di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Trigonometric Functions


Question: $已知\frac{\pi }{4}<\alpha <\frac{\pi }{2}, f(\alpha )=\frac{2\cos \left(\frac{\pi }{2}+\alpha \right) \cdot \sqrt{1-\sin  2\alpha }}{\tan (\alpha +\pi ) \cdot \sqrt{2+2\cos  2\alpha }}.$
$若f(\alpha)=-\frac{1}{5},求\tan 2\alpha的值.$

Solution:
$\because  \frac{\pi }{4} < \alpha  < \frac{\pi }{2}, \therefore \sin  \alpha  > \cos  \alpha  > 0.$
$由 \left\{\begin{matrix}\cos  \alpha -\sin  \alpha =-\frac{1}{5},\\ \sin ^2\alpha +\cos ^2\alpha =1,\end{matrix}\right. 可得 \left\{\begin{matrix}\sin  \alpha =\frac{4}{5},\\ \cos  \alpha =\frac{3}{5},\end{matrix}\right. 所以 \tan  \alpha  = \frac{\sin  \alpha }{\cos  \alpha } = \frac{4}{3}， 所以 \tan  2\alpha  = \frac{2\tan  \alpha }{1-\tan ^2\alpha } = \frac{2 \times \frac{4}{3}}{1-\frac{16}{9}} = -\frac{24}{7}.$

Final answer: $-\frac{24}{7}$
Subfield: Trigonometric Functions


Question: $已知\cos  \alpha =\frac{1}{3}，且\alpha 是第四象限角.$
$求\sin  2\alpha 和\cos  2\alpha 的值；$

Solution:
$由cos \alpha = \frac{1}{3}，sin^2\alpha + cos^2\alpha = 1得，sin^2\alpha = 1 - cos^2\alpha = \frac{8}{9}，又\because \alpha是第四象限角，\therefore sin \alpha=-\sqrt{\frac{8}{9}} = - \frac{2\sqrt{2}}{3}，\therefore sin 2\alpha = 2sin \alpha cos \alpha= -\frac{4\sqrt{2}}{9}，cos 2\alpha = cos^2\alpha -sin^2\alpha = -\frac{7}{9}.$

Final answer: $sin 2\alpha = -\frac{4\sqrt{2}}{9}，cos 2\alpha = -\frac{7}{9}.$
Subfield: Trigonometric Functions


Question: $已知\cos  \alpha =\frac{1}{3}，且\alpha 是第四象限角.$
$求tan(\alpha -\frac{\pi }{4})的值.$

Solution:
$由(1)可知\tan  \alpha = \frac{\sin  \alpha }{\cos  \alpha }=-2\sqrt{2}，$

$\therefore \tan \left(\alpha -\frac{\pi }{4}\right)=\frac{\tan  \alpha -\tan  \frac{\pi }{4}}{1+\tan  \alpha  \cdot \tan  \frac{\pi }{4}}=\frac{-2\sqrt{2}-1}{1+(-2\sqrt{2}) \times 1}=\frac{9+4\sqrt{2}}{7}.$

Final answer: \frac{9+4\sqrt{2}}{7}
Subfield: Trigonometric Functions


Question: $已知 2\sin \alpha=2\sin^2\frac{\alpha}{2}-1$
$求sin \alpha cos \alpha + cos 2\alpha的值；$

Solution:
$由已知得2sin \alpha =-cos \alpha，所以tan \alpha =-\frac{1}{2}，$

$则sin \alpha cos \alpha +cos 2\alpha = \frac{\sin  \alpha \cos  \alpha +{\cos }^2\alpha -{\sin }^2\alpha }{{\sin }^2\alpha +{\cos }^2\alpha } = \frac{\tan  \alpha +1-{\tan }^2\alpha }{{\tan }^2\alpha +1} = \frac{1}{5}.$

Final answer: $\frac{1}{5}$
Subfield: Trigonometric Functions


Question: $已知 2\sin \alpha=2\sin^2\frac{\alpha}{2}-1$
$已知\alpha\in (0,\pi )，\beta\in (0,\frac{\pi }{2})，且tan^{2}\beta-6tan \beta=1，求\alpha+2\beta的值.$

Solution:
$由tan^2\beta -6\tan  \beta =1，可得tan 2\beta =\frac{2\tan  \beta  }{1-{\tan }^2\beta  }=-\frac{1}{3}，$

$则tan(\alpha +2\beta )=\frac{\tan  \alpha  +\tan  2\beta  }{1-\tan  \alpha  \tan  2\beta  }=\frac{-\frac{1}{2}-\frac{1}{3}}{1-\left(-\frac{1}{2}\right) \times \left(-\frac{1}{3}\right)}=-1.$

$因为\beta \in \left(0,\frac{\pi }{2}\right)，所以2\beta \in (0,\pi )，又tan 2\beta =-\frac{1}{3}＞-\frac{\sqrt{3}}{3}，则2\beta \in \left(\frac{5\pi }{6},\pi \right)，$

$因为\alpha \in (0,\pi )，tan \alpha =-\frac{1}{2}＞-\frac{\sqrt{3}}{3}，所以\alpha \in \left(\frac{5\pi }{6},\pi \right)，$

$则\alpha +2\beta \in \left(\frac{5\pi }{3},2\pi \right)，所以\alpha +2\beta =\frac{7\pi }{4}.$

Final answer: \frac{7\pi }{4}
Subfield: Trigonometric Functions


Question: $已知\alpha,\beta\in (0,\frac{\pi }{2})，\cos  \alpha=\frac{3}{5}，\cos (\alpha+\beta)=\frac{5}{13}.$
$求\sin  \beta 的值；$

Solution:
$因为 \alpha  ，\beta  均为锐角，\cos  \alpha  = \frac{3}{5} ，所以0 < \alpha  + \beta  < \pi ，\sin  \alpha  = \sqrt{1-\cos^2\alpha } = \frac{4}{5} ，又\cos (\alpha  + \beta ) = \frac{5}{13} > 0, 所以0 < \alpha  + \beta  < \frac{\pi}{2}，\sin (\alpha  + \beta ) = \sqrt{1-\cos^2(\alpha  + \beta )} = \frac{12}{13}. $

$所以\sin  \beta  = \sin  [\alpha  + \beta  -\alpha ] = \sin (\alpha  + \beta ) \cos  \alpha  - \cos (\alpha  + \beta )・\sin  \alpha  = \frac{12}{13} \times  \frac{3}{5} - \frac{5}{13} \times  \frac{4}{5} = \frac{16}{65}.$

Final answer: $\frac{16}{65}$
Subfield: Trigonometric Functions


Question: $已知\alpha,\beta\in (0,\frac{\pi }{2})，\cos  \alpha=\frac{3}{5}，\cos (\alpha+\beta)=\frac{5}{13}.$
$求 cos(\alpha + 2\beta) 的值.$

Solution:
$因为\sin  \beta =\frac{16}{65},且\beta 为锐角，$
$所以\cos  \beta =\sqrt{1-{\sin }^2\beta  }=\frac{63}{65}，$
$所以\cos (\alpha +2\beta )=\cos [(\alpha +\beta )+\beta ]$
$=\cos (\alpha +\beta )\cos  \beta -\sin (\alpha +\beta )\sin  \beta =\frac{5}{13}\times \frac{63}{65}-\frac{12}{13}\times \frac{16}{65}=\frac{123}{845}.$

Final answer: \frac{123}{845}
Subfield: Trigonometric Functions


Question: $设函数f(x)=\sin\left(\omega x-\frac{\pi }{6}\right)+\sin\left(\omega x-\frac{\pi }{2}\right)，其中0<\omega<3.已知f\left(\frac{\pi }{6}\right)=0.$
$求\omega；$

Solution:
$f(x) = \sin\left(\omega x-\frac{\pi }{6}\right)+\sin\left(\omega x-\frac{\pi }{2}\right)$
$=\frac{\sqrt{3}}{2}\sin \omega x-\frac{1}{2}\cos \omega x-\cos \omega x$
$=\frac{\sqrt{3}}{2}\sin \omega x-\frac{3}{2}\cos \omega x=\sqrt{3}\left(\frac{1}{2}\sin  \omega x-\frac{\sqrt{3}}{2}\cos  \omega x\right)$
$=\sqrt{3}\sin\left(\omega x-\frac{\pi }{3}\right)$
$因为f\left(\frac{\pi }{6}\right)=0，所以\frac{\omega \pi }{6}-\frac{\pi }{3}=k\pi，k\in Z.$
$故\omega=6k+2，k\in Z，又0<\omega<3，所以\omega=2.$

Final answer: $\omega=2$
Subfield: Elementary Functions


Question: $设函数f(x)=\sin\left(\omega x-\frac{\pi }{6}\right)+\sin\left(\omega x-\frac{\pi }{2}\right)，其中0<\omega<3.已知f\left(\frac{\pi }{6}\right)=0.$
$将函数y=f(x)的图象上各点的横坐标伸长为原来的2倍(纵坐标不变)，再将得到的图象向左平移 \frac{\pi }{4} 个单位，得到函数y=g(x)的图象，求g(x)在 \left[-\frac{\pi }{4},\frac{3\pi }{4}\right] 上的最小值.$

Solution:
$由(1)得 f(x)=\sqrt{3}sin\left(2x-\frac{\pi }{3}\right)，$

$所以 g(x)=\sqrt{3}sin\left(x+\frac{\pi }{4}-\frac{\pi }{3}\right)=\sqrt{3}sin\left(x-\frac{\pi }{12}\right).$

$因为 x\in \left[-\frac{\pi }{4},\frac{3\pi }{4}\right]，所以 x-\frac{\pi }{12}\in \left[-\frac{\pi }{3},\frac{2\pi }{3}\right]，$

$当 x-\frac{\pi }{12}=-\frac{\pi }{3}，即 x=-\frac{\pi }{4} 时， g(x) 取得最小值 -\frac{3}{2}.$

Final answer: $-\frac{3}{2}$
Subfield: Trigonometric Functions


Question: $已知函数f(x)= \sin(2\pi - x) \sin(\frac{3\pi}{2} - x) - \sqrt{3} \cos^{2} x + \sqrt{3}$
$当x\in [0,\frac{7\pi }{12}]时，求f(x)的最小值．$

Solution:
$由(1)可得 f(x) = \sin(2x - \frac{\pi}{3}) + \frac{\sqrt{3}}{2}.$
$由 x \in [0, \frac{7\pi}{12}]，可得 2x \in [0, \frac{7\pi}{6}]，2x - \frac{\pi}{3} \in [-\frac{\pi}{3}, \frac{5\pi}{6}]，$
$则 \sin(2x - \frac{\pi}{3}) \in [-\frac{\sqrt{3}}{2}, 1]，\sin(2x - \frac{\pi}{3}) + \frac{\sqrt{3}}{2} \in [0, \frac{\sqrt{3} + 2}{2}]，故 x \in [0, \frac{7\pi}{12}] 时，f(x) 的最小值为 0.$

Final answer: $0$
Subfield: Trigonometric Functions


Question: $已知向量a=(1,-\sqrt{3})，b=(\sin  x,\cos  x)，f(x)=a\cdot b.$
$若f(\theta )=0，求\frac{2\cos ^2\frac{\theta }{2}-\sin  \theta -1}{\sqrt{2}\sin \left(\theta +\frac{\pi }{4}\right)}的值；$

Solution:
$因为a=(1,-\sqrt{3})，b=(\sin  x,\cos  x)，所以f(x)=a\cdot b=\sin  x-\sqrt{3}\cos  x，因为f(\theta )=0，所以\sin  \theta -\sqrt{3}\cos  \theta =0，所以\tan  \theta =\sqrt{3}，所以\frac{2\cos ^2\frac{\theta }{2}-\sin  \theta -1}{\sqrt{2}\sin \left(\theta +\frac{\pi }{4}\right)}=\frac{\cos  \theta -\sin  \theta }{\sin  \theta +\cos  \theta }=\frac{1-\tan  \theta }{\tan  \theta +1}=\frac{1-\sqrt{3}}{\sqrt{3}+1}=-2+\sqrt{3}.$

Final answer: $-2+\sqrt{3}$
Subfield: Trigonometric Functions


Question: $已知函数f(x)=\cos^2{x}+\sqrt{3} \sin x \cos x - \frac{1}{2} (x \in \mathbb{R}).$
$求f(x)的最小正周期；$

Solution:
$f(x) = \cos ^2 x + \sqrt{3} \sin  x \cos  x - \frac{1}{2} = \frac{1+\cos  2x}{2} + \frac{\sqrt{3}}{2} \sin  2x - \frac{1}{2} = \sin \left(2x+\frac{\pi }{6}\right), 所以 T = \frac{2\pi }{|\omega |} = \pi .$

Final answer: $\pi$
Subfield: Trigonometric Functions


Question: $在\triangle ABC中，内角A，B，C所对的边分别为a，b，c.已知a \sin  A=4b \sin  B，ac=\sqrt{5}(a^2-b^2-c^2).$
$求\cos  A的值；$

Solution:
$由a sinA=4b sinB及\frac{a}{\sin A}= \frac{b}{\sin B}，得a=2b.$

$由ac=\sqrt{5}(a^2-b^2-c^2)及余弦定理的推论，得\cos  A=\frac{b^2+c^2-a^2}{2bc}=\frac{-\frac{\sqrt{5}}{5}ac}{ac}=-\frac{\sqrt{5}}{5}。$

Final answer: $-\frac{\sqrt{5}}{5}$
Subfield: Trigonometric Functions


Question: $在\triangle ABC中，内角A，B，C所对的边分别为a，b，c.已知a \sin  A=4b \sin  B，ac=\sqrt{5}(a^2-b^2-c^2).$
$求\sin (2B-A)的值.$

Solution:
$由(1)可得\sin  A=\frac{2\sqrt{5}}{5}，代入a\sin  A=4b\sin  B，得\sin  B=\frac{a\sin  A}{4b}=\frac{\sqrt{5}}{5}.由(1)知，A为钝角，所以\cos  B=\sqrt{1-\sin ^2B}=\frac{2\sqrt{5}}{5}.于是\sin  2B=2\sin  B\cos  B=\frac{4}{5}， \cos  2B=1-2\sin ^2B=\frac{3}{5}，故\sin (2B-A)=\sin  2B\cos  A-\cos  2B\sin  A=\frac{4}{5}\times \left(-\frac{\sqrt{5}}{5}\right)-\frac{3}{5}\times \frac{2\sqrt{5}}{5}=-\frac{2\sqrt{5}}{5}.$

Final answer: $-\frac{2\sqrt{5}}{5}$
Subfield: Trigonometric Functions


Question: $在\triangle ABC中，3sin A=2sin B, tan C=\sqrt{35}.$
求cos 2C；

Solution:
$\because tan C = \sqrt{35}，$
$\therefore cos C = \frac{1}{6}，$
$\therefore cos 2C = 2\times (\frac{1}{6})^{2} -1 = -\frac{17}{18}.$

Final answer: $-\frac{17}{18}$
Subfield: Trigonometric Functions


Question: $在\triangle ABC中，3sin A=2sin B, tan C=\sqrt{35}.$
$若AC-BC=1，求\triangle ABC的周长.$

Solution:
设\triangle ABC的内角A, B, C的对边分别为a, b, c.

\because 3sin A=2sin B，\therefore 3a=2b，

\because AC-BC=b-a=1，\therefore a=2，b=3.

$由余弦定理可得 c^{2}=a^{2}+b^{2}-2ab \cos  C=4+9-2\times 2\times 3\times \frac{1}{6}=11，则c= \sqrt{11}，$

$故\triangle ABC的周长为5+ \sqrt{11}.$

Final answer: $5+ \sqrt{11}$
Subfield: Trigonometric Functions


Question: $\triangle ABC的内角A、B、C的对边分别为a、b、c,已知\triangle ABC的面积为  \frac{a^2}{3\\sin  A} .$
$求sin B sin C;$

Solution:
由题设得
$\frac{1}{2}ac\sin B=\frac{a^2}{3\sin A}，$
$即\frac{1}{2}c\sin B=\frac{a}{3\sin A}.$
$由正弦定理得\frac{1}{2}\sin C\sin B=\frac{\sin A}{3\sin A}，$
$故\sin  B\sin  C=\frac{2}{3}.$

Final answer: $\frac{2}{3}$
Subfield: Trigonometric Functions


Question: $\triangle ABC的内角A、B、C的对边分别为a、b、c,已知\triangle ABC的面积为  \frac{a^2}{3\\sin  A} .$
$若6\cos  B \cos  C = 1,a=3,求\triangle ABC的周长.$

Solution:
$由题设及(1)得 \cos  B \cos  C - \sin  B \sin  C = -\frac{1}{2}，$
$即 \cos (B+C) = -\frac{1}{2}. 所以 B+C = \frac{2\pi }{3}，故 A = \frac{\pi }{3}.$
$由题设得 \frac{1}{2} bc \sin  A = \frac{a^2}{3 \sin  A}，即 bc=8.$
$由余弦定理得 b^2 + c^2 - bc = 9，即 (b + c)^2 - 3bc = 9，得 b + c = \sqrt{33}. 故\triangle ABC的周长为 3 + \sqrt{33}.$

Final answer: $3 + \sqrt{33}$
Subfield: Trigonometric Functions


Question: $在平面四边形ABCD中，\angle ADC=90^\circ ，\angle A=45^\circ ，AB=2，BD=5。$
求$cos\angle ADB$；

Solution:
在$\triangle ABD$中，由正弦定理得

$\frac{BD}{\sin \angle A} = \frac{AB}{\sin \angle ADB}$

由题设知，

$\frac{5}{\sin  45 ^{\circ} } = \frac{2}{\sin \angle ADB}$

$所以 \\sin  \angle  ADB = \frac{\sqrt{2}}{5} $

$由题设知，\angle ADB＜90^\circ ，所以 \\cos  \angle  ADB = \sqrt{1-\frac{2}{25}} = \frac{\sqrt{23}}{5}$

Final answer: $\frac{\sqrt{23}}{5}$
Subfield: Trigonometric Functions


Question: $在平面四边形ABCD中，\angle ADC=90^\circ ，\angle A=45^\circ ，AB=2，BD=5。$
$若DC=2\sqrt{2}，求BC.$

Solution:
$由题设及(1)知，\cos \angle BDC = \sin \angle ADB = \frac{\sqrt{2}}{5}. $

$在\triangle BCD中，由余弦定理得 BC^2 = BD^2 + DC^2 - 2\cdot BD\cdot DC\cdot \cos \angle BDC = 25 + 8 -2\times 5\times 2\sqrt{2}\times \frac{\sqrt{2}}{5} = 25. $

$所以BC = 5.$

Final answer: $BC = 5$
Subfield: Trigonometric Functions


Question: $江西某中学校园内有块扇形空地OPQ，经测量其半径为60 m，圆心角为\frac{\pi }{3}，学校准备在此扇形空地上修建一所矩形室内篮球场ABCD，初步设计方案1如图1所示。$

<img_2716>
$取弧PQ的中点E，连接OE，设\angle BOE=\alpha ，试用\alpha 表示方案1中矩形ABCD的面积，并求其最大值；$

Solution:
$如图所示，设OE交AD于点M，交BC于点N，显然矩形ABCD关于OE所在直线对称，点M、N分别为AD、BC的中点，\angle BOE=\alpha ，0<\alpha <\frac{\pi }{6}.  $

在Rt\triangle ONB中，BN=60sin \alpha ，ON=60cos \alpha .  

$OM=\frac{DM}{\tan \frac{\pi }{6}}=\sqrt{3}DM=\sqrt{3}CN=60\sqrt{3}sin \alpha ，  $

$\therefore MN=ON-OM=60cos \alpha -60\sqrt{3}sin \alpha ，  $

$即AB=60cos \alpha -60\sqrt{3}sin \alpha ，  $

$又BC=2CN=120sin \alpha ，故矩形ABCD的面积S=AB\cdot BC=3 600(cos \alpha -\sqrt{3}sin \alpha )\cdot 2sin \alpha =3 600(2sin \alpha cos \alpha -2\sqrt{3}sin^2\alpha )=3 600[sin 2\alpha -\sqrt{3}(1-cos 2\alpha )]=3 600(sin 2\alpha +\sqrt{3}cos 2\alpha -\sqrt{3})=7 200sin\left(2\alpha +\frac{\pi }{3}\right)-3 600\sqrt{3}.  $

$\because 0<\alpha <\frac{\pi }{6}，\therefore 0<2\alpha <\frac{\pi }{3}，\therefore \frac{\pi }{3}<2\alpha +\frac{\pi }{3}<\frac{2\pi }{3}. 故当2\alpha +\frac{\pi }{3}=\frac{\pi }{2}，即\alpha =\frac{\pi }{12}时，S取得最大值3 600(2-\sqrt{3})，$
$\therefore 矩形ABCD面积的最大值为3 600(2-\sqrt{3})m^2.  $

<img_2712>

Final answer: 3600(2-\sqrt{3})m^2
Subfield: Trigonometric Functions


Question: $江西某中学校园内有块扇形空地OPQ，经测量其半径为60 m，圆心角为\frac{\pi }{3}，学校准备在此扇形空地上修建一所矩形室内篮球场ABCD，初步设计方案1如图1所示。$

![图1](https://cdntk.53zaixian.com/tq/1632611650396233731/img/compression/M_5839487b9709daa6fc03df22d74a3e13.jpg)
$取弧PQ的中点E，连接OE，设\angle BOE=\alpha ，试用\alpha 表示方案1中矩形ABCD的面积，并求其最大值；$

Solution:
$如图所示，设OE交AD于点M，交BC于点N，显然矩形ABCD关于OE所在直线对称，点M、N分别为AD、BC的中点，\angle BOE=\alpha ，0<\alpha <\frac{\pi }{6}.  $

在Rt\triangle ONB中，BN=60sin \alpha ，ON=60cos \alpha .  

$OM=\frac{DM}{\tan \frac{\pi }{6}}=\sqrt{3}DM=\sqrt{3}CN=60\sqrt{3}sin \alpha ，  $

$\therefore MN=ON-OM=60cos \alpha -60\sqrt{3}sin \alpha ，  $

$即AB=60cos \alpha -60\sqrt{3}sin \alpha ，  $

$又BC=2CN=120sin \alpha ，故矩形ABCD的面积S=AB\cdot BC=3 600(cos \alpha -\sqrt{3}sin \alpha )\cdot 2sin \alpha =3 600(2sin \alpha cos \alpha -2\sqrt{3}sin^2\alpha )=3 600[sin 2\alpha -\sqrt{3}(1-cos 2\alpha )]=3 600(sin 2\alpha +\sqrt{3}cos 2\alpha -\sqrt{3})=7 200sin\left(2\alpha +\frac{\pi }{3}\right)-3 600\sqrt{3}.  $

$\because 0<\alpha <\frac{\pi }{6}，\therefore 0<2\alpha <\frac{\pi }{3}，\therefore \frac{\pi }{3}<2\alpha +\frac{\pi }{3}<\frac{2\pi }{3}. 故当2\alpha +\frac{\pi }{3}=\frac{\pi }{2}，即\alpha =\frac{\pi }{12}时，S取得最大值3 600(2-\sqrt{3})，$
$\therefore 矩形ABCD面积的最大值为3 600(2-\sqrt{3})m^2.  $

![24AXGK5SX333.TIF](https://cdntk.53zaixian.com/tq/1632611651277037570/img/compression/M_e044160cffcec5f0a3082dfc7d6ea9a1.jpg)

Final answer: 3600(2-\sqrt{3})m^2
Subfield: Trigonometric Functions


Question: $在\triangle ABC中，内角A，B，C所对的边分别为a，b，c，且 \sin (2A+B)=\sin  B-\sin  A。$
求C的大小；

Solution:
$因为 \sin (2A+B)=\sin  B-\sin  A，$

$所以 \sin (A+B+A)=\sin (C+A)-\sin  A，$

$故 \sin (\pi+A-C)=\sin (C+A)-\sin  A，$

$则 \sin (C-A)=\sin (C+A)-\sin  A，$

$$\sin $ C $\cos $ A- $\cos $ C $\sin $ A=$\sin $ C $\cos $ A+ $\cos $ C $\sin $ A- $\sin $ A$，

$2 \cos  C \sin  A= \sin  A，由于 0< A,C < \pi，所以 \sin  A>0，所以 \cos  C= \frac{1}{2}，$

$则 C 为锐角，且 C= \frac{\pi }{3}.$

Final answer: $\frac{\pi }{3}$
Subfield: Trigonometric Functions


Question: $在\triangle ABC中，内角A，B，C所对的边分别为a，b，c，且 \sin (2A+B)=\sin  B-\sin  A。$
$若CD平分\angle ACB交AB于D且CD=\sqrt{3}，求\triangle ABC面积的最小值.$

Solution:
$在\triangle ACD中，由正弦定理得\frac{\sqrt{3}}{\\sin   A}=\frac{AD}{\\sin   \frac{\pi }{6}}，$
$在\triangle BCD中，由正弦定理得\frac{\sqrt{3}}{\\sin   B}=\frac{BD}{\\sin   \frac{\pi }{6}}，$
$所以AD\cdot sin A=BD\cdot sin B，由正弦定理得\frac{AD}{BD}=\frac{b}{a}.$
$在\triangle ACD中，由余弦定理得AD^2=b^2+3-2\sqrt{3}b\cdot \\cos  \frac{\pi }{6}=b^2-3b+3，$
$在\triangle BCD中，由余弦定理得BD^2=a^2+3-2\sqrt{3}a\cdot \\cos  \frac{\pi }{6}=a^2-3a+3，$
$所以\frac{AD^2}{BD^2}=\frac{b^2-3b+3}{a^2-3a+3}=\frac{b^2}{a^2}，整理得(a+b-ab)(a-b)=0，所以a=b或a+b=ab.$
$当a=b时，\triangle ABC是等边三角形，CD\perp AB，AD=BD=1，$
$AB=AC=BC=2，所以S_{\triangle ABC}=\frac{1}{2}\times2\times2\times \\sin  \frac{\pi }{3}=\sqrt{3}.$
$当a+b=ab时，ab=a+b\geq2\sqrt{ab},\sqrt{ab}\geq2,ab\geq4，当且仅当a=b=2时等号成立，$
$所以S_{\triangle ABC}=\frac{1}{2}ab \\sin  C\geq\frac{1}{2}\times4\times\frac{\sqrt{3}}{2}=\sqrt{3}.$
$综上所述，\triangle ABC面积的最小值为\sqrt{3}.$

Final answer: $\sqrt{3}$
Subfield: Trigonometric Functions


Question: $已知函数f(x)=2\sqrt{3}sin xcos x+2\cos ^2x+m，其中m为实常数.$
$设集合A=\left\{x\left|\right.-\frac{\pi }{12}\leq x\leq \frac{\pi }{6}\right\}，已知当x\in A时，f(x)的最小值为2，当x\in A时，求f(x)的最大值.$

Solution:
$由(1)得f(x)=2\sin \left(2x+\frac{\pi }{6}\right)+1+m，当-\frac{\pi }{12} \leq x \leq \frac{\pi }{6}时，0 \leq 2x+\frac{\pi }{6} \leq \frac{\pi }{2}，所以0 \leq \sin \left(2x+\frac{\pi }{6}\right) \leq 1，所以 f(x)_{min}=2\times 0+1+m=2，解得m=1，所以 f(x)_{max}=2\times 1+1+1=4，所以当x \in A时，f(x)的最大值为4.$

Final answer: $4$
Subfield: Trigonometric Functions


Question: $已知函数f(x)=\frac{\sqrt{3}}{3}\left[\\cos  \left(2x+\frac{\pi }{6}\right)+4\\sin   x\\cos   x\right]+1，x\in \mathbb{R}．$
$若 f\left(\frac{x_0}{2}\right)=\frac{3}{5}，x_0\in [-\frac{\pi}{2},0]，求cos 2x_0；$

Solution:
$f(x)=\frac{\sqrt{3}}{3}\left[\\cos  \left(2x+\frac{\pi }{6}\right)+4\\sin   x\\cos   x\right]+1$
$=\frac{\sqrt{3}}{3}\left[\frac{\sqrt{3}}{2}\\cos   2x-\frac{1}{2}\\sin   2x+2\\sin   2x\right]+1$
$=\frac{\sqrt{3}}{2}\sin  2x+\frac{1}{2}\cos  2x+1=\sin \left(2x+\frac{\pi }{6}\right)+1$
$由f\left(\frac{x_0}{2}\right)=\frac{\sqrt{3}}{2}\sin  x_0+\frac{1}{2}\cos  x_0+1=\frac{3}{5}，得\sqrt{3}\sin  x_0+\cos  x_0=-\frac{4}{5}$
$即 \cos  x_0=-\frac{4}{5}-\sqrt{3}\sin  x_0，又因为 \\sin ^2 x_0+\cos ^2x_0=1，可得 4\sin ^2x_0+\frac{8\sqrt{3}}{5}\sin  x_0-\frac{9}{25}=0$
$由x_0\in\left[-\frac{\pi }{2},0\right]，解得 \sin  x_0=\frac{-\sqrt{21}-2\sqrt{3}}{10}$
$则有 \cos  2x_0=1-2\\sin ^2x_0=\frac{17-12\sqrt{7}}{50}．$

Final answer: $\frac{17-12\sqrt{7}}{50}$
Subfield: Sequence


Question: $已知集合A=\{x|x=2n-1,n\in N^\}，B=\{x|x=3^{n},n\in N^\}$，$将A与B中的所有元素按从小到大的顺序排列构成数列\{a_n\}(若有相同元素，按重复方式计入排列)为1，3，3，5，7，9，9，11，\ldots .设数列\{a_n\}的前n项和为S_n.$
$求S_{50}的值.$

Solution:
$2\times 50-1=99，3^4=81<99，3^5=243>99，$

$因此数列{a_n}的前50项中含有B中的元素为3，9，27，81，共有4项，含有A中的元素为1，3，5，7，9，\ldots ，27，29，\ldots ，79，81，83，\ldots ，2\times 46-1=91，共有46项，$

$\therefore S_{50}=\frac{46 \times (1+91)}{2}+(3+9+27+81)=2 116+120=2 236.$

Final answer: 2236
Subfield: Sequence


Question: $已知数列a_n的前n项和为S_n，a_1=-11，a_2=-9，且S_{n+1}+S_{n-1}=2S_n+2 (n\geq 2)．$
$已知b_n = \frac{1}{a_na_{n+1}}，求数列b_n的前n项和T_n。$

Solution:
$由题知，b_n = \frac{1}{(2n-13)(2n-11)} = \frac{1}{2} \left(\frac{1}{2n-13}-\frac{1}{2n-11}\right),$

$则 T_n = \frac{1}{2} \left[ \left(-\frac{1}{11}+\frac{1}{9}\right) + \left(-\frac{1}{9}+\frac{1}{7}\right) +\ldots + \left(\frac{1}{2n-13}-\frac{1}{2n-11}\right) \right] = \frac{1}{2} \left(-\frac{1}{11}-\frac{1}{2n-11}\right) = \frac{n}{121-22n}.$

Final answer: $\frac{n}{121-22n}$
Subfield: Sequence


Question: $已知数列{a_n}满足a_na_{n+1}=2^{2n},a_1=1.$
$求满足a_1+a_2+\ldots +a_{2n}<2022的最大的正整数n的值.$

Solution:
$令 S_{2n} =a_1+a_2+\ldots +a_{2n}，所以 S_{2n} = \frac{1-4^n}{1-4} + \frac{4(1-4^n)}{1-4} = \frac{5(4^n-1)}{3}，易知 f(x) = \frac{5(4^x-1)}{3} 在定义域上单调递增，且 f(4) =425， f(5) = 1705， f(6) = 6825，因为 a_1+a_2+\ldots +a_{2n} < 2022，所以 n < 6，又因为 n 为正整数，所以 n 的最大值为5。$

Final answer: 5
Subfield: Sequence


Question: $记S_n为数列{a_n}的前n项和，已知S_n=na_n-n^2+n.$
$若a_1，a_4，a_6成等比数列，求\frac{S_n+9}{n}的最小值.$

Solution:
$由(1)可得a_4=a_1+6，a_6=a_1+10.$
$\because a_1，a_4，a_6成等比数列，\therefore a^2_4=a_1a_6，即(a_1+6)^2=a_1(a_1+10)，解得a_1=-18,$
$\therefore S_n=-18n+\frac{n(n-1)}{2}\times 2=n^2-19n,$
$\therefore \frac{S_n+9}{n}=\frac{n^2-19n+9}{n}=n+\frac{9}{n}-19\geq 2\sqrt{n \cdot \frac{9}{n}}-19=-13,$
$当且仅当n=\frac{9}{n}，即n=3时取等号，\therefore \frac{S_n+9}{n}的最小值为-13.$

Final answer: $-13$
Subfield: Sequence


Question: $等差数列{a_n}的前n项和为S_n，已知a_{10}=30，a_{20}=50.$
$若S_n=242，求n。$

Solution:
$由(1)可得 S_n = 12n + \frac{n(n-1)}{2} \times 2 = n^2 + 11n, 令 n^2 + 11n = 242，解得t = -22 (舍去)或 n=11，故 n=11.$

Final answer: $n=11$
Subfield: Trigonometric Functions


Question: $已知函数f(x)=2\sin x \cdot \left(\cos^2\frac{x}{2} - \sin^2\frac{x}{2}\right) - \sqrt{3}\cos 2x．$
$求f(x)的最小正周期；$

Solution:
$f(x)=2sinx(cos^2\frac{x}{2}-sin^2\frac{x}{2})-\sqrt{3}cos2x=2sinxcosx-\sqrt{3}cos2x=sin2x-\sqrt{3}cos2x=2sin\left(2x-\frac{\pi }{3}\right).$

$所以f(x)的最小正周期为\pi ．$

Final answer: $\pi$
Subfield: Trigonometric Functions


Question: $在\triangle ABC中，\sin  2C=\sqrt{3}\sin  C.$
$求\angle C；$

Solution:
$\because \sin  2 C = \sqrt{3} \sin   C , \therefore 2\sin   C  \cos   C = \sqrt{3} \sin   C , 又\sin   C \neq 0, \therefore \cos   C = \frac{\sqrt{3}}{2}. $

$\because \angle  C \in (0,\pi ), \therefore \angle  C = \frac{\pi }{6}.$

Final answer: $\frac{\pi }{6}$
Subfield: Trigonometric Functions


Question: $在\triangle ABC中，\sin  2C=\sqrt{3}\sin  C.$
$若b=6，且\triangle ABC的面积为6\sqrt{3}，求\triangle ABC的周长.$

Solution:
$\because S_{\triangle ABC}=\frac{1}{2}ab\sin  C=\frac{1}{2}a\times 6\times sin\frac{\pi }{6}=\frac{3}{2}a=6\sqrt{3},$
$\therefore a=4\sqrt{3}.$
$由余弦定理得c^2=(4\sqrt{3})^2+6^2-2\times 4\sqrt{3}\times 6\times \frac{\sqrt{3}}{2}=12,$
$\therefore c=2\sqrt{3},\therefore \triangle ABC的周长为a+b+c=6+6\sqrt{3}.$

Final answer: $6+6\sqrt{3}$
Subfield: Trigonometric Functions


Question: $在\triangle ABC中，a=3，b-c=2，\cos  B=-\frac{1}{2}.$
求b，c的值；

Solution:
$由余弦定理 b^2=a^2+c^2-2ac \cos B 及已知，得 b^2=3^2+c^2-2 \times 3 \times c \times \left(-\frac{1}{2}\right).$

$因为 b=c+2 ,所以 (c+2)^2=3^2+c^2-2 \times 3 \times c \times \left(-\frac{1}{2}\right).$

$解得 c=5. 所以 b=7.$

Final answer: c=5, b=7
Subfield: Trigonometric Functions


Question: $在\triangle ABC中，a=3，b-c=2，\cos  B=-\frac{1}{2}.$
$求\sin (B-C)的值.$

Solution:
$由\cos  B=-\frac{1}{2}得\sin  B=\frac{\sqrt{3}}{2}.$
$由正弦定理得\sin  C=\frac{c}{b}\sin  B=\frac{5\sqrt{3}}{14}.$
$在\triangle ABC中，\angle B是钝角，所以\angle C为锐角.$
$所以\cos  C=\sqrt{1-\sin ^2C}=\frac{11}{14}.$
$所以\sin (B-C)=\sin  B\cos  C-\cos  B\sin  C=\frac{4\sqrt{3}}{7}.$

Final answer: $\frac{4\sqrt{3}}{7}$
Subfield: Trigonometric Functions


Question: $已知在\triangle ABC中，A + B = 3C, 2sin(A - C) = sin B.$
$求\sin  A;$

Solution:
$\because A+B+C=\pi ,A+B=3C,$

$\therefore C=\frac{\pi }{4},B=\frac{3\pi }{4}-A,$

$又\because 2\sin (A-C)=\sin  B,$

$\therefore 2\sin \left(A-\frac{\pi }{4}\right)=\sin \left(\frac{3\pi }{4}-A\right),$

$即2\left(\frac{\sqrt{2}}{2}\sin  A-\frac{\sqrt{2}}{2}\cos  A\right)=\frac{\sqrt{2}}{2}\cos  A-\left(-\frac{\sqrt{2}}{2}\right)\sin  A，$

$整理得\sin  A=3\cos  A，$

$又\because \sin ^2A+cos^2A=1,A\in \left(0,\frac{3\pi }{4}\right),\therefore \sin  A=\frac{3\sqrt{10}}{10}.$
$\because A+B+C=\pi ，A+B=3C，\therefore C=\frac{\pi }{4}.$

$又\because 2\sin (A-C)=\sin  B,\therefore 2\sin (A-C)=\sin (A+C),$

$即2\sin  A\cos  C-2\cos  A\sin  C=\sin  A\cos  C+\cos  A\sin  C,$

$化简得\sin  A\cos  C=3\cos  A\sin  C,$

$\therefore \tan  A=3\tan  C=3,\therefore \frac{\sin  A}{\cos  A}=3,$

$又\because \sin ^2A+cos^2A=1,A\in \left(0,\frac{3\pi }{4}\right),\therefore \sin  A=\frac{3\sqrt{10}}{10}.$

Final answer: \frac{3\sqrt{10}}{10}
Subfield: Trigonometric Functions


Question: $已知在\triangle ABC中，A + B = 3C, 2sin(A - C) = sin B.$
$设AB=5, 求AB边上的高.$

Solution:
$解法一：过C作CD\perp AB，垂足为D，如图$

<img_74>

$在\triangle ABC中，由正弦定理得\frac{AB}{\sin \angle ACB}=\frac{BC}{\sin  A}，即\frac{5}{\sin \frac{\pi }{4}}=\frac{BC}{\frac{3\sqrt{10}}{10}}，\therefore BC=3\sqrt{5}。$

$由(1)知\cos  A=\frac{\sqrt{10}}{10}，$

$\therefore \sin  B=\sin \left(\frac{3\pi }{4}-A\right)=\frac{\sqrt{2}}{2}\cos  A+\frac{\sqrt{2}}{2}\sin  A=\frac{2\sqrt{5}}{5}。$

$在Rt\triangle BCD中，CD=BC\cdot \sin  B=3\sqrt{5}\times \frac{2\sqrt{5}}{5}=6，即AB边上的高为6。$

$解法二：由(1)知C=\frac{\pi }{4}，\sin  A=\frac{3\sqrt{10}}{10}，\cos  A=\frac{\sqrt{10}}{10}，则\sin  B=\sin \left(\frac{3\pi }{4}-A\right)=\frac{\sqrt{2}}{2}\cos  A+\frac{\sqrt{2}}{2}\sin  A=\frac{2\sqrt{5}}{5}。$

$在\triangle ABC中，由正弦定理得\frac{AB}{\sin  C}=\frac{AC}{\sin  B}=\frac{BC}{\sin  A}，$

$\therefore \frac{5}{\frac{\sqrt{2}}{2}}=\frac{AC}{\frac{2\sqrt{5}}{5}}=\frac{BC}{\frac{3\sqrt{10}}{10}}，\therefore AC=2\sqrt{10}，BC=3\sqrt{5}，$

$\therefore S_{\triangle ABC}=\frac{1}{2}AC\cdot BC\cdot \sin  C=\frac{1}{2}\times 2\sqrt{10}\times 3\sqrt{5}\times \frac{\sqrt{2}}{2}=15。$

$设AB边上的高为h，则\frac{1}{2}\times 5h=15，\therefore h=6。$

Final answer: 6
Subfield: Trigonometric Functions


Question: $记\triangle ABC的内角A，B，C的对边分别为a，b，c，已知\triangle ABC面积为\sqrt{3}，D为BC的中点，且AD=1.$
$若\angle ADC=\frac{\pi }{3}，求\tan  B;$

Solution:
$由题意知 S_{\triangle ABC} = \sqrt{3}, BD=DC,\therefore  S_{\triangle ADC} = \frac{\sqrt{3}}{2}.$
$\because  S_{\triangle ADC} = \frac{1}{2}DA \cdot DC \cdot sin\angle ADC = \frac{\sqrt{3}}{2}, DA=1, \angle ADC = \frac{\pi}{3}, \therefore  \frac{1}{2}DCsin\frac{\pi}{3} = \frac{\sqrt{3}}{2}, \therefore  DC=2,$
$\therefore  BD=2, 易知 \angle ADB = \frac{2\pi}{3},$
$在\triangle ADB中，由余弦定理可知，AB^{2} = BD^{2} + DA^{2} - 2DA \cdot DB \cdot cos \angle ADB, 即 AB^{2} = 2^2 + 1 - 2 \cdot 1 \cdot 2 \cdot \left(-\frac{1}{2}\right) = 7,$
$\therefore  AB = \sqrt{7},$
$\therefore  cos B = \frac{AB^{2} + BD^{2} - AD^{2}}{2AB \cdot BD} = \frac{7 + 4 - 1}{2\sqrt{7} \cdot 2} = \frac{5\sqrt{7}}{14},$
$\therefore  sin B = \sqrt{1 - cos^{2}B} = \sqrt{1 - \frac{25}{28}} = \frac{\sqrt{21}}{14},$
$\therefore  tan B = \frac{\sin B}{\cos B} = \frac{\sqrt{3}}{5}.$

Final answer: $\frac{\sqrt{3}}{5}$