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• 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>
  11 KB (1,794 words) - 15:32, 14 January 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>.
  35 KB (5,215 words) - 23:08, 29 October 2023
• 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>).
  13 KB (2,046 words) - 18:33, 28 October 2023
• 2020 AMC 12A Problems/Problem 9
  How many solutions does the equation <math>\tan(2x)=\cos(\tfrac{x}{2})</math> have on the interval <math>[0,2\pi]?</math> We count the intersections of the graphs of <math>y=\tan(2x)</math> and <math>y=\cos\left(\frac x2\right):</math>
  4 KB (615 words) - 04:07, 8 July 2022
• 2020 AMC 12A Problems/Problem 12
  ...\frac{3}{5}</math>, and the angle we are rotating around is x, then <math>\tan x = \frac{3}{5}</math> <math>\tan(x+45^{\circ}) = \frac{\tan x + \tan(45^{\circ})}{1-\tan x*\tan(45^{\circ})} = \frac{0.6+1}{1-0.6} = \frac{1.6}{0.4} = 4</math>
  7 KB (1,145 words) - 20:27, 5 November 2023
• 2020 AMC 12A Problems/Problem 14
  ...}{2}</math>, the area of the octagon is then <math>\frac{1}{2} \cdot \text{tan}(67.5) \cdot \frac{8}{2}</math>. = 4 \cdot \frac{\text{sin}^2(67.5)}{2\cdot \text{tan}(67.5)}
  11 KB (1,654 words) - 02:01, 17 September 2023
• 2020 AMC 12B Problems/Problem 7
  ==Solution 4 (tan)== ...45^{\circ}+\theta)=\frac{\tan(45^{\circ})+\tan(\theta)}{1-\tan(45^{\circ})\tan(\theta)} = \frac{1+a}{1-a}</math>. Since the slope of one line is <math>6</
  5 KB (895 words) - 15:03, 8 June 2023

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