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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 (315 words) - 21:13, 28 February 2022
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=

2 KB (356 words) - 17:10, 4 April 2020
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>.

9 KB (1,523 words) - 12:23, 7 September 2022
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>?

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) - 19:38, 12 November 2023
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

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