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User:I equal tan 90

107 bytes (23 words) - 00:57, 10 January 2020

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Trigonometric substitution
...we make use of the identity <math>\tan^2x+1=\sec^2x</math>. Set <math>x=a\tan\theta</math> and the radical will go away. However, the <math>dx</math> wil Since <math>\sec^2(\theta)-1=\tan^2(\theta)</math>, let <math>x=a\sec\theta</math>.

1 KB (173 words) - 18:42, 30 May 2021
Trigonometric identities
* <math>\tan^2x + 1 = \sec^2x</math> * <math>\tan(x + y) = \frac{\tan (x) + \tan (y)}{1 - \tan (x) \tan (y)} </math>

8 KB (1,397 words) - 21:55, 20 January 2024
Trigonometry
...de opposite <math>A</math> to the side adjacent to <math>A</math>. <cmath>\tan (A) = \frac{\textrm{opposite}}{\textrm{adjacent}} = \frac{a}{b}.</cmath> ...e reciprocal of the tangent of <math>A</math>. <cmath>\cot (A) = \frac{1}{\tan (x)} = \frac{\textrm{adjacent}}{\textrm{opposite}} = \frac{b}{a}.</cmath>

8 KB (1,217 words) - 20:15, 7 September 2023
AoPS Wiki:AoPS Community Awards
* Michael Tan (2009)

20 KB (2,025 words) - 01:33, 11 June 2024
Mock AMC
| Nathan Liu, Vincent Wang, Drake Tan, Channing Yang

51 KB (6,220 words) - 21:22, 17 July 2024
2000 AMC 12 Problems
<math>\textbf {(A)}\ \sec^2 \theta - \tan \theta \qquad \textbf {(B)}\ \frac 12 \qquad \textbf {(C)}\ \frac{\cos^2 \t

13 KB (1,948 words) - 10:35, 16 June 2024
2003 AMC 12B Problems
real r = 5/dir(54).x, h = 5 tan(54*pi/180);

13 KB (1,987 words) - 18:53, 10 December 2022
2004 AMC 12B Problems
...of <math>\tan \angle CBE</math>, <math>\tan \angle DBE</math>, and <math>\tan \angle ABE</math> form a [[geometric progression]], and the values of <math

13 KB (2,049 words) - 13:03, 19 February 2020
2006 AMC 12A Problems/Problem 19
...>L_2</math> and the x-axis, so <math>m=\tan{2\theta}=\frac{2\tan\theta}{1-\tan^2{\theta}}=\frac{120}{119}</math>. We also know that <math>L_1</math> and <

2 KB (253 words) - 22:52, 29 December 2021
2005 AMC 12B Problems/Problem 24
...e positive x- axis, the answer is <math>\dfrac{\tan(BOJ) + \tan(BOJ+60) + \tan(BOJ-60)}{2}</math>. Using <math>\tan(BOJ) = 2</math>, and the tangent addition formula, this simplifies to <math

4 KB (761 words) - 09:10, 1 August 2023
2005 AIME II Problems/Problem 12
...BG</math>). Then <math>\tan \angle EOG = \frac{x}{450}</math>, and <math>\tan \angle FOG = \frac{y}{450}</math>. ...frac{y}{450}}{1 - \frac{x}{450} \cdot \frac{y}{450}}.</cmath> Since <math>\tan 45 = 1</math>, this simplifies to <math>1 - \frac{xy}{450^2} = \frac{x + y}

13 KB (2,080 words) - 13:14, 23 July 2024
2005 AIME II Problems/Problem 15
...y find that <math>\tan \angle OF_1T=\sqrt{69}/10</math>. Therefore, <math>\tan\angle XOT</math>, which is the desired slope, must also be <math>\sqrt{69}/ ...rac{\sqrt3\cdot\sin\theta}{2\cos\theta}=\frac65\sec\theta-\frac{\sqrt3}{2}\tan\theta</math>

12 KB (2,001 words) - 20:26, 23 July 2024
2005 AIME II Problems/Problem 14
...5}</math>. Therefore, <math>\overline{AF} = \frac{52}{5}</math>, so <math>\tan{(\alpha)} = \frac{6}{13}</math>. Our goal now is to use tangent <math>\angl ...}</math> or <math>\frac{126}{137}</math>. Now we solve the equation <math>\tan{\angle EAG} = \frac{126}{137} = \frac{\frac{60-4x}{5}}{\frac{3x+25}{5}}</ma

14 KB (2,340 words) - 16:38, 21 August 2024
2004 AIME II Problems/Problem 7
...B'EF=\theta</math>, so <math>\angle B'EA = \pi-2\theta</math>. Then <math>\tan(\pi-2\theta)=\frac{15}{8}</math>, or <cmath>\frac{2\tan(\theta)}{\tan^2(\theta)-1}=\frac{15}{8}</cmath> using supplementary and double angle iden

9 KB (1,501 words) - 05:34, 30 October 2023
1986 AIME Problems
...tan x+\tan y=25</math> and <math>\cot x + \cot y=30</math>, what is <math>\tan(x+y)</math>?

5 KB (847 words) - 15:48, 21 August 2023
1988 AIME Problems
In triangle <math>ABC</math>, <math>\tan \angle CAB = 22/7</math>, and the altitude from <math>A</math> divides <mat

6 KB (902 words) - 08:57, 19 June 2021
1991 AIME Problems
Suppose that <math>\sec x+\tan x=\frac{22}7</math> and that <math>\csc x+\cot x=\frac mn,</math> where <ma draw(Circle(((1/cos(pi/12))*cos(a*pi/12), (1/cos(pi/12))*sin(a*pi/12)), tan(pi/12)));

7 KB (1,106 words) - 22:05, 7 June 2021
1996 AIME Problems
Find the smallest positive integer solution to <math>\tan{19x^{\circ}}=\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\si

6 KB (931 words) - 17:49, 21 December 2018
1999 AIME Problems
Given that <math>\sum_{k=1}^{35}\sin 5k=\tan \frac mn,</math> where angles are measured in degrees, and <math>m_{}</math

7 KB (1,094 words) - 13:39, 16 August 2020
1983 AIME Problems/Problem 4
...an{(\angle OAB)}=\frac{2-\frac{1}{3}}{1+2\cdot \frac{1}{3}}</cmath><cmath>\tan{(\angle OAB)}=1</cmath><cmath>\cos{(\angle OAB)}=\frac{1}{\sqrt{2}}</cmath>

11 KB (1,741 words) - 22:40, 23 November 2023
1983 AIME Problems/Problem 15
.../math>, we have <math>OM = \sqrt{OB^2 - BM^2} =4</math>. This gives <math>\tan \angle BOM = \frac{BM}{OM} = \frac 3 4</math>. ...efore, since <math>\angle AOM</math> is clearly acute, we see that <cmath>\tan \angle AOM =\frac{\sqrt{1 - \cos^2 \angle AOM}}{\cos \angle AOM} = \frac{\s

20 KB (3,497 words) - 15:37, 27 May 2024
1984 AIME Problems/Problem 13
...y the addition formula, <math>\tan(x+y) = \frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}</math>. Let <math>a = \cot^{-1}(3)</math>, <math>b=\cot^{-1}(7)</math>, ...an(a)=\frac{1}{3},\quad\tan(b)=\frac{1}{7},\quad\tan(c)=\frac{1}{13},\quad\tan(d)=\frac{1}{21}</math>,</p></center>

3 KB (473 words) - 12:06, 18 December 2018
1986 AIME Problems/Problem 15
...ective medians; in other words, <math>\tan \theta_1 = 1</math>, and <math>\tan \theta_2 =2</math>. ...ta_2 - \theta_1) = \frac{\tan \theta_2 - \tan \theta_1}{1 + \tan \theta_1 \tan \theta_2} = \frac{2-1}{1 + 2 \cdot 1 } = \frac{1}{3}. </cmath>

11 KB (1,722 words) - 09:49, 13 September 2023
1986 AIME Problems/Problem 3
...tan x+\tan y=25</math> and <math>\cot x + \cot y=30</math>, what is <math>\tan(x+y)</math>? Since <math>\cot</math> is the reciprocal function of <math>\tan</math>:

3 KB (545 words) - 23:44, 12 October 2023
1987 AIME Problems/Problem 15
Let <math>\tan\angle ABC = x</math>. Now using the 1st square, <math>AC=21(1+x)</math> and ...ving, we get <math>\sin{2\theta} = \frac{1}{10}</math>. Now to find <math>\tan{\theta}</math>, we find <math>\cos{2\theta}</math> using the Pythagorean

5 KB (838 words) - 18:05, 19 February 2022
1988 AIME Problems/Problem 7
In [[triangle]] <math>ABC</math>, <math>\tan \angle CAB = 22/7</math>, and the [[altitude]] from <math>A</math> divides ...lpha + \tan \beta}{1 - \tan \alpha \cdot \tan \beta}</math>, we get <math>\tan (\alpha + \beta) = \dfrac {\frac {20}{h}}{\frac {h^2 - 51}{h^2}} = \frac {2

1 KB (178 words) - 23:25, 20 November 2023
1989 AIME Problems/Problem 10
...\beta)^2-\tan \alpha \tan \beta}{\tan^2 \alpha + 2\tan \alpha \tan \beta +\tan^2 \beta}</math> ...sqrt{995}</math>. We see that <math>\tan \beta = \infty</math>, and <math>\tan \alpha = \sqrt{994}</math>.

8 KB (1,401 words) - 21:41, 20 January 2024
1990 AIME Problems/Problem 7
...th>-axis and <math>PR</math>. The equation of the angle bisector is <math>\tan\left(\frac{\alpha+\beta}{2}\right)</math>. ...}}{\frac{18}{25}}}=\pm\sqrt{\frac{16}{9}}=\pm\frac{4}{3}</math> and <math>\tan\left(\frac{\beta}{2}\right)=\pm\sqrt{\frac{1-\cos(\beta)}{1+\cos(\beta)}}=\

8 KB (1,319 words) - 11:34, 22 November 2023
1991 AIME Problems/Problem 15
Let <math>a_{i} = (2i - 1) \tan{\theta_{i}}</math> for <math>1 \le i \le n</math> and <math>0 \le \theta_{i ...that that <math>S_{n} + 17 = \sum_{k = 1}^{n}(2k - 1)(\sec{\theta_{k}} + \tan{\theta_{k}})</math>.

4 KB (658 words) - 16:58, 10 November 2023
1991 AIME Problems/Problem 11
draw(Circle(((1/cos(pi/12))*cos(a*pi/12), (1/cos(pi/12))*sin(a*pi/12)), tan(pi/12))); ...h>OA</math> and <math>m \angle MOA = 15^\circ</math>. Thus <math>AM = (1) \tan{15^\circ} = 2 - \sqrt {3}</math>, which is the radius of one of the circles

4 KB (740 words) - 17:46, 24 May 2024
1991 AIME Problems/Problem 9
Suppose that <math>\sec x+\tan x=\frac{22}7</math> and that <math>\csc x+\cot x=\frac mn,</math> where <ma ...s#Pythagorean Identities|trigonometric Pythagorean identities]] <math>1 + \tan^2 x = \sec^2 x</math> and <math>1 + \cot^2 x = \csc^2 x</math>.

10 KB (1,590 words) - 14:04, 20 January 2023
1993 AIME Problems/Problem 13
Since <math>PC=100</math>, <math>PX=200</math>. So, <math>\tan(\angle OXP)=\frac{OP}{PX}=\frac{50}{200}=\frac{1}{4}</math>. Thus, <math>\tan(\angle BXA)=\tan(2\angle OXP)=\frac{2\tan(\angle OXP)}{1- \tan^2(\angle OXP)} = \frac{2\cdot \left(\frac{1}{4}\right)}{1-\left(\frac{1}{4}

8 KB (1,388 words) - 04:38, 27 August 2024
1995 AIME Problems/Problem 9
...le sum identity gives <cmath>\tan 3x=\tan(2x+x)=\frac{3\tan x-\tan^3x}{1-3\tan^2x}.</cmath> Thus, <math>\frac{3-\tan^2x}{1-3\tan^2x}=11</math>. Solving, we get <math>\tan x= \frac 12</math>. Hence, <math>CM=\frac{11}2</math> and <math>AC= \frac{1

7 KB (1,181 words) - 13:47, 3 February 2023
1996 AIME Problems/Problem 10
Find the smallest positive integer solution to <math>\tan{19x^{\circ}}=\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\si ...2\sin{141^{\circ}}\cos{45^{\circ}}}{2\cos{141^{\circ}}\sin{45^{\circ}}} = \tan{141^{\circ}}</math>.

4 KB (503 words) - 15:46, 3 August 2022
1999 AIME Problems/Problem 14
\begin{align*}DP&=z\tan\theta\\ EP&=x\tan\theta\\

7 KB (1,184 words) - 13:25, 22 December 2022
1999 AIME Problems/Problem 12
\begin{eqnarray*} \tan \alpha & = & \frac {21}{27} \\ \tan \beta & = & \frac {21}{23} \\

3 KB (472 words) - 19:03, 21 June 2024
1999 AIME Problems/Problem 11
Given that <math>\sum_{k=1}^{35}\sin 5k=\tan \frac mn,</math> where angles are measured in degrees, and <math>m_{}</math ...ath>, we get <cmath>s = \frac{1 - \cos 175}{\sin 175} \Longrightarrow s = \tan \frac{175}{2},</cmath> and our answer is <math>\boxed{177}</math>.

5 KB (816 words) - 13:39, 29 June 2024
2000 AIME I Problems/Problem 13
...rrow AB=\sqrt{OB^2-AO^2}=\sqrt{5^2-1.4^2}=\frac{24}{5}</math>. Then <math>\tan(\angle ABO)=\frac{OA}{AB}=\frac{7}{24}</math>, so the [[slope]] of line <ma

3 KB (571 words) - 00:38, 13 March 2014
2001 AIME I Problems/Problem 5
Note that the slope of <math>\overline{AC}</math> is <math>\tan 60^\circ = \sqrt {3}.</math> Hence, the equation of the line containing <ma

6 KB (1,043 words) - 10:09, 15 January 2024
2003 AIME I Problems/Problem 11
<cmath>2 > \tan 2x \Longrightarrow x < \frac 12 \arctan 2.</cmath> ...math>. We can divide both sides by <math>\cos 2x</math> and we have <math>\tan 2x > 2</math>. The solutions to this occur when <math>45 \geq x > \frac{\a

5 KB (895 words) - 15:24, 21 June 2024
2003 AIME I Problems/Problem 10
pair A=(0,0), B=(2,0), C=(1,Tan(37)), M=IP(A--(2Cos(30),2Sin(30)),B--B+(-2,2Tan(23))); pair A=(0,0), B=(2,0), C=(1,Tan(37)), M=IP(A--(2Cos(30),2Sin(30)),B--B+(-2,2Tan(23)));

7 KB (1,058 words) - 01:41, 6 December 2022
2003 AIME II Problems/Problem 7
Hence <math>x=25\sin\theta=50\cos\theta</math>. Solving <math>\tan\theta=2</math>, <math>\sin\theta=\frac{2}{\sqrt{5}}, \cos\theta=\frac{1}{\s

2 KB (323 words) - 09:56, 16 September 2022
2002 AIME II Problems/Problem 15
...we have that <math>\frac{y}{x}=\tan{\frac{\theta}{2}}</math>. Let <math>\tan{\frac{\theta}{2}}=m_1</math>, for convenience. Therefore if <math>(x,y)</ma <cmath>\tan{\frac{\theta}{2}}=\sqrt{\frac{1-\cos{\theta}}{1+\cos{\theta}}}</cmath>

7 KB (1,188 words) - 17:46, 23 June 2024
2002 AIME II Problems/Problem 14
We have that <math>\tan(\angle AMO)=\frac{19}{x},</math> so <cmath>\tan(\angle M)=\tan (2\cdot \angle AMO)=\frac{38x}{x^{2}-361}.</cmath>

4 KB (658 words) - 19:15, 19 December 2021
2000 AIME II Problems/Problem 10
...</math> to get <cmath>\tan(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+\tan(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=n\cdot0=0.</cmath> Use the identity for <math>\tan(A+B)</math> again to get <cmath>\frac{\tfrac{45}{r}}{1-19\cdot\tfrac{26}{r^

2 KB (399 words) - 17:37, 2 January 2024
Law of Tangents
<cmath> \frac{a-b}{a+b}=\frac{\tan [\frac{1}{2}(A-B)]}{\tan [\frac{1}{2}(A+B)]} . </cmath> ...2\sin s \cos d} = \frac{\tan d}{\tan s} = \frac{\tan [\frac{1}{2} (A-B)]}{\tan[ \frac{1}{2} (A+B)]} </cmath>

2 KB (306 words) - 16:11, 21 February 2023
University of South Carolina High School Math Contest/1993 Exam/Problem 18
...}{\sqrt{1 - \cos^2 (x)}} + \frac{\cos(x)}{\sqrt{1 - \sin^2 (x) }} + \frac{\tan(x)}{\sqrt{\sec^2 (x) - 1}} + \frac{\cot (x)}{\sqrt{\csc^2 (x) - 1}}</math>< | <center><math> \tan^2 x + 1 = \sec^2 x </math> </center>

2 KB (331 words) - 00:37, 26 January 2023
University of South Carolina High School Math Contest/1993 Exam/Problems
...}{\sqrt{1 - \cos^2 (x)}} + \frac{\cos(x)}{\sqrt{1 - \sin^2 (x) }} + \frac{\tan(x)}{\sqrt{\sec^2 (x) - 1}} + \frac{\cot (x)}{\sqrt{\csc^2 (x) - 1}}</cmath>

14 KB (2,102 words) - 22:03, 26 October 2018
Mock AIME 1 2006-2007 Problems/Problem 14
...f <math>AB</math>. Let <math>f(m,n)</math> denote the maximum value <math>\tan^{2}\angle AMP</math> for fixed <math>m</math> and <math>n</math> where <mat <math>\tan{\angle{OAB}} = \dfrac{OT}{AT} = \dfrac{r}{m}</math>

3 KB (541 words) - 17:32, 22 November 2023
Mock AIME 1 2006-2007/Problems
...f <math>AB</math>. Let <math>f(m,n)</math> denote the maximum value <math>\tan^{2}\angle AMP</math> for fixed <math>m</math> and <math>n</math> where <mat

8 KB (1,355 words) - 14:54, 21 August 2020
Mock AIME 2 2006-2007 Problems/Problem 9
..., <math>\frac{AY}{CY}=\sqrt 3,</math> and <math>CY=CX-BX</math>. If <math>\tan \angle APB= -\frac{a+b\sqrt{c}}{d},</math> where <math>a,b,</math> and <mat ...angle DPB)=270^\circ</math>, we have <cmath>\begin{align*}\tan\angle APB&=\tan[270^\circ-(\angle APE+\angle BPD)]\\&=\cot (\angle APE+\angle BPD)\\&=-\dfr

2 KB (358 words) - 23:22, 3 May 2014
Mock AIME 2 2006-2007 Problems/Problem 6
If <math>\tan 15^\circ \tan 25^\circ \tan 35^\circ =\tan \theta</math> and <math>0^\circ \le \theta \le 180^\circ, </math> find <mat ...rc}=\frac{\sin 5^\circ(1+2\cos 20^\circ)}{\cos 5^\circ(1+2\cos 20^\circ)}=\tan 5^\circ</cmath>

1 KB (157 words) - 10:51, 4 April 2012
Mock AIME 2 2006-2007 Problems
If <math>\tan 15^\circ \tan 25^\circ \tan 35^\circ =\tan \theta</math> and <math>0^\circ \le \theta \le 180^\circ, </math> find <mat ...c{BX}{CX}=\frac23</math> and <math>\frac{AY}{CY}=\sqrt 3.</math> If <math>\tan \angle APB= \frac{a+b\sqrt{c}}{d},</math> where <math>a,b,</math> and <math

5 KB (848 words) - 23:49, 25 February 2017
1959 IMO Problems/Problem 4
...>. Thus, <math>\frac{a}{b} = \tan 15^\circ</math> and <math>\frac{a}{b} = \tan 75^\circ</math>, and so one of the angles of the triangle must be <math>15^

6 KB (939 words) - 17:31, 15 July 2023
Derivative/Formulas
| <math>\frac d{dx} \tan x = \sec^2 x</math> | <math>\frac d{dx} \sec x = \sec x \tan x</math>

3 KB (506 words) - 16:23, 11 March 2022
Integral
*<math>\int\tan x\,dx = \ln |\cos x| + C</math> *<math>\int \sec x\,dx = \ln |\sec x + \tan x| + C</math>

5 KB (909 words) - 14:16, 31 May 2022
1970 IMO Problems/Problem 1
& = &q \left[ \tan\left(\frac{A}{2}\right) + \tan\left(\frac{B}{2}\right) \right]\; . \end{matrix}</math> <math>\frac{r}{q} = \tan (A/2) \tan (B/2)</math>.

2 KB (380 words) - 22:12, 19 May 2015
2007 iTest Problems
...}{4}</math> and <math>\tan{y}=\frac{1}{6}</math>, find the value of <math>\tan{x}</math>.

30 KB (4,794 words) - 23:00, 8 May 2024
Qq808casino
...opular games like baccarat, blackjack, roulette, dragon tiger, sic bo, fan tan and more. Besides, there is a selection of providers where you can expect t

2 KB (276 words) - 03:46, 9 December 2019
Apothem
...side length, <math>s</math>, the length of the apothem is <math>\frac{s}{2\tan\left(\frac{\pi}{n}\right)}</math>.

1 KB (171 words) - 18:33, 3 September 2024
Euler line
...vec BD}{\vec DA} = n = \frac {\tan \alpha – \tan \gamma}{\tan \beta – \tan \gamma} > 0.</math> ...ac {\vec AE}{\vec EC} = \frac {\tan \beta – \tan \gamma}{\tan \beta – \tan \alpha} > 0.</math>

59 KB (10,203 words) - 04:47, 30 August 2023
1985 IMO Problems/Problem 1
\begin{matrix} {CE} & = & r \tan(COE) \\

4 KB (684 words) - 07:28, 3 October 2021
Asymptote (geometry)
...the vertical asymptotes of 1) <math>y = \frac{1}{x^2-5x}</math> 2) <math>\tan 3x</math>. 2) Since <math>\tan 3x = \frac{\sin 3x}{\cos 3x}</math>, we need to find where <math>\cos 3x =

4 KB (664 words) - 11:44, 8 May 2020
Mock AIME 3 Pre 2005 Problems
The value of <math>\tan\left(\Omega\right)</math> can be expressed as <math>\frac{m}{n}</math>, whe

7 KB (1,135 words) - 23:53, 24 March 2019
Mock AIME 3 Pre 2005 Problems/Problem 15
The value of <math>\tan\left(\Omega\right)</math> can be expressed as <math>\frac{m}{n}</math>, whe ...ric substitution; namely, define <math>\theta</math> such that <math> x = \tan{\theta}</math>. Then the RHS becomes

2 KB (312 words) - 10:38, 4 April 2012
1960 IMO Problems/Problem 3
\tan{\alpha}=\frac{4nh}{(n^2-1)a}. ...c}{b}\cdot\frac{n-1}{n+1}</math>, and <math>\text{slope}</math><math>(QA)=\tan{\angle QAB}=\frac{c}{b}\cdot\frac{n+1}{n-1}</math>.

3 KB (501 words) - 00:14, 17 May 2015
1960 IMO Problems
\tan{\alpha}=\frac{4nh}{(n^2-1)a}.

3 KB (511 words) - 21:21, 20 August 2020
2007 AIME I Problems/Problem 9
...{1 - \cos \theta}{1 + \cos \theta}}</math>). We see that <math>\frac rx = \tan \frac{180 - \theta}{2} = \sqrt{\frac{1 - \cos (180 - \theta)}{1 + \cos (180 ...We see that <math>\frac rx = \tan \left(\frac{180 - 2\theta}{2}\right) = \tan (90 - \theta)</math>. In terms of <math>r</math>, we find that <math>x = \f

11 KB (1,853 words) - 20:10, 21 July 2024
2007 AIME I Problems/Problem 12
...th>[\triangle EFB'] = \frac{1}{2} (FB' \cdot EF) = \frac{1}{2} (FB') (FB' \tan 75^{\circ})</math>. With some horrendous [[algebra]], we can calculate [\triangle EFB'] &= \frac{1}{2}\tan (30 + 45) \cdot (20 - 5\sqrt{2} - 5\sqrt{6})^2 \\

10 KB (1,458 words) - 20:50, 3 November 2023
2007 AIME II Problems/Problem 15
...2}{3}</math> according to half angle formula. Similarly, we can find <math>tan\angle NCK=\frac{1}{2}</math>. So we can see that <math>JK=ON=14-\frac{7x}{2

11 KB (2,099 words) - 17:51, 4 January 2024
1961 IMO Problems/Problem 5
<math>b \tan{\frac{\omega}{2}} \le c < b</math> ...we require <math>AX \geqslant AC > AB</math>. But <math>\frac{AB}{AX} = \tan{\frac{\omega}{2}}</math>, so we get the condition in the question

1 KB (205 words) - 04:12, 7 June 2021
User:Azjps/1951 AHSME Problems/Problem 3
...ments <math>\overline{AO}, \overline{BO}</math> will have slope <math>\pm \tan{60^{\circ}} = \pm \sqrt{3}</math>. [[Without loss of generality]] consider

1 KB (184 words) - 13:10, 27 August 2020
LaTeX:Commands
...||\cos||<math>\textstyle \sin</math>||\sin||<math>\textstyle \tan</math>||\tan

12 KB (1,796 words) - 13:14, 12 September 2024
User:Temperal/sandbox
(a) <math>\tan(45^\circ)</math> (g) <math>\tan(21\pi)</math>

2 KB (338 words) - 13:28, 24 March 2020
2006 AIME II Problems/Problem 6
...Also note that <cmath>AB = 1 = \overline{AA'} + \overline{A'B} = \frac{x}{\tan(15)} + x</cmath> Using the fact <math>\tan(15) = 2-\sqrt{3}</math>, this yields <cmath>x = \frac{1}{3+\sqrt{3}} = \fra

7 KB (1,067 words) - 12:23, 8 April 2024
User:Temperal/The Problem Solver's Resource1
<math>\tan (-A)=-\tan A</math> <math>\tan (\pi+A) = \tan A</math>

6 KB (1,097 words) - 18:18, 21 January 2016
1961 IMO Problems
<math>b \tan{\frac{\omega}{2}} \le c < b</math>

3 KB (425 words) - 21:18, 20 August 2020
2004 AMC 12A Problems/Problem 18
...e CGF</math>. Thus, <math>\angle EGF=\angle FCG</math> and <math>\tan EGF=\tan FCG=\frac{1}{2}</math>. Solving <math>EF=\frac{1}{2}</math>. Adding, the an

5 KB (811 words) - 15:30, 27 June 2024
2004 AMC 10A Problems/Problem 20
E = (0,Tan(15)); F = (1 - Tan(15),1);

4 KB (710 words) - 18:04, 6 September 2024
1999 AHSME Problems
...al number such that <math>\sec x - \tan x = 2</math>. Then <math>\sec x + \tan x =</math>

13 KB (1,945 words) - 18:28, 19 June 2023
1960 IMO Problems/Problem 6
<cmath>\tan\left(\frac{\theta}{2}\right) = \frac{1}{x} = \frac{\sqrt{2}}{4}</cmath> ...3</math> and <math>V_1 = \frac{\pi a^2 \times H_1 H_2}{3} = \frac{\pi a^3 \tan (\angle A_1 A H_1) }{3}</math> .

7 KB (1,214 words) - 18:49, 29 January 2018
1992 USAMO Problems/Problem 2
Consider the points <math>M_k = (1, \tan k^\circ)</math> in the coordinate plane with origin <math>O=(0,0)</math>, f ...hen the left hand side of the equation simplifies to <math>\tan 89-\tan 0=\tan 89=\frac{\sin 89}{\cos 89}=\frac{\cos 1}{\sin 1}</math> as desired.

4 KB (628 words) - 07:41, 19 July 2016
2000 AMC 12 Problems/Problem 17
<math>\text {(A)}\ \sec^2 \theta - \tan \theta \qquad \text {(B)}\ \frac 12 \qquad \text {(C)}\ \frac{\cos^2 \theta ...<cmath> \frac{OB}{OC} = \frac{AB}{AC} \Longrightarrow AC \sec \theta = OC \tan \theta </cmath> We multiply both sides by <math>\cos \theta</math> to simpl

7 KB (1,072 words) - 00:38, 20 August 2024
Construction
21. Construct <math>sin C, cos C, tan C</math> given unit segment <math>1</math> and acute angle <math>C</math>.

3 KB (443 words) - 20:52, 28 August 2014
1995 USAMO Problems
..., </math> <math>\tan, \; \sin^{-1}, \; \cos^{-1}, \,</math> and <math>\, \tan^{-1} \,</math> buttons. The display initially shows 0. Given any positive

3 KB (540 words) - 13:31, 4 July 2013
2004 AMC 12B Problems/Problem 24
...of <math>\tan \angle CBE</math>, <math>\tan \angle DBE</math>, and <math>\tan \angle ABE</math> form a [[geometric progression]], and the values of <math ...a)\tan(DBE + \alpha) = \frac {\tan^2 DBE - \tan^2 \alpha}{1 - \tan ^2 DBE \tan^2 \alpha},

2 KB (302 words) - 19:59, 3 July 2013
1995 USAMO Problems/Problem 2
..., </math> <math>\tan, \; \sin^{-1}, \; \cos^{-1}, \,</math> and <math>\, \tan^{-1} \,</math> buttons. The display initially shows 0. Given any positive <cmath> \cos \tan^{-1} \sqrt{(n-m)/m} = \sqrt{m/n} . </cmath>

3 KB (516 words) - 00:18, 6 April 2020
2008 AMC 12A Problems
...midpoint of <math>BC</math>. What is the largest possible value of <math>\tan{\angle BAD}</math>?

13 KB (2,025 words) - 13:56, 2 February 2021
2008 AMC 12A Problems/Problem 24
...dpoint]] of <math>BC</math>. What is the largest possible value of <math>\tan{\angle BAD}</math>? ..., and since <math>\tan\angle BAF = \frac{2\sqrt{3}}{x-2}</math> and <math>\tan\angle DAE = \frac{\sqrt{3}}{x-1}</math>, we have

3 KB (513 words) - 14:35, 7 June 2018
2008 AMC 12B Problems/Problem 24
...a</math> with the x-axis and passes through the origin has equation <math>\tan(\theta)x</math>, so the line through <math>A_0</math> and <math>B_1</math>

9 KB (1,482 words) - 13:52, 4 April 2024
2008 AIME I Problems/Problem 8
Since we are dealing with acute angles, <math>\tan(\arctan{a}) = a</math>. Note that <math>\tan(\arctan{a} + \arctan{b}) = \dfrac{a + b}{1 - ab}</math>, by tangent additio

3 KB (490 words) - 22:36, 28 November 2023
2008 AIME II Problems/Problem 5
<cmath>\begin{align*}\tan{37}\times (1008-x) &= \tan{53} \times x\\ \frac{(1008-x)}{x} &= \frac{\tan{53}}{\tan{37}} = \frac{\sin{53}}{\cos{53}} \times\frac{\cos{37}}{\sin{37}}\end{align*

8 KB (1,338 words) - 00:23, 21 August 2024
2008 AIME II Problems/Problem 11
...om the [[trigonometric identity|half-angle identity]], we find that <math>\tan(\theta) = \frac {3}{4}</math>. Therefore, <math>XC = \frac {64}{3}</math>. ...the points as shown in the first solution's diagram ; realise that <math>\tan\theta = r/x = 16/y = r + 16/(x+y)</math> where <math>x = BY</math> and <mat

6 KB (1,092 words) - 22:22, 18 August 2024
2008 AIME II Problems/Problem 14
...tarrow (2-\sqrt{3}k)\cos x\le k\sin x\rightarrow \frac{2-\sqrt{3}k}{k}\le \tan x,</cmath>

8 KB (1,387 words) - 11:56, 29 January 2024
2008 AMC 10B Problems/Problem 24
<cmath>\cot(\theta)=\tan(5^\circ)</cmath>

12 KB (1,944 words) - 18:31, 28 August 2024
1998 AHSME Problems/Problem 28
...c{AC(\tan 3\theta - \tan 2\theta)}{AC \tan 2\theta} = \frac{\tan 3\theta}{\tan 2\theta} - 1.</math></center> ...\tan ^2 \theta},\ \tan 3\theta = \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta}</math>, and

4 KB (662 words) - 00:51, 3 October 2023
2000 USAMO Problems/Problem 2
...>. Denote <math>x=\tan{(A/2)}</math>, <math>x=\tan{(B/2)}</math>, <math>z=\tan{(C/2)}</math>, then we have, <cmath>z = \tan{(C/2)} = \tan{(90- (A+B)/2))} = \frac{1-xy}{x+y} </cmath>

4 KB (703 words) - 18:40, 3 January 2019
2008 iTest Problems
...ath> in the interval <math>[0,2\pi)</math> that satisfy <math>\tan^2 x - 2\tan x\sin x=0</math>. Compute <math>\lfloor10S\rfloor</math>. Let a and b be the two possible values of <math>\tan\theta</math> given that <math>\sin\theta + \cos\theta = \dfrac{193}{137}</m

71 KB (11,749 words) - 01:31, 2 November 2023
1989 IMO Problems/Problem 2
...he other triangles. Thus, the area of triangle <math>A_1BC=\frac{1}{4}a^2\tan\frac{A}{2}=\frac{1}{4}a^2\left(\frac{2r}{b+c-a}\right)</math> and similarly

3 KB (568 words) - 11:50, 30 January 2021
2001 USAMO Problems/Problem 3
...\tan\frac{A}{2}\sin B\tan\frac{B}{2}} = 2\sqrt{\sin A\tan\frac{B}{2}\sin B\tan\frac{A}{2}} \\ &\leq \sin A\tan\frac{B}{2} + \sin B\tan\frac{A}{2} \\

4 KB (807 words) - 10:45, 9 April 2023
2009 AMC 12A Problems/Problem 25
...ce <math>\{\theta_1, \theta_2, \theta_3...\}</math> such that <math>a_n = \tan{\theta_n}</math>, and <math>0 \leq \theta_n < 180</math>. ...+ 2}} & = \frac {\tan{\theta_n} + \tan{\theta_{n + 1}}}{1 - \tan{\theta_n}\tan{\theta_{n + 1}}} \\

7 KB (994 words) - 16:57, 9 July 2024
2009 AMC 10A Problems/Problem 10
...get <math>xy = 3</math>, and using the fact that <math>\cot{a} = \frac{1}{\tan{a}}</math>, <math>\frac{x}{y} = 4</math>. Solving for x and y(using substit

5 KB (879 words) - 18:57, 30 April 2024
2001 AMC 12 Problems/Problem 24
...<math>\angle ACH</math> can be simplified. Indeed, if you know that <math>\tan(75)=2+\sqrt{3}</math> or even take a minute or two to work out the sine and ...= 2 + \sqrt{3}</math>. Looking that the answer options we see that <math>\tan{75^\circ} = 2 + \sqrt{3}</math>. This means the answer is <math>D</math>.

8 KB (1,316 words) - 22:48, 7 March 2024
2007 AMC 12B Problems/Problem 22
...}</math> and the <math>x</math>-axis is <math>30^{\circ}</math>, and <math>tan(30) = \frac{\sqrt{3}}{3}</math>.

4 KB (707 words) - 16:36, 15 February 2021
2009 AMC 10B Problems/Problem 20
...e BAD = \angle DAC</math>. Notice <math>\tan \theta = BD</math> and <math>\tan 2 \theta = 2</math>. By the double angle identity, <cmath>2 = \frac{2 BD}{1

2 KB (376 words) - 12:50, 10 August 2024
2009 AIME I Problems/Problem 12
...t <math>O</math> be the center of <math>\omega</math>; notice that <cmath>\tan(\angle DAO)=\dfrac{DO}{AD}=\dfrac{210/37}{144/37}=\dfrac{35}{24}</cmath> so Since we have <math>\tan OAB = \frac {35}{24}</math> and <math>\tan OBA = \frac{6}{35}</math> , we have <math>\sin {(OAB + OBA)} = \frac {1369}

13 KB (2,170 words) - 23:35, 3 September 2024
2009 AIME I Problems/Problem 15
...BOP</math> and <math>COP</math>, with <math>BO=CO=7</math> and <math>OP=7 \tan 15=7(2-\sqrt{3})=14-7\sqrt3</math>. Then, the area of [<math>\triangle BPC<

6 KB (1,048 words) - 19:35, 2 January 2023
2009 AIME II Problems/Problem 15
<cmath>\frac{NV}{MV} = \frac{\sin (\alpha)}{\sin (90^\circ - \alpha)} = \tan (\alpha)</cmath> ...math>VW = NW + MV - 1 = \frac{1}{1+\frac{3}{4}\cot(\alpha)} + \frac{1}{1+\tan (\alpha)} - 1</math>. Taking the derivative of <math>VW</math> with respect

11 KB (1,862 words) - 21:23, 23 May 2024
Alternating permutation
\sum_{n = 0}^\infty \frac{E_n}{n!} x^n = \sec x + \tan x .

2 KB (246 words) - 12:50, 6 August 2009
1998 USAMO Problems/Problem 3
...rac {\pi}{4}\right)} + \tan{\left(a_1 - \frac {\pi}{4}\right)} + \cdots + \tan{\left(a_n - \frac {\pi}{4}\right)}\ge n - 1</cmath> Prove that <math>\tan{\left(a_0\right)}\tan{\left(a_1\right)}\cdots \tan{\left(a_n\right)}\ge n^{n + 1}</math>.

2 KB (322 words) - 13:31, 23 August 2023
Differentiation Rules
If <math>y(x) = \tan x</math>, then <math>\frac{dy}{dx} = \sec^2 x</math>. Note that this follow

2 KB (309 words) - 10:50, 4 June 2024
2010 AMC 12A Problems/Problem 8
<cmath>\frac{\sqrt{3}}{2}\cos(x)=\frac{3}{2}\sin(x)\implies \tan(x)=\frac{\sqrt{3}}{3}</cmath>

4 KB (697 words) - 20:33, 19 August 2024
2010 AMC 12B Problems
...ath> has <math>a_1=\sin x</math>, <math>a_2=\cos x</math>, and <math>a_3= \tan x</math> for some real number <math>x</math>. For what value of <math>n</ma

12 KB (1,845 words) - 13:00, 19 February 2020
Mock AIME 1 2010 Problems
...sqrt{159}}{7}</math>, and <math>ED^2 + EB^2 = 3050</math>, and that <math>\tan m \angle ACB</math> can be expressed in the form <math>\frac{a \sqrt{b}}{c}

8 KB (1,377 words) - 13:20, 2 August 2024
2010 AIME I Problems/Problem 15
<cmath>r = AX \tan(A/2) = CY \tan(C/2)</cmath> ...^2}{2\cdot 12\cdot 15} = \frac{200}{30\cdot 12}=\frac{5}{9}</math>, <math>\tan(A/2) = \sqrt{\frac{1-\cos A}{1+\cos A}} = \sqrt{\frac{9-5}{9+5}} = \frac{2}

14 KB (2,210 words) - 13:14, 11 January 2024
Mock AIME 2 2010 Answer Key
Hence, <math>(4 \sin^2(x+y) - 7 \cos^2(x+y)) \leq 0</math>, yielding <math>\tan^2(x+y) = \frac{7}{4}</math>, or <math>x + y = \pm \arctan{\frac{\sqrt{7}}{2

36 KB (6,214 words) - 20:22, 13 July 2023
2010 USAMO Problems/Problem 1
<math>AY \cos(\delta)\sin(\alpha)\tan(\delta) = AY \sin(\delta)\sin(\alpha)</math>

13 KB (2,271 words) - 14:08, 23 May 2024
1966 IMO Problems
...se sides. Prove that if <cmath> a+b=\tan{\frac{\gamma}{2}}(a\tan{\alpha}+b\tan{\beta}) </cmath> the triangle is isosceles.

3 KB (509 words) - 12:39, 29 January 2021
2010 AMC 12B Problems/Problem 20
...ath> has <math>a_1=\sin x</math>, <math>a_2=\cos x</math>, and <math>a_3= \tan x</math> for some real number <math>x</math>. For what value of <math>n</ma ...ometric sequence, we have <math>\cos^2x=\sin x \tan x</math>. Since <math>\tan x=\frac{\sin x}{\cos x}</math>, we can rewrite this as <math>\cos^3x=\sin^2

2 KB (375 words) - 19:52, 24 June 2022
Mock USAJMO
| Tan

3 KB (295 words) - 20:01, 4 March 2020
2011 AMC 12A Problems/Problem 24
...\frac{C}{2}=90^\circ-\angle \frac{A}{2}</math>, so <math>\cot\frac{C}{2}=\tan\frac{A}{2}</math>, and we find that <math>\frac{r}{x}=\frac{x-5}{r}</math>. ...rac{B}{2}=\frac{14-x}{r}</math>, and <math>\cot\frac{D}{2}=\frac{12-x}{r}=\tan\frac{B}{2}=\frac{r}{14-x}</math>. Therefore, <math>r^2=\frac{12-x}{14-x}</m

5 KB (826 words) - 13:49, 4 September 2024
2011 AMC 10B Problems/Problem 18
...- 12x^3 + 54x^2 - 108x + 405} \cdot \cos (\theta) = 36</cmath> and <cmath>\tan (\theta) = \frac{3}{x}</cmath> YAY!!! We have two equations for two variabl Finally, since <math>\tan (\theta) = \frac{3}{6 - 3 \sqrt{3}} = 2 + \sqrt{3}</math>, <math>\theta = \

5 KB (777 words) - 19:18, 2 September 2024
User:Jonjoseph
label("$\tan \theta$", (1,.375), E);

7 KB (964 words) - 21:43, 28 March 2022
2012 IMO Problems/Problem 1
<cmath>a = R \tan \frac{B}{2} + R \tan \frac{C}{2} = R \frac{\sin \frac{B+C}{2}}{\cos \frac{B}{2} \cos \frac{C}{2}

7 KB (1,189 words) - 01:22, 19 November 2023
2011 AIME I Problems/Problem 14
Suppose the common slope of the lines is <math>m</math> and let <math>m=\tan\theta</math>. Then, we want to find <cmath>\cos 2\left(90-\theta\right)=2\c ...rt{2}\right)m^{2}&=4 \\ m^{2}&=\frac{4}{8+8\sqrt{2}} \\ \Rightarrow m^{2}=\tan^{2}\theta=\frac{\sin^{2}\theta}{\cos^{2}\theta}&=\frac{1}{2+2\sqrt{2}}.\end

8 KB (1,365 words) - 16:07, 2 September 2024
2011 AIME II Problems/Problem 10
<cmath>\textrm{tan }^{2} \alpha = \frac{\frac{9z-16}{9z}}{\frac{16}{9z}} = \frac{9z-16}{16}.< <cmath>\textrm{tan } \alpha = \textrm{tan } \angle MOP = \frac{MP}{OM} = \frac{14-y}{20}.</cmath>

11 KB (1,720 words) - 03:12, 18 December 2023
2011 AIME II Problems/Problem 13
<cmath>AP = AB \cdot (\sin 45^\circ + \cos 45^\circ \cdot \tan 30^\circ),</cmath>

13 KB (2,055 words) - 05:25, 9 September 2022
1998 USAMO Problems
...eft(a_0-\frac{\pi}{4}\right)+ \tan \left(a_1-\frac{\pi}{4}\right)+\cdots +\tan \left(a_n-\frac{\pi}{4}\right)\geq n-1. </cmath> <cmath> \tan a_0\tan a_1 \cdots \tan a_n\geq n^{n+1}. </cmath>

3 KB (486 words) - 06:11, 24 November 2020
Mock Geometry AIME 2011 Problems
...nt to <math>OB,OC,</math> and arc <math>BC</math>. It is known that <math>\tan AOC=\frac{24}{7}</math>. The ratio <math>\frac{r_2} {r_1}</math> can be exp

8 KB (1,349 words) - 19:10, 14 June 2022
2011 USAJMO Problems/Problem 3
...n (\theta + 120)+\tan (\theta-120)}{6}=\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}</cmath> and <cmath>\begin{align*} ...{3}&=\frac{\tan\theta (\tan(\theta-120)+\tan(\theta+120))+\tan(\theta-120)\tan(\theta+120)}{12}\\

15 KB (2,593 words) - 13:37, 29 January 2021
LaTeX Cheat Sheet
\sin, \cos, \tan represents <math>\sin, \cos, \tan</math>

2 KB (318 words) - 13:47, 4 May 2024
1999 AHSME Problems/Problem 15
...number such that <math> \sec x - \tan x = 2</math>. Then <math> \sec x + \tan x =</math> ...(\sec x + \tan x) = \sec^{2} x - \tan^{2} x = 1</math>, so <math>\sec x + \tan x = \boxed{\textbf{(E)}\ 0.5}</math>.

931 bytes (144 words) - 19:36, 1 May 2023
1984 AHSME Problems
...}) \qquad \mathrm{(D) \ }\tan{15^\circ} \qquad \mathrm{(E) \ } \frac{1}{4}\tan{60^\circ} </math>

13 KB (1,879 words) - 14:00, 19 February 2020
Trigonometric Equations
<math>sin^2(x) + tan^2(x) = -cos^2(x) + \frac{1}{sin^2(x) + cos^2(x)}</math> <math>sin^2(x) + cos^2(x) + tan^2(x) = \frac{1}{sin^2(x) + cos^2(x)}</math>

8 KB (1,351 words) - 20:30, 10 July 2016
1984 AHSME Problems/Problem 29
The slope we are looking for is equivalent to <math>\tan (\theta + 45)</math> where <math>\angle AOX = \theta</math>. Using tangent <cmath> \tan (\theta + 45)= \frac{\tan \theta + \tan 45}{1-\tan\theta\tan 45} = \frac{\frac{1}{\sqrt{2}}+1}{1-\frac{1}{\sqrt{2}}}=3+2\sqrt{2}</cmath>

4 KB (614 words) - 20:09, 12 September 2022
1984 AHSME Problems/Problem 15
...times\sin{3x}}{\cos{2x}\times\cos{3x}}=1 </math>, or <math> \tan{2x}\times\tan{3x}=1 </math>. ...identity]] <math> -\tan{x}=\tan{-x} </math>, we have <math> \tan{2x}\times\tan{-3x}=-1 </math>.

3 KB (493 words) - 18:16, 4 June 2021
1966 IMO Problems/Problem 2
<cmath> a+b=\tan{\frac{\gamma}{2}}(a\tan{\alpha}+b\tan{\beta}) </cmath> ...at as <math>\gamma = \pi -\alpha-\beta</math> then and the identity <math>\tan\left(\frac \pi 2 - x \right)=\cot x</math> our equation becomes:

2 KB (416 words) - 17:54, 13 January 2022
1984 AHSME Problems/Problem 23
...}) \qquad \mathrm{(D) \ }\tan{15^\circ} \qquad \mathrm{(E) \ } \frac{1}{4}\tan{60^\circ} </math> ...irc}}{2\cos{15^\circ}\cos{5^\circ}}=\frac{\sin{15^\circ}}{\cos{15^\circ}}=\tan{15^\circ}, \boxed{\text{D}} </math>.

1 KB (159 words) - 12:52, 5 July 2013
1984 AHSME Problems/Problem 26
...can also use trig manipulation on <math>BCE</math> to get that <math>CE=a\tan{\beta}</math>. <math>[BED]=\frac{BD\cdot CE}{2}=\frac{ac\cos{\beta}\tan{\beta}}{4}=\frac{ac\sin{\beta}}{4}</math>

2 KB (303 words) - 20:28, 2 October 2023
1985 AHSME Problems
...= 25 \degree</math>, then the value of <math>\left(1+\tan A\right)\left(1+\tan B\right)</math> is ...\qquad \mathrm{(C) \ } 1+\sqrt{2} \qquad \mathrm{(D) \ } 2\left(\tan A+\tan B\right) \qquad \mathrm{(E) \ }\text{none of these} </math>

17 KB (2,488 words) - 03:26, 20 March 2024
1985 AHSME Problems/Problem 16
...= 25 \degree</math>, then the value of <math>\left(1+\tan A\right)\left(1+\tan B\right)</math> is ...\qquad \mathrm{(C) \ } 1+\sqrt{2} \qquad \mathrm{(D) \ } 2\left(\tan A+\tan B\right) \qquad \mathrm{(E) \ }\text{none of these} </math>

5 KB (904 words) - 22:25, 19 March 2024
2012 AMC 12B Problems/Problem 25
...counter-clockwise order and right angle at <math>A</math>, let <math>f(t)=\tan(\angle{CBA})</math>. What is <cmath>\prod_{t\in T} f(t)?</cmath> ...on <math>A'B'C'</math> labeled that way will give us <math>\tan CBA \cdot \tan C'B'A' = 1</math>. First we consider the reflection about the line <math>y=

5 KB (715 words) - 00:28, 3 August 2024
Mock Geometry AIME 2011 Problems/Problem 1
Now we have <math> BE=BA\cdot\tan\angle EAB=1\cdot\tan30^\circ=\frac{\sqrt{3}}{3} </math>. Finally, <math> [A

2 KB (376 words) - 22:06, 23 December 2022
Mock Geometry AIME 2011 Problems/Problem 11
...nt to <math>OB,OC,</math> and arc <math>BC</math>. It is known that <math>\tan AOC=\frac{24}{7}</math>. The ratio <math>\frac{r_2} {r_1}</math> can be exp ...h>, so <math> \cos AOC=\frac{7}{25} </math>. Now, we have <math> \tan FOD=\tan\frac{AOC}{2}=\sqrt{\frac{1-\cos AOC}{1+\cos AOC}}=\sqrt{\frac{1-\frac{7}{25

3 KB (432 words) - 14:12, 2 January 2012
Mock Geometry AIME 2011 Problems/Problem 8
...t triangles <math>\Delta PO_1H, \Delta O_1S_1O_2</math>. Similarly, <math>\tan{\angle PO_2O_1}=\frac{h}{b}=\frac{52}{39}</math> using right triangles <mat

3 KB (522 words) - 21:25, 3 January 2012
1989 AHSME Problems
<math> \cot 10+\tan 5 = </math> Find the sum of the roots of <math>\tan^2x-9\tan x+1=0</math> that are between <math>x=0</math> and <math>x=2\pi</math> radi

15 KB (2,247 words) - 13:44, 19 February 2020
1991 AHSME Problems/Problem 21
\text{(C) } tan^2\theta\quad

738 bytes (126 words) - 21:56, 17 October 2016
1989 AHSME Problems/Problem 28
Find the sum of the roots of <math>\tan^2x-9\tan x+1=0</math> that are between <math>x=0</math> and <math>x=2\pi</math> radi ...</math> are positive and distinct, so by considering the graph of <math>y=\tan x</math>, the smallest two roots of the original equation <math>x_1,\ x_2</

2 KB (282 words) - 21:14, 2 March 2019
1989 AHSME Problems/Problem 14
<math>\cot 10+\tan 5=</math> We have <cmath>\cot 10 +\tan 5=\frac{\cos 10}{\sin 10}+\frac{\sin 5}{\cos 5}=\frac{\cos10\cos5+\sin10\si

534 bytes (69 words) - 16:11, 25 February 2022
2012 AMC 12B Problems
...counter-clockwise order and right angle at <math>A</math>, let <math>f(t)=\tan(\angle{CBA})</math>. What is <cmath>\prod_{t\in T} f(t)?</cmath>

20 KB (2,681 words) - 09:47, 29 June 2023
2012 AMC 12B Problems/Problem 17
...</math> so that <math>\cos\theta=\sin(90-\theta)=s/6</math>. Then <math>m=\tan\theta=3</math>. Substituting into <math>\left(\frac{4m^2+10}{m^2+1},\frac{6 ...</math>. Setting <math>6\sin\theta=-2\cos\theta</math>, we get that <math>\tan\theta=-1/3</math>. This means <math>-1/3</math> is the slope of line <math>

12 KB (2,183 words) - 21:05, 23 December 2023
2012 AIME I Problems
...\angle C.</math> If <math>\frac{DE}{BE} = \frac{8}{15},</math> then <math>\tan B</math> can be written as <math>\frac{m \sqrt{p}}{n},</math> where <math>m

10 KB (1,617 words) - 14:49, 2 June 2023
2012 AIME I Problems/Problem 12
...\angle C.</math> If <math>\frac{DE}{BE} = \frac{8}{15},</math> then <math>\tan B</math> can be written as <math>\frac{m \sqrt{p}}{n},</math> where <math>m ...<math>FD = 4a\sqrt{3}</math>, and <math>FB = 11a</math>. Finally, <math>\tan{B} = \tfrac{DF}{FB}=\tfrac{4\sqrt{3}a}{11a} = \tfrac{4\sqrt{3}}{11}</math>.

10 KB (1,682 words) - 01:26, 12 September 2024
Mock AIME II 2012 Problems/Problem 6
pair tan = reflect(origin,(5,y))*(0,y); draw((5,y)--tan,linetype("4 4"));

3 KB (478 words) - 03:06, 5 April 2012
2006 SMT/Calculus Problems
Evaluate: <math> \int(x\tan^{-1}x)dx </math> <cmath> \frac{d}{dx}\tanh^{-1}\tan x </cmath>

3 KB (525 words) - 13:59, 27 May 2012
2006 SMT/Team Problems
...exist two positive numbers <math> x </math> such that <math> \sin(\arccos(\tan(\arcsin(x))))=x </math>. Find the product of the two possible <math> x </ma

6 KB (910 words) - 17:32, 27 May 2012
1953 AHSME Problems/Problem 50
...rac{2}{3}</math>, <math>\tan{\frac{B}{2}} = \frac{1}{2}</math>, and <math>\tan{\frac{C}{2}} = \frac{4}{x},</math> so we have the equation <math>\frac{1}{2

2 KB (314 words) - 21:17, 31 December 2023
2013 AMC 12B Problems
..., Q=(\cot x, \cot^2 x), R=(\sin x, \sin^2 x)</math> and <math>S =(\tan x, \tan^2 x)</math> are the vertices of a trapezoid. What is <math>\sin(2x)</math>?

15 KB (2,458 words) - 23:05, 4 July 2024
2013 AMC 12B Problems/Problem 20
..., Q=(\cot x, \cot^2 x), R=(\sin x, \sin^2 x)</math> and <math>S =(\tan x, \tan^2 x)</math> are the vertices of a trapezoid. What is <math>\sin(2x)</math>? Let <math>f,g,h,j</math> be <math>\sin, \cos, \tan, \cot</math> (not respectively). Then we have four points <math>(f,f^2),(g,

2 KB (375 words) - 00:54, 28 September 2021
1982 USAMO Problems/Problem 3
...} \cdot \frac{r}{s-b} \cdot \frac{r}{s-c} = \frac{1}{4} \tan A/2 \tan B/2 \tan C/2.</cmath> Lemma. <math>\tan x \tan (A - x)</math> is increasing on <math>0 < x < \frac{A}{2}</math>, where <ma

2 KB (376 words) - 23:29, 18 May 2015
2013 AIME I Problems/Problem 14
...tter. Only the numerator, because we are trying to find <math>\frac{P}{Q}=\tan\text{arg}(\Sigma)</math> a PROPORTION of values. So denominators would canc

10 KB (1,654 words) - 01:18, 21 August 2024
1991 AHSME Problems
\textbf{(C) } \tan^2\theta\qquad

16 KB (2,451 words) - 04:27, 6 September 2021
2003 AMC 12B Problems/Problem 10
real r = 5/dir(54).x, h = 5 tan(54*pi/180);

1 KB (237 words) - 23:06, 3 February 2020
Hyperbolic trig functions
<math>\tanh(x)= -1\tan{iz}</math>

423 bytes (78 words) - 23:33, 22 May 2013
2013 USAJMO Problems/Problem 5
<cmath>\frac{XC}{CY}=\tan {\angle CYZ}=\tan (90-\alpha)</cmath> <cmath>\frac{CQ}{CY}=\tan {\angle CYQ}=\tan (\alpha+\beta).</cmath>

7 KB (1,250 words) - 18:05, 1 October 2021
2014 AIME II Problems/Problem 12
<math>\tan{\frac{3A}{2}}\tan{\frac{3B}{2}}=1</math> Note that <math>\tan{x}=\frac{1}{\tan(90-x)}</math>, or <math>\tan{x}\tan(90-x)=1</math>

5 KB (875 words) - 17:56, 2 October 2023
2013 IMO Problems/Problem 4
...an{\angle{C}}-1)m}{i\tan{\angle{C}}}.</cmath> We wish to simplify <math>(i\tan{\angle{C}}-1)m</math> first. Note that <cmath>m=\frac{|CM|}{|CA|}\cdot(a)=\ (i\tan{\angle{C}}-1)m&=(i\tan{\angle{C}}-1)((|BC|)\cos{\angle{C}}(\cos{\angle{C}}+i\sin{\angle{C}}))\\

11 KB (1,991 words) - 01:31, 19 November 2023
User talk:Bobthesmartypants/Sandbox
label("$\frac{1}{\tan\varphi}$",(sqrt(3)/2,0),dir(-90));

17 KB (2,269 words) - 21:25, 6 November 2014
1967 IMO Problems/Problem 4
...se ratio is the <math>\sin</math> (or the <math>\cos</math>, or the <math>\tan</math>)

9 KB (1,689 words) - 02:54, 4 September 2024
User talk:Dwnadoa
Bifu Tan Yan, now Hublot sales inside Chinese market, a compact proportion, comprisi

2 KB (352 words) - 01:52, 20 December 2013
2014 AMC 10A Problems/Problem 22
...ow EC=20-10 \sqrt 3</math>. (It is important to memorize the sin, cos, and tan values of <math>15^\circ</math> and <math>75^\circ</math>.) Therefore, we h

13 KB (1,983 words) - 12:39, 7 June 2024
2014 AMC 12A Problems/Problem 25
...an angle <math>\theta</math> relative to the coordinate axis, where <math>\tan\theta = \tfrac 34</math>. We rotate the coordinate axis by angle <math>\the

4 KB (661 words) - 16:18, 2 September 2022
2014 AMC 12B Problems/Problem 19
...tan {\theta})^4 - 3(\tan {\theta})^2 + 1 = 0.</math> This gives us <math>(\tan {\theta})^2 = \dfrac{3+\sqrt{5}}{2}\longrightarrow \boxed{E}</math>

4 KB (703 words) - 18:38, 15 June 2024
2014 AIME I Problems
...\left(\tfrac{w-z}{z}\right) </math>. The maximum possible value of <math>\tan^2 \theta</math> can be written as <math>\tfrac{p}{q}</math>, where <math>p<

9 KB (1,472 words) - 13:59, 30 November 2021
2014 AIME I Problems/Problem 7
...\left(\tfrac{w-z}{z}\right) </math>. The maximum possible value of <math>\tan^2 \theta</math> can be written as <math>\tfrac{p}{q}</math>, where <math>p< We know that <math>\tan{\theta}</math> is equal to the imaginary part of the above expression divid

5 KB (782 words) - 20:25, 10 October 2023
1973 AHSME Problems
...nd <math> \sin \frac12 \theta = \sqrt{\frac{x-1}{2x}}</math>, then <math> \tan \theta</math> equals

18 KB (2,788 words) - 19:25, 22 August 2024
1980 AHSME Problems/Problem 12
..._2)}{1-\tan^2(\theta_2)} = 4\tan(\theta_2)</math>. Solving, we have <math>\tan(\theta_2) = 0, \dfrac{\sqrt{2}}{2}</math>. But line <math>L_1</math> is not

2 KB (237 words) - 18:02, 20 March 2018
1987 AHSME Problems
...c})+\log_{10}(\tan 3^{\circ})+\cdots+\log_{10}(\tan 88^{\circ})+\log_{10}(\tan 89^{\circ}). </math>

16 KB (2,291 words) - 13:45, 19 February 2020
1986 AHSME Problems
<math>\textbf{(A)}\ \tan \theta = \theta\qquad \textbf{(B)}\ \tan \theta = 2\theta\qquad

17 KB (2,512 words) - 18:30, 12 October 2023
1983 AHSME Problems
...angle at <math>C</math>. If <math>\sin A = \frac{2}{3}</math>, then <math>\tan B</math> is If <math>\tan{\alpha}</math> and <math>\tan{\beta}</math> are the roots of <math>x^2 - px + q = 0</math>, and <math>\co

15 KB (2,309 words) - 23:43, 2 December 2021
1978 AHSME Problems
If <math>\sin x+\cos x=1/5</math> and <math>0\le x<\pi</math>, then <math>\tan x</math> is

15 KB (2,432 words) - 01:06, 22 February 2024
1972 AHSME Problems
If <math>\tan x=\dfrac{2ab}{a^2-b^2}</math> where <math>a>b>0</math> and <math>0^\circ <x real x = 6-h*tan(t);

17 KB (2,725 words) - 00:21, 26 May 2024
2007 iTest Problems/Problem 17
...}{4}</math> and <math>\tan{y}=\frac{1}{6}</math>, find the value of <math>\tan{x}</math>. <math>\tan(x+\arctan\frac{1}{6})=\tan\frac{\pi}{4}=1</math>

2 KB (266 words) - 21:30, 4 February 2023
2007 IMO Problems/Problem 4
...PK = \dfrac{1}{2}a \tan \dfrac{1}{2}C</math> and <math>QL = \dfrac{1}{2}b \tan \dfrac{1}{2}C</math> from right triangles <math>\triangle PKC</math> and <m <cmath>= \dfrac{\frac{1}{2}a\tan\frac{1}{2}C \cdot (a + b)}{a\sin\frac{1}{2}C} </cmath>

8 KB (1,480 words) - 14:52, 5 August 2022
1988 AHSME Problems/Problem 13
...ation by <math>\cos{(x)}</math> to get <cmath>\frac{\sin{(x)}}{\cos{(x)}}=\tan{(x)}=3.</cmath>

2 KB (402 words) - 20:53, 24 August 2021
1987 AHSME Problems/Problem 20
...c})+\log_{10}(\tan 3^{\circ})+\cdots+\log_{10}(\tan 88^{\circ})+\log_{10}(\tan 89^{\circ}). </math> ...b}</math>, the answer is <math>\log_{10} {\tan 1^\circ \tan 2^\circ \dots \tan 89^\circ} = \log_{10} 1 = 0.</math> <math>\boxed{\textbf{(A)}}.</math>

1 KB (164 words) - 12:42, 31 March 2018
1986 AHSME Problems/Problem 21
<math>\textbf{(A)}\ \tan \theta = \theta\qquad \textbf{(B)}\ \tan \theta = 2\theta\qquad

2 KB (301 words) - 18:50, 1 April 2018
1986 AHSME Problems/Problem 28
...}{DP}, DP = \frac{1}{\tan 54}</cmath>Therefore, <math>AB = 2DP = \frac{2}{\tan 54}</math>. ...efore, <math>AO + AQ + AR = AO + 2AQ = \frac{1}{\sin 54}+\frac{4 \sin 72}{\tan 54} = \frac{1}{\sin 54} + 8 \sin 36 \cos 54 = \frac{1}{\cos 36} + 8-8\cos^2

4 KB (702 words) - 17:13, 17 April 2020
Angle Addition Formulas (Trigonometry)
..., we get<cmath>\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}.</cmath>

2 KB (458 words) - 19:24, 2 February 2020
2015 AMC 10A Problems/Problem 19
...> is an isosceles <math>30 - 75 - 75</math> triangle. Thus, <math>DF = CF \tan 15^\circ = \dfrac{5}{\sqrt{2}} (2 - \sqrt{3})</math> by the Half-Angle form

9 KB (1,522 words) - 19:54, 15 September 2024
2015 AIME II Problems/Problem 15
Setting the two areas equal, we get <cmath>\tan A = \frac{\sin A}{\cos A} = 8 \iff \sin A = \frac{8}{\sqrt{65}}, \cos A = \ <cmath>\tan{\angle{CYE}} = \frac{1}{8}</cmath>

31 KB (5,086 words) - 19:15, 20 December 2023
2015 USAMO Problems/Problem 2
Let angle <math>\angle XAB=A</math>, which is an acute angle, <math>\tan{A}=t</math>, then <math>X=(1-a,at)</math>.

5 KB (902 words) - 09:58, 20 August 2021
2015 USAJMO Problems/Problem 3
Let angle <math>\angle XAB=A</math>, which is an acute angle, <math>\tan{A}=t</math>, then <math>X=(1-a,at)</math>.

4 KB (760 words) - 16:45, 29 April 2020
2016 AMC 12B Problems/Problem 13
From Alice's point of view, <math>\tan(\theta)=\frac{z}{y}</math>. <math>\tan{30}=\frac{\sin{30}}{\cos{30}}=\frac{1}{\sqrt{3}}</math>. So, <math>y=z*\sqr From Bob's point of view, <math>\tan(\theta)=\frac{z}{x}</math>. <math>\tan{60}=\frac{\sin{60}}{\cos{60}}=\sqrt{3}</math>. So, <math>x = \frac{z}{\sqrt

6 KB (803 words) - 00:37, 2 November 2023
2016 AIME I Problems/Problem 6
...ath> is half the perimeter of <math>\triangle ABC</math>) and since <math>\tan\left(\frac{\gamma}{2}\right)=\frac{\frac{3}{2}r}{\frac{1}{2}c}</math> from

15 KB (2,694 words) - 13:36, 16 July 2024
2016 AIME I Problems/Problem 9
<cmath>\frac{31}{40} \geq \tan\theta</cmath> However, <math>\tan\theta = \tan(\frac{90-A}{2}) = \frac{\sin(90-A)}{\cos(90-A)+1} = \frac{\cos A}{\sin A +

9 KB (1,526 words) - 02:31, 29 December 2021
2016 AIME II Problems/Problem 14
...and <math>KQ=y</math>, assuming WLOG <math>x>y</math>, we must have <math>\tan(120^{\circ})=-\sqrt{3}=\dfrac{\dfrac{x+y}{100 \sqrt{3}}}{\dfrac{30000-xy}{3 ...<math>\sqrt{3}\tan{\left(\alpha\right)}</math>, we can set <math>\sqrt{3}\tan{\left(\alpha\right)}=a</math> for convenience since we really only care abo

15 KB (2,560 words) - 01:44, 1 July 2023
1983 AHSME Problems/Problem 20
If <math>\tan{\alpha}</math> and <math>\tan{\beta}</math> are the roots of <math>x^2 - px + q = 0</math>, and <math>\co ...cot\theta=\frac{1}{\tan\theta}</math>, we have <math>\frac{1}{\tan(\alpha)\tan(\beta)}=\frac{1}{q}=s</math>.

1 KB (222 words) - 00:58, 20 February 2019
2016 USAMO Problems/Problem 3
...= \frac {VI_A}{VO} = \frac {R \sin \psi + 2R \sin \alpha}{R \cos \psi} = \tan \psi + \frac{2 \sin\alpha}{\cos \psi}.</cmath> ...cot \angle UTW = \frac {TW}{UW} = \frac {AW \cdot \tan \psi}{AU – AW} = \tan \psi \cdot \frac {2a +b+c}{b+c} =</cmath>

6 KB (998 words) - 21:36, 17 October 2022
2016 USAMO Problems/Problem 5
...s clear that <math>I = \left(\frac{b + c – a}{2} , \frac{b + c – a}{2}\tan(A / 2)\right)</math>. ...and the <math>y</math> coordinate of <math>O</math> is <math>-\frac{b}{2} \tan{B-90}</math>. From this, <math>(5)</math> follows in this case as well.

8 KB (1,449 words) - 00:09, 12 October 2023
1983 AHSME Problems/Problem 5
...angle at <math>C</math>. If <math>\sin A = \frac{2}{3}</math>, then <math>\tan B</math> is so <math>\tan{B} = \frac{x \sqrt{5}}{2x} = \frac{\sqrt{5}}{2}</math>, which is choice <ma

1 KB (171 words) - 00:42, 20 February 2019
1979 AHSME Problems/Problem 20
...{2}+\frac{1}{3}}{1-(\frac{1}{2})(\frac{1}{3})}</math>. Simplifying, <math>\tan(\theta_a + \theta_b) = 1</math>, so <math>\theta_a + \theta_b</math> in rad

2 KB (363 words) - 13:33, 16 July 2024
2017 AMC 12A Problems/Problem 15
Let <math>f(x) = \sin{x} + 2\cos{x} + 3\tan{x}</math>, using radian measure for the variable <math>x</math>. In what in ...tan function. Upon further examination, it is clear that the positive the tan function creates will balance the other two functions, and thus the first s

3 KB (564 words) - 14:12, 23 October 2021
2017 AMC 12A Problems
Let <math>f(x) = \sin{x} + 2\cos{x} + 3\tan{x}</math>, using radian measure for the variable <math>x</math>. In what in

15 KB (2,418 words) - 16:58, 7 November 2022
2013 Mock AIME I Problems/Problem 12
...7{AE/2} = \tfrac{14}{AE}</math>. Equating these two expressions for <math>\tan\theta</math>, we see that <math>AE \cdot CE = 7 \cdot 14 = 98</math>. From

6 KB (1,035 words) - 10:42, 31 July 2024
2017 AIME I Problems/Problem 15
<cmath>x(\frac {\sin \beta}{\tan{\alpha}} - \cos \beta) +x (\frac {\cos\beta}{2} +\frac{\sin\beta \sqrt{3}}{

22 KB (3,622 words) - 17:11, 6 January 2024
Reaper Archives
| 67 || Senpai-Tan || 80 || 8151.479 || 101.893

187 KB (10,824 words) - 18:27, 3 February 2022
1972 AHSME Problems/Problem 20
If <math>\tan x=\dfrac{2ab}{a^2-b^2}</math> where <math>a>b>0</math> and <math>0^\circ <x We start by letting <math>\tan x = \frac{\sin x}{\cos x}</math> so that our equation is now: <cmath>\frac{

1 KB (177 words) - 19:14, 2 January 2024
1972 AHSME Problems/Problem 30
real x = 6-h*tan(t); real y = x*tan(2*t);

3 KB (431 words) - 19:52, 23 June 2021
2020 AMC 12A Problems
How many solutions does the equation <math>\tan{(2x)} = \cos{(\tfrac{x}{2})}</math> have on the interval <math>[0, 2\pi]?</

14 KB (2,073 words) - 15:15, 21 October 2021
2017 AMC 8 Problems/Problem 22
...le <math> ABD </math>, the radius of its inscribed circle <math> r = \frac{tan \frac{\angle BAD}{2}\cdot (AB + AD - BD)}{2} = \frac{\frac{5}{12} \cdot 16}

2 KB (318 words) - 07:58, 27 August 2024
MIE 96/97
If <math>\tan a</math> and <math>\tan b</math> are the roots of <math>x^2+px+q=0</math>, then compute, in terms o

2 KB (377 words) - 14:52, 7 January 2018
MIE 2016/Day 1/Problem 4
&\tan(2b)= &\frac{1}{4}\\&

571 bytes (90 words) - 05:19, 17 June 2021
MIE 2015
<math>\sin x\left(1+\tan x\tan\frac{x}{2}\right)=4-\cot x</math>

7 KB (1,127 words) - 18:23, 11 January 2018
British Mathematical Olympiad
...> height of the cone be <math>h,</math> radius of the cone be <math>r = h \tan \theta.</math> <cmath>BO = a, BC = \frac {a}{\sqrt {2}}, AO = h, DO = r = h \tan \theta.</cmath>

6 KB (1,034 words) - 10:12, 7 June 2023
2018 AMC 12A Problems/Problem 23
...\frac{2\sin(46)\cos(10)}{-2\sin(46)\sin({-10})}=\frac{\sin(80)}{\cos(80)}=\tan(80)</cmath>

12 KB (1,878 words) - 00:53, 2 September 2024
2018 AMC 12A Problems/Problem 22
...c{7\pi}{6}</math> without the loss of generality. Since <math>\tan(2\phi)>\tan\frac{\pi}{3},</math> we deduce that <math>2\phi>\frac{\pi}{3},</math> from

10 KB (1,662 words) - 12:45, 13 September 2021
2018 AMC 12B Problems/Problem 21
...nd keeping the idea that cos(θ) = <math>\frac{5}{13}</math>, we find that tan(θ/2) = sqrt((1-cosθ)/(1+cosθ)) = <math>\sqrt{\frac{4}{9}}</math> = <math

7 KB (1,166 words) - 23:14, 8 September 2024
1969 IMO Problems/Problem 2
<cmath>\tan{x_1}=\frac{\cos{a_1}+\frac{1}{2}\cos{a_2}+\frac{1}{4}\cos{a_3}+\cdots+\frac ...ath> is <math>\pi</math>, this means that <math>\tan{x_1}=\tan{(x_1+\pi)}=\tan{(x_1+m\pi)}</math> for any natural number <math>m</math>. That implies that

5 KB (1,005 words) - 00:55, 9 August 2024
2018 AIME I Problems/Problem 13
<cmath>\frac {(r-r_1)\cdot (r-r_2)}{r_1 \cdot r_2} =\tan\beta \tan\gamma.</cmath> <cmath>1 -\frac{2r}{h} = \frac {b+c-a}{b+c+a} = \frac {r}{r_a} = \tan\beta \tan\gamma .</cmath>

13 KB (2,200 words) - 21:36, 6 January 2024
1973 AHSME Problems/Problem 17
...nd <math> \sin \frac12 \theta = \sqrt{\frac{x-1}{2x}}</math>, then <math> \tan \theta</math> equals <cmath>\tan \frac{\theta}{2} = \sqrt{\frac{x-1}{2x}} \div \sqrt{\frac{x+1}{2x}}</cmath>

1 KB (184 words) - 14:00, 20 February 2020
1983 AHSME Problems/Problem 27
...}{2}\right )\right )=\tan \left (\frac{1}{2} \right )</math>. Since <math>\tan \left(\frac{\theta}{2} \right ) = \frac{1-\cos \left(\theta \right )}{\sin

1 KB (245 words) - 14:00, 29 January 2023
2008 iTest Problems/Problem 46
...ath> in the interval <math>[0,2\pi)</math> that satisfy <math>\tan^2 x - 2\tan x\sin x=0</math>. Compute <math>\lfloor10S\rfloor</math>. ...0</math>. By the Zero Product Property, <math>\tan x = 0</math> or <math>\tan x = 2\sin x</math>.

969 bytes (158 words) - 19:00, 12 July 2018
2008 iTest Problems/Problem 57
Let a and b be the two possible values of <math>\tan\theta</math> given that <math>\sin\theta + \cos\theta = \dfrac{193}{137}</m ...the sum formula for tangent, the sum of the two possible values of <math>\tan \theta</math> is

2 KB (343 words) - 20:35, 4 August 2018
2006 Indonesia MO Problems
...all triangles <math> ABC</math> which have property: <math> \tan A,\tan B,\tan C</math> are positive integers. Prove that all triangles in <math> S</math>

3 KB (439 words) - 12:39, 4 September 2018
User:Danhngonds
...úi lửa phun trào tạo ra nhiều khói bụi và các loại khí mê-tan, sunfua độc hại.

3 KB (788 words) - 00:10, 5 December 2018
2008 iTest Problems/Problem 96
\tan(2x) &= \frac{\sqrt{3}}{4}.

3 KB (558 words) - 20:13, 4 January 2019
2008 iTest Problems/Problem 98
\tan(\angle AIS + \angle DIS) &= -\tan(\angle BIT + \angle CIT) \\ ...ngle DIS} &= \frac{\tan \angle BIT + \tan \angle CIT}{1 - \tan \angle BIT \tan \angle CIT}

3 KB (586 words) - 23:47, 8 January 2019
2019 AIME II Problems/Problem 10
...th> such that for nonnegative integers <math>n</math>, the value of <math>\tan{\left(2^{n}\theta\right)}</math> is positive when <math>n</math> is a multi Note that if <math>\tan \theta</math> is positive, then <math>\theta</math> is in the first or thir

8 KB (1,172 words) - 18:21, 8 August 2024
2019 USAJMO Problems/Problem 4
<cmath>FG = AG - FG = \frac{r_a}{\tan \left( \frac{A}{2} \right)} - b \cos (A)</cmath> <cmath>FH = AH - AF = \frac{r_a}{\tan \left( \frac{A}{2} \right)} - c \cos (A)</cmath>

10 KB (1,536 words) - 20:27, 12 April 2021
2006 iTest Problems/Problem U6
...ouble Angle Identity yields <math>\tan 2\theta = \frac34</math>, so <math>\tan (90 - 2\theta) = \frac43</math>.

4 KB (722 words) - 20:53, 27 March 2019
2019 AMC 10B Problems/Problem 23
...olving, we obtain <math>\tan{x}=\frac{1}{4}</math>. Then, note that <math>\tan{x}=r/{BC}</math>, so <math>r=\frac{1}{4}*\sqrt{170}</math>. Finishing off,

6 KB (967 words) - 10:25, 20 December 2023
2019 AIME I Problems/Problem 6
...65}{28}=\frac{28\sin\alpha}{8\cos\alpha}=\frac{7}{2}\tan\alpha\Rightarrow \tan\alpha=\frac{65}{98}</cmath> Thus, <math>MK=\frac{MN}{\tan\alpha}=98</math>, so <math>MO=MK-KO=\boxed{090}</math>.

11 KB (1,717 words) - 20:11, 19 January 2024
2019 AIME II Problems
...th> such that for nonnegative integers <math>n,</math> the value of <math>\tan(2^n\theta)</math> is positive when <math>n</math> is a multiple of <math>3<

7 KB (1,254 words) - 14:45, 21 August 2023
2019 AIME II Problems/Problem 1
<cmath>\tan a = \frac{8}{15}</cmath> <cmath>A = \frac{1}{2}* 9*\frac{9}{2}\tan a = \frac{54}{5}</cmath>

9 KB (1,508 words) - 14:02, 7 September 2024
Euler's formula and It's history
x = tan-1 ( 4 / 3 ) = 0.927 (to 3 decimals)

3 KB (543 words) - 15:24, 13 June 2019
Mock AIME 2 Pre 2005 Problems
...lpha) + \tan^{-1}(\beta) + \tan^{-1} (\gamma).</cmath> The value of <math>\tan(\omega)</math> can be written as <math>\tfrac{m}{n}</math> where <math>m</m

6 KB (1,052 words) - 13:52, 9 June 2020
2006 SMT/Calculus Problems/Problem 5
Evaluate: <math> \int(x\tan^{-1}x)dx </math> \int(x\tan^{-1}x)dx &= \frac{x^2}{2}\tan^{-1}x-\int\frac{x^2}{2(x^2+1)}dx\\

670 bytes (116 words) - 18:31, 14 January 2020
2006 SMT/Advanced Topics Problems/Problem 10
Using the identity that <math>\tan(x) = -\tan(-x)</math>

1 KB (175 words) - 18:30, 14 January 2020
2006 Indonesia MO Problems/Problem 3
...all triangles <math> ABC</math> which have property: <math> \tan A,\tan B,\tan C</math> are positive integers. Prove that all triangles in <math> S</math> ...B+C = 180^\circ</math>, so <math>z = \tan C = \tan (180^\circ - (A+B)) = -\tan(A+B)</math>.

3 KB (465 words) - 12:00, 26 September 2019
2020 AIME I Problems/Problem 13
We compute that <math>\cos{\angle{ABC}}=\frac{1}{8}</math>, so <math>\tan{\angle{ABC}}=3\sqrt{7}</math>. ...n \angle ABC}{1 + \cos \angle ABC} = \frac{\sqrt{7}}{3}</math>, and <math>\tan \angle DAF = \frac{\sqrt{7}}{7}</math>.

37 KB (5,512 words) - 06:56, 3 August 2024
User:Melindahsgateas
...y ra. Cùng công ty[http://icalshare.com/calendars/10983 Rut ham cau Quan Tan Binh] tìm hiểu nguyên nhân cũng như cách xử lý phù hợp. ...ttps://www.facebook.com/ruthamcauphuloc/posts/1184756188402107 Rut ham cau Tan Binh] sẽ giúp giải quyết triệt để sự cố tắc nghẽn nhà

6 KB (1,452 words) - 04:53, 15 October 2019
2023 AIME I Problems/Problem 12
...erty that <cmath>\angle AEP = \angle BFP = \angle CDP.</cmath> Find <math>\tan^2(\angle AEP).</math> ...ath>D=(0, 0)</math>, and <math>C=(48, 0)</math>, where we will find <math>\tan^{2}\left(\measuredangle CDP\right)</math> with <math>P=(BFD)\cap(CDE)</math

16 KB (2,592 words) - 15:40, 13 April 2024
2000 Pan African MO Problems/Problem 1
<cmath> \sin^3{x}(1+\cot{x})+\cos^3{x}(1+\tan{x})=\cos{2x} </cmath> ...= 0</math>, we can divide both sides by <math>\cos{x}</math> to get <math>\tan{x} = -1</math>. Thus, <math>x = \frac{3 \pi}{4} + \pi n</math>, where <mat

2 KB (305 words) - 06:07, 23 February 2023
2006 iTest Problems/Problem U7
...opposite angle to <math>x</math> be <math>\theta</math>, and let <math>t:=\tan\frac{\theta}{2}</math>; let the [[area]] be <math>A</math> and the [[semipe

4 KB (674 words) - 16:03, 25 February 2021
2000 Pan African MO Problems
<cmath> \sin^3{x}(1+\cot{x})+\cos^3{x}(1+\tan{x})=\cos{2x} </cmath>

2 KB (393 words) - 13:39, 4 December 2019
2000 SMT/Calculus Problems
...t value of <math>x</math> <math>(0 < x < \frac{\pi}{2})</math> does <math>\tan x + \cot x</math> achieve its minimum?

3 KB (413 words) - 13:10, 21 January 2020
2000 SMT/Team Problems
...-\beta) = 0</math>, <math>\tan(\beta) = \frac{1}{2000}</math>, find <math>\tan(\alpha)</math>.

4 KB (618 words) - 13:33, 21 January 2020
Unit Circle
...t can be used to demonstrate trigonometric functions such as sin, cos, and tan, but it is most commonly used to visualize the complex numbers. This is don

741 bytes (131 words) - 11:50, 22 January 2020
2020 AMC 10A Problems/Problem 20
...lve the quadratic, taking the positive solution (C is acute) to get <math>\tan{C} = \frac{1}{3}.</math> So if <math>AB = a,</math> then <math>BC = 3a</mat ...n angle bisector of <math>\triangle ABC</math> (because we will get <math>\tan(x) = 1</math>).

14 KB (2,120 words) - 13:36, 25 August 2024

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