sqrt{2}}{10}$.$\sin(\alpha+2\beta)=\sin\alpha\cos2\beta+\cos\alpha\sin2\beta=\frac{\sqrt{2}}{10}\times\frac{4}{5}+\frac{7\sqrt{2}}{10}\times\frac{3}{5}=\frac{4\sqrt{2}+21\sqrt{2}}{50}=\frac{25\sqrt{2}}{50}=\frac{\sqrt{2}}{2}$.(2)由$\alpha+\beta+\gamma=\frac{\pi}{2}$，得$\gamma=\frac{\pi}{2}-(\alpha+\beta)$，所以$\tan\gamma=\tan(\frac{\pi}{2}-\alpha-\beta)=\cot(\alpha+\beta)=\frac{1}{\tan(\alpha+\beta)}$.$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\frac{\frac{1}{7}+\frac{1}{3}}{1-\frac{1}{21}}=\frac{10}{20}=\frac{1}{2}$，則$\tan\gamma=2$.解：(1)由題意知，函數(shù)$f(x)$的圖象經(jīng)過點$(\frac{\pi}{6},\frac{1}{2})$和$(\frac{\pi}{2},-1)$，且在區(qū)間$(\frac{\pi}{6},\frac{\pi}{2})$上單調(diào)遞減.設(shè)函數(shù)周期為T，則$\frac{T}{4}=\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3}$，所以$T=\frac{4\pi}{3}$，$\omega=\frac{2\pi}{T}=\frac{3}{2}$.又$f(\frac{\pi}{2})=\sin(\frac{3}{2}\times\frac{\pi}{2}+\varphi)=\sin(\frac{3\pi}{4}+\varphi)=-1$，則$\frac{3\pi}{4}+\varphi=\frac{3\pi}{2}+2k\pi(k\in\mathbb{Z})$，解得$\varphi=\frac{3\pi}{4}+2k\pi$.因為$0<\varphi<\pi$，所以$\varphi=\frac{3\pi}{4}$.則$f(x)=\sin(\frac{3}{2}x+\frac{3\pi}{4})$.驗證$f(\frac{\pi}{6})=\sin(\frac{3}{2}\times\frac{\pi}{6}+\frac{3\pi}{4})=\sin(\frac{\pi}{4}+\frac{3\pi}{4})=\sin\pi=0\neq\frac{1}{2}$，不符合題意.因此周期應(yīng)為$T=2\times(\frac{\pi}{2}-\frac{\pi}{6})=\frac{2\pi}{3}$，$\omega=3$.$f(\frac{\pi}{2})=\sin(3\times\frac{\pi}{2}+\varphi)=\sin(\frac{3\pi}{2}+\varphi)=-1$，則$\frac{3\pi}{2}+\varphi=\frac{3\pi}{2}+2k\pi$，得$\varphi=2k\pi$.因為$0<\varphi<\pi$，所以$\varphi=\pi$.$f(x)=\sin(3x+\pi)=-\sin3x$.驗證$f(\frac{\pi}{6})=-\sin\frac{\pi}{2}=-1\neq\frac{1}{2}$，仍不符合題意.重新設(shè)函數(shù)周期$T=4\times(\frac{\pi}{2}-\frac{\pi}{6})=\frac{4\pi}{3}$，$\omega=\frac{3}{2}$.$f(\frac{\pi}{6})=\sin(\frac{3}{2}\times\frac{\pi}{6}+\varphi)=\sin(\frac{\pi}{4}+\varphi)=\frac{1}{2}$，則$\frac{\pi}{4}+\varphi=\frac{\pi}{6}+2k\pi$或$\frac{5\pi}{6}+2k\pi$.因為函數(shù)在$(\frac{\pi}{6},\frac{\pi}{2})$上單調(diào)遞減，所以$\frac{\pi}{4}+\varphi=\frac{5\pi}{6}+2k\pi$，得$\varphi=\frac{5\pi}{6}-\frac{\pi}{4}+2k\pi=\frac{7\pi}{12}+2k\pi$.取$\varphi=\frac{7\pi}{12}$，則$f(x)=\sin(\frac{3}{2}x+\frac{7\pi}{12})$.驗證$f(\frac{\pi}{2})=\sin(\frac{3}{2}\times\fr