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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>?

16 KB (2,459 words) - 02:46, 30 January 2021
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 (1,641 words) - 20:03, 3 January 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
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

12 KB (1,821 words) - 18:16, 29 October 2023
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) - 16:24, 9 September 2022
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) - 13:55, 20 February 2020
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 (2,732 words) - 13:54, 20 February 2020
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 (1,513 words) - 23:44, 13 May 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 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) - 12:46, 10 May 2022
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
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
...an use the famous mnemonic SOH CAH TOA. <math>AD=AB-DB=13-5=8 \Rightarrow \tan \angle BAC = \frac{5}{12}=\frac{r}{8} \Rightarrow 12r=40 \Rightarrow r= \fr

5 KB (762 words) - 03:46, 22 April 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) - 22:11, 23 October 2021
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
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 .../math> 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 e

1 KB (269 words) - 11:29, 4 April 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
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

7 KB (1,109 words) - 00:40, 28 January 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>

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
2020 AIME I Problems/Problem 1
A = (0, tan(3 * pi / 7));

6 KB (968 words) - 15:01, 24 January 2024
AMC 12C 2020
How many solutions does the trigonometric equation <math>tan(cos(x)) = cos((x\pi^{2} - sin(x))</math> have in the interval <math>[-\pi,

9 KB (1,450 words) - 18:33, 21 April 2020
Pythagorean identities
<math>\tan^2x + 1 = \sec^2x</math>

975 bytes (164 words) - 14:07, 3 January 2024
Angle addition identities
<math>\tan(x + y) = \frac{\tan (x) + \tan (y)}{1 - \tan (x) \tan (y)}</math> ...cos \alpha \cos \beta}} = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}</math>

2 KB (372 words) - 20:46, 13 January 2024
Double angle identities
* <math>\tan (2x) = \frac{2\tan (x)}{1-\tan^2 (x)}</math>

357 bytes (54 words) - 20:21, 7 September 2023
Half angle identities
* <math>\tan \left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 - \cos (x)}{1+\cos (x)}} = \fr

956 bytes (149 words) - 20:23, 7 September 2023
2020 AIME II Problems/Problem 4
...n</math> angle formula, <math>\tan{(a-b)}=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}</math> to find the slope of line <math>CP</math>. We know that line <ma

10 KB (1,542 words) - 13:29, 19 January 2024
2020 AIME II Problems/Problem 13
...<math>\tan \left(\frac{\angle B}{2}\right) = \frac{r}{2}</math> and <math>\tan \left(\frac{\angle E}{2}\right) = \frac{r}{4}</math>, so <cmath>\cos (\angle B) =\frac{1-\tan^2 \left(\frac{\angle B}{2}\right)}{1+\tan^2 \left(\frac{\angle B}{2}\right)} = \frac{4-r^2}{4+r^2}</cmath>and

13 KB (2,197 words) - 23:00, 8 January 2024
2021 AIME I Problems/Problem 2
...ngles from the larger rectangle, we get Area = <math>33-3BG=33-\frac{9}{\tan(\angle DAE)}</math>. <math>\alpha=\tan^{-1}\left(\frac{3}{11}\right)</math>

8 KB (1,294 words) - 00:59, 23 August 2022
2021 AIME I Problems/Problem 11
The angle <math>\theta</math> between diagonals satisfies <cmath>\tan{\frac{\theta}{2}}=\sqrt{\frac{(s-b)(s-d)}{(s-a)(s-c)}}</cmath> (see https:/ ...\tan{\frac{\theta}{2}}=\sqrt{\frac{(11-4)(11-6)}{(11-5)(11-7)}}\text{ or }\tan{\frac{\theta}{2}}=\sqrt{\frac{(11-5)(11-7)}{(11-4)(11-6)}}.</cmath>

16 KB (2,635 words) - 19:56, 24 December 2023

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