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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 (1,182 words) - 09:56, 7 February 2022
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>.

33 KB (5,177 words) - 21:05, 4 February 2023
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 (169 words) - 18:22, 9 March 2014
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 (1,851 words) - 12:31, 21 December 2021
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
LaTeX:Commands
...||\cos||<math>\textstyle \sin</math>||\sin||<math>\textstyle \tan</math>||\tan

12 KB (1,898 words) - 15:31, 22 February 2024
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
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 (738 words) - 13:11, 27 March 2023
2004 AMC 10A Problems/Problem 20
E = (0,Tan(15)); F = (1 - Tan(15),1);

4 KB (710 words) - 02:47, 18 April 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

6 KB (979 words) - 12:50, 17 July 2022
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{\sin{37}}{\cos{37}}\end{align*

8 KB (1,338 words) - 23:15, 28 November 2023
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>. ...now drop altitude AY to solve for tan2A ; now since we know tan2A we know tan A = r/x in terms of r hence solve the resulting equation in r

6 KB (1,065 words) - 20:12, 9 August 2022
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) - 17:15, 20 January 2024

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