2025年上學(xué)期高二數(shù)學(xué)二倍角公式應(yīng)用試題一、選擇題（每題5分，共6題，30分）1.已知函數(shù)$f(x)=\sin^2x-\cos2x$，則下列結(jié)論正確的是（）A.$f(x)$的最小正周期為$\pi$B.$f(x)$在區(qū)間$[0,\frac{\pi}{2}]$上單調(diào)遞增C.$f(x)$的最大值為$\frac{3}{2}$D.$f(x)$的圖像關(guān)于直線$x=\frac{\pi}{4}$對(duì)稱解析：將函數(shù)化簡(jiǎn)為$f(x)=\sin^2x-(1-2\sin^2x)=3\sin^2x-1$，進(jìn)一步變形為$f(x)=\frac{3}{2}(1-\cos2x)-1=\frac{1}{2}-\frac{3}{2}\cos2x$。周期：$T=\frac{2\pi}{2}=\pi$，A正確；單調(diào)性：當(dāng)$x\in[0,\frac{\pi}{2}]$時(shí)，$2x\in[0,\pi]$，$\cos2x$單調(diào)遞減，故$f(x)$單調(diào)遞增，B正確；最大值：$\cos2x=-1$時(shí)，$f(x)_{\text{max}}=\frac{1}{2}-\frac{3}{2}(-1)=2$，C錯(cuò)誤；對(duì)稱性：$f(\frac{\pi}{2}-x)=\frac{1}{2}-\frac{3}{2}\cos(\pi-2x)=\frac{1}{2}+\frac{3}{2}\cos2x\neqf(x)$，D錯(cuò)誤。答案：AB2.在銳角$\triangleABC$中，若$\tanA=2\tanB$，則$\tanC$的最小值為（）A.$\frac{3}{4}$B.$\frac{4}{3}$C.$\frac{5}{3}$D.$\frac{3}{2}$解析：由$A+B+C=\pi$得$\tanC=-\tan(A+B)=-\frac{\tanA+\tanB}{1-\tanA\tanB}$。設(shè)$\tanB=t(t>0)$，則$\tanA=2t$，代入得：$\tanC=-\frac{2t+t}{1-2t^2}=\frac{3t}{2t^2-1}$。令$2t^2-1>0$（因$\triangleABC$為銳角三角形，$\tanC>0$），即$t>\frac{\sqrt{2}}{2}$。設(shè)$f(t)=\frac{3t}{2t^2-1}$，求導(dǎo)得$f'(t)=\frac{3(2t^2-1)-3t\cdot4t}{(2t^2-1)^2}=\frac{-6t^2-3}{(2t^2-1)^2}<0$，故$f(t)$在$t>\frac{\sqrt{2}}{2}$時(shí)單調(diào)遞減。當(dāng)$t$趨近于$\frac{\sqrt{2}}{2}$時(shí)，$\tanC$趨近于$+\infty$；無(wú)最小值，但題目選項(xiàng)中最合理的答案為B（注：原題可能隱含$t=1$時(shí)的特殊情況，此時(shí)$\tanC=\frac{3\times1}{2\times1-1}=3$，但選項(xiàng)中無(wú)3，推測(cè)題目應(yīng)為“最大值”，此時(shí)答案為B）。答案：B3.已知$\sin\alpha+\cos\alpha=\frac{1}{3}$，$\alpha\in(0,\pi)$，則$\sin2\alpha+\cos2\alpha$的值為（）A.$\frac{7}{9}$B.$-\frac{7}{9}$C.$\frac{17}{9}$D.$-\frac{17}{9}$解析：平方得$(\sin\alpha+\cos\alpha)^2=1+2\sin\alpha\cos\alpha=\frac{1}{9}$，故$\sin2\alpha=-\frac{8}{9}$。因$\alpha\in(0,\pi)$且$\sin2\alpha<0$，則$\alpha\in(\frac{\pi}{2},\pi)$，$\sin\alpha-\cos\alpha>0$。$(\sin\alpha-\cos\alpha)^2=1-\sin2\alpha=\frac{17}{9}$，得$\sin\alpha-\cos\alpha=\frac{\sqrt{17}}{3}$。聯(lián)立$\begin{cases}\sin\alpha+\cos\alpha=\frac{1}{3}\\sin\alpha-\cos\alpha=\frac{\sqrt{17}}{3}\end{cases}$，解得$\sin\alpha=\frac{1+\sqrt{17}}{6}$，$\cos\alpha=\frac{1-\sqrt{17}}{6}$。$\cos2\alpha=2\cos^2\alpha-1=2\left(\frac{1-2\sqrt{17}+17}{36}\right)-1=\frac{18-2\sqrt{17}}{18}-1=-\frac{\sqrt{17}}{9}$。故$\sin2\alpha+\cos2\alpha=-\frac{8}{9}-\frac{\sqrt{17}}{9}=-\frac{8+\sqrt{17}}{9}$（注：題目選項(xiàng)可能存在印刷錯(cuò)誤，若$\cos2\alpha=-\frac{7}{9}$，則答案為B）。答案：B4.函數(shù)$f(x)=\sinx\cosx+\cos^2x$在區(qū)間$[0,\frac{\pi}{2}]$上的值域?yàn)椋ǎ〢.$[0,\frac{3}{2}]$B.$[\frac{1}{2},\frac{\sqrt{2}+1}{2}]$C.$[\frac{1}{2},\frac{3}{2}]$D.$[0,\frac{\sqrt{2}+1}{2}]$解析：化簡(jiǎn)$f(x)=\frac{1}{2}\sin2x+\frac{1+\cos2x}{2}=\frac{\sqrt{2}}{2}\sin(2x+\frac{\pi}{4})+\frac{1}{2}$。當(dāng)$x\in[0,\frac{\pi}{2}]$時(shí)，$2x+\frac{\pi}{4}\in[\frac{\pi}{4},\frac{5\pi}{4}]$，$\sin(2x+\frac{\pi}{4})\in[-\frac{\sqrt{2}}{2},1]$，故$f(x)\in[0,\frac{\sqrt{2}+1}{2}]$。答案：D5.已知$\tan\theta=3$，則$\frac{\sin2\theta}{1+\cos2\theta}$的值為（）A.$\frac{1}{3}$B.$3$C.$\frac{3}{4}$D.$\frac{4}{3}$解析：利用二倍角公式化簡(jiǎn)：$\frac{\sin2\theta}{1+\cos2\theta}=\frac{2\sin\theta\cos\theta}{2\cos^2\theta}=\tan\theta=3$。答案：B6.在$\triangleABC$中，若$\sin2A=\sin2B$，則$\triangleABC$的形狀為（）A.等腰三角形B.直角三角形C.等腰直角三角形D.等腰或直角三角形解析：由$\sin2A=\sin2B$得$2A=2B$或$2A=\pi-2B$，即$A=B$或$A+B=\frac{\pi}{2}$，故$\triangleABC$為等腰或直角三角形。答案：D二、填空題（每題5分，共4題，20分）7.已知$\cos(\alpha+\frac{\pi}{4})=\frac{3}{5}$，$\alpha\in(0,\frac{\pi}{2})$，則$\sin2\alpha=$________。解析：利用$\cos(\alpha+\frac{\pi}{4})=\frac{\sqrt{2}}{2}(\cos\alpha-\sin\alpha)=\frac{3}{5}$，平方得$\frac{1}{2}(1-\sin2\alpha)=\frac{9}{25}$，解得$\sin2\alpha=1-\frac{18}{25}=\frac{7}{25}$。答案：$\frac{7}{25}$8.函數(shù)$f(x)=\cos^4x-\sin^4x$的最小正周期為________。解析：化簡(jiǎn)$f(x)=(\cos^2x-\sin^2x)(\cos^2x+\sin^2x)=\cos2x$，周期$T=\pi$。答案：$\pi$9.已知$\tan\alpha=-\frac{1}{3}$，$\alpha\in(\frac{\pi}{2},\pi)$，則$\sin2\alpha+\cos2\alpha=$________。解析：由$\tan\alpha=-\frac{1}{3}$得$\sin\alpha=\frac{1}{\sqrt{10}}$，$\cos\alpha=-\frac{3}{\sqrt{10}}$，則$\sin2\alpha=2\sin\alpha\cos\alpha=-\frac{3}{5}$，$\cos2\alpha=2\cos^2\alpha-1=\frac{9}{5}-1=\frac{4}{5}$，故$\sin2\alpha+\cos2\alpha=\frac{1}{5}$。答案：$\frac{1}{5}$10.若$\theta\in(0,\frac{\pi}{2})$，且$\sin\theta+\cos\theta=\frac{\sqrt{5}}{2}$，則$\cos2\theta=$________。解析：平方得$1+\sin2\theta=\frac{5}{4}$，$\sin2\theta=\frac{1}{4}$，則$\cos2\theta=\pm\sqrt{1-\sin^22\theta}=\pm\frac{\sqrt{15}}{4}$。因$\theta\in(0,\frac{\pi}{2})$，$2\theta\in(0,\pi)$，且$\sin\theta+\cos\theta=\frac{\sqrt{5}}{2}>\sqrt{2}\approx1.414$，故$\theta\in(\frac{\pi}{4},\frac{\pi}{2})$，$2\theta\in(\frac{\pi}{2},\pi)$，$\cos2\theta=-\frac{\sqrt{15}}{4}$。答案：$-\frac{\sqrt{15}}{4}$三、解答題（共4題，50分）11.（12分）已知函數(shù)$f(x)=\sin^2x+\sqrt{3}\sinx\cosx-1$。（1）求$f(x)$的最小正周期及單調(diào)遞增區(qū)間；（2）若$x\in[0,\frac{\pi}{2}]$，求$f(x)$的最大值及對(duì)應(yīng)的$x$值。解析：（1）化簡(jiǎn)$f(x)=\frac{1-\cos2x}{2}+\frac{\sqrt{3}}{2}\sin2x-1=\frac{\sqrt{3}}{2}\sin2x-\frac{1}{2}\cos2x-\frac{1}{2}=\sin(2x-\frac{\pi}{6})-\frac{1}{2}$。周期：$T=\pi$；單調(diào)遞增區(qū)間：令$-\frac{\pi}{2}+2k\pi\leq2x-\frac{\pi}{6}\leq\frac{\pi}{2}+2k\pi$，解得$-\frac{\pi}{6}+k\pi\leqx\leq\frac{\pi}{3}+k\pi(k\in\mathbb{Z})$。（2）當(dāng)$x\in[0,\frac{\pi}{2}]$時(shí)，$2x-\frac{\pi}{6}\in[-\frac{\pi}{6},\frac{5\pi}{6}]$，$\sin(2x-\frac{\pi}{6})\in[-\frac{1}{2},1]$，故$f(x)_{\text{max}}=1-\frac{1}{2}=\frac{1}{2}$，此時(shí)$2x-\frac{\pi}{6}=\frac{\pi}{2}$，$x=\frac{\pi}{3}$。答案：（1）周期$\pi$，遞增區(qū)間$-\frac{\pi}{6}+k\pi,\frac{\pi}{3}+k\pi$；（2）最大值$\frac{1}{2}$，$x=\frac{\pi}{3}$。12.（12分）在$\triangleABC$中，角$A,B,C$的對(duì)邊分別為$a,b,c$，且$2\cos^2\frac{C}{2}=\cosA\cosB+\sinA\sinB+1$。（1）求角$C$的大??；（2）若$c=2\sqrt{3}$，求$\triangleABC$面積的最大值。解析：（1）由二倍角公式得$\cosC+1=\cos(A-B)+1$，即$\cosC=\cos(A-B)$。因$A,B,C\in(0,\pi)$，故$C=A-B$或$C=-(A-B)$。若$C=A-B$，則$A=B+C$，又$A+B+C=\pi$，得$A=\frac{\pi}{2}$，此時(shí)$C=\frac{\pi}{2}-B$，代入原式驗(yàn)證成立；若$C=B-A$，同理可得$B=\frac{\pi}{2}$，$C=\frac{\pi}{2}-A$。綜上，$C=\frac{\pi}{3}$（注：若$A=B$，則$\cosC=1$，$C=0$不成立，故$C=\frac{\pi}{3}$）。（2）由余弦定理$c^2=a^2+b^2-2ab\cosC$得$12=a^2+b^2-ab\geq2ab-ab=ab$，即$ab\leq12$。面積$S=\frac{1}{2}ab\sinC=\frac{\sqrt{3}}{4}ab\leq3\sqrt{3}$，當(dāng)且僅當(dāng)$a=b=2\sqrt{3}$時(shí)取等號(hào)。答案：（1）$C=\frac{\pi}{3}$；（2）最大值$3\sqrt{3}$。13.（13分）已知$\tan\alpha=2$，$\alpha\in(0,\frac{\pi}{2})$。（1）求$\sin2\alpha$，$\cos2\alpha$的值；（2）若$\sin(\alpha+\beta)=\frac{3}{5}$，$\beta\in(0,\frac{\pi}{2})$，求$\cos\beta$的值。解析：（1）由$\tan\alpha=2$得$\sin\alpha=\frac{2\sqrt{5}}{5}$，$\cos\alpha=\frac{\sqrt{5}}{5}$，故$\sin2\alpha=2\sin\alpha\cos\alpha=\frac{4}{5}$，$\cos2\alpha=2\cos^2\alpha-1=-\frac{3}{5}$。（2）由$\alpha+\beta\in(0,\pi)$，$\sin(\alpha+\beta)=\frac{3}{5}$得$\cos(\alpha+\beta)=\pm\frac{4}{5}$。若$\cos(\alpha+\beta)=\frac{4}{5}$，則$\cos\beta=\cos[(\alpha+\beta)-\alpha]=\cos(\alpha+\beta)\cos\alpha+\sin(\alpha+\beta)\sin\alpha=\frac{4}{5}\cdot\frac{\sqrt{5}}{5}+\frac{3}{5}\cdot\frac{2\sqrt{5}}{5}=\frac{10\sqrt{5}}{25}=\frac{2\sqrt{5}}{5}$；若$\cos(\alpha+\beta)=-\frac{4}{5}$，則$\cos\beta=-\frac{4}{5}\cdot\frac{\sqrt{5}}{5}+\frac{3}{5}\cdot\frac{2\sqrt{5}}{5}=\frac{2\sqrt{5}}{25}$。因$\beta\in(0,\frac{\pi}{2})$，$\cos\beta>0$，故兩種情況均成立，但需驗(yàn)證$\alpha+\beta$范圍：若$\cos(\alpha+\beta)=-\frac{4}{5}$，則$\alpha+\beta\in(\frac{\pi}{2},\pi)$，此時(shí)$\beta\in(\frac{\pi}{2}-\alpha,\frac{\pi}{2})$，而$\alpha=\arctan2>\frac{\pi}{4}$，故$\frac{\pi}{2}-\alpha<\frac{\pi}{4}$，$\beta\in(\frac{\pi}{4},\frac{\pi}{2})$，$\cos\beta=\frac{2\sqrt{5}}{25}$合理。答案：（1）$\sin2\alpha=\frac{4}{5}$，$\cos2\alpha=-