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  • ...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
  • * <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
  • ...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
  • * Michael Tan (2009)
    20 KB (2,025 words) - 01:33, 11 June 2024
  • | Nathan Liu, Vincent Wang, Drake Tan, Channing Yang
    51 KB (6,220 words) - 21:22, 17 July 2024
  • <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
  • real r = 5/dir(54).x, h = 5 tan(54*pi/180);
    13 KB (1,987 words) - 18:53, 10 December 2022
  • ...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
  • ...>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
  • ...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
  • ...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
  • ...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
  • ...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
  • ...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
  • ...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
  • 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
  • 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
  • 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
  • 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
  • ...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
  • .../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
  • ...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
  • ...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
  • ...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
  • 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
  • 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
  • ...\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
  • ...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
  • 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
  • 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
  • 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
  • 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
  • ...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
  • 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
  • \begin{align*}DP&=z\tan\theta\\ EP&=x\tan\theta\\
    7 KB (1,184 words) - 13:25, 22 December 2022
  • \begin{eqnarray*} \tan \alpha & = & \frac {21}{27} \\ \tan \beta & = & \frac {21}{23} \\
    3 KB (472 words) - 19:03, 21 June 2024
  • 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
  • ...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
  • 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
  • <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
  • 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
  • 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
  • ...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
  • 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
  • ...</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
  • <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
  • ...}{\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
  • ...}{\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
  • ...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
  • ...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
  • ..., <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
  • 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
  • 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
  • ...>. 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
  • | <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
  • *<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
  • & = &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
  • ...}{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
  • ...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
  • ...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
  • ...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
  • \begin{matrix} {CE} & = & r \tan(COE) \\
    4 KB (684 words) - 07:28, 3 October 2021
  • ...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
  • 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
  • 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
  • \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
  • \tan{\alpha}=\frac{4nh}{(n^2-1)a}.
    3 KB (511 words) - 21:21, 20 August 2020
  • ...{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
  • ...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
  • ...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
  • <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
  • ...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
  • ...||\cos||<math>\textstyle \sin</math>||\sin||<math>\textstyle \tan</math>||\tan
    12 KB (1,796 words) - 13:14, 12 September 2024
  • (a) <math>\tan(45^\circ)</math> (g) <math>\tan(21\pi)</math>
    2 KB (338 words) - 13:28, 24 March 2020
  • ...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
  • <math>\tan (-A)=-\tan A</math> <math>\tan (\pi+A) = \tan A</math>
    6 KB (1,097 words) - 18:18, 21 January 2016
  • <math>b \tan{\frac{\omega}{2}} \le c < b</math>
    3 KB (425 words) - 21:18, 20 August 2020
  • ...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
  • E = (0,Tan(15)); F = (1 - Tan(15),1);
    4 KB (710 words) - 18:04, 6 September 2024
  • ...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
  • <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
  • 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
  • <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
  • 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
  • ..., </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
  • ...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
  • ..., </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
  • ...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
  • ...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
  • ...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
  • 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
  • <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
  • ...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
  • ...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
  • <cmath>\cot(\theta)=\tan(5^\circ)</cmath>
    12 KB (1,944 words) - 18:31, 28 August 2024
  • ...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
  • ...>. 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
  • ...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
  • ...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
  • ...\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
  • ...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
  • ...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
  • ...<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
  • ...}</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
  • ...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
  • ...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
  • ...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
  • <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
  • \sum_{n = 0}^\infty \frac{E_n}{n!} x^n = \sec x + \tan x .
    2 KB (246 words) - 12:50, 6 August 2009
  • ...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
  • If <math>y(x) = \tan x</math>, then <math>\frac{dy}{dx} = \sec^2 x</math>. Note that this follow
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  • <cmath>\frac{\sqrt{3}}{2}\cos(x)=\frac{3}{2}\sin(x)\implies \tan(x)=\frac{\sqrt{3}}{3}</cmath>
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  • ...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
  • ...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}
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  • <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}
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  • 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
  • <math>AY \cos(\delta)\sin(\alpha)\tan(\delta) = AY \sin(\delta)\sin(\alpha)</math>
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  • ...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
  • ...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
  • | Tan
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  • ...\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
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  • ...- 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 = \
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  • label("$\tan \theta$", (1,.375), E);
    7 KB (964 words) - 21:43, 28 March 2022
  • <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
  • 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
  • <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
  • <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
  • ...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
  • ...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
  • ...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
  • \sin, \cos, \tan represents <math>\sin, \cos, \tan</math>
    2 KB (318 words) - 13:47, 4 May 2024
  • ...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
  • ...}) \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
  • <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
  • 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
  • ...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
  • <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
  • ...}) \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
  • ...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
  • ...= 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
  • ...= 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
  • ...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=
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  • 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
  • ...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
  • ...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
  • <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
  • \text{(C) } tan^2\theta\quad
    738 bytes (126 words) - 21:56, 17 October 2016
  • 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
  • <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
  • ...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
  • ...</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
  • ...\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
  • ...\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
  • pair tan = reflect(origin,(5,y))*(0,y); draw((5,y)--tan,linetype("4 4"));
    3 KB (478 words) - 03:06, 5 April 2012
  • 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
  • ...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
  • ...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
  • ..., 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
  • ..., 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
  • ...} \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
  • ...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
  • \textbf{(C) } \tan^2\theta\qquad
    16 KB (2,451 words) - 04:27, 6 September 2021
  • real r = 5/dir(54).x, h = 5 tan(54*pi/180);
    1 KB (237 words) - 23:06, 3 February 2020
  • <math>\tanh(x)= -1\tan{iz}</math>
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  • <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
  • <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
  • ...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
  • label("$\frac{1}{\tan\varphi}$",(sqrt(3)/2,0),dir(-90));
    17 KB (2,269 words) - 21:25, 6 November 2014
  • ...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
  • Bifu Tan Yan, now Hublot sales inside Chinese market, a compact proportion, comprisi
    2 KB (352 words) - 01:52, 20 December 2013
  • ...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
  • ...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
  • ...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
  • ...\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
  • ...\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
  • ...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
  • ..._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
  • ...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
  • <math>\textbf{(A)}\ \tan \theta = \theta\qquad \textbf{(B)}\ \tan \theta = 2\theta\qquad
    17 KB (2,512 words) - 18:30, 12 October 2023
  • ...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
  • 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
  • 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
  • ...}{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
  • ...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
  • ...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
  • ...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
  • <math>\textbf{(A)}\ \tan \theta = \theta\qquad \textbf{(B)}\ \tan \theta = 2\theta\qquad
    2 KB (301 words) - 18:50, 1 April 2018
  • ...}{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
  • ..., 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
  • ...> 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
  • 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
  • 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
  • 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
  • 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
  • ...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
  • <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
  • ...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
  • 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
  • ...= \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
  • ...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
  • ...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
  • ...{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
  • 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
  • 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
  • ...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
  • <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
  • | 67 || Senpai-Tan || 80 || 8151.479 || 101.893
    187 KB (10,824 words) - 18:27, 3 February 2022
  • 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
  • real x = 6-h*tan(t); real y = x*tan(2*t);
    3 KB (431 words) - 19:52, 23 June 2021
  • 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
  • ...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
  • 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
  • &\tan(2b)= &\frac{1}{4}\\&
    571 bytes (90 words) - 05:19, 17 June 2021
  • <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
  • ...> 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
  • ...\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
  • ...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
  • ...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
  • <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
  • <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
  • ...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
  • ...}{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
  • ...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
  • 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
  • ...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
  • ...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
  • \tan(2x) &= \frac{\sqrt{3}}{4}.
    3 KB (558 words) - 20:13, 4 January 2019
  • \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
  • ...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
  • <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
  • ...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
  • ...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
  • ...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
  • ...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
  • <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
  • x = tan-1 ( 4 / 3 ) = 0.927 (to 3 decimals)
    3 KB (543 words) - 15:24, 13 June 2019
  • ...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
  • 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
  • Using the identity that <math>\tan(x) = -\tan(-x)</math>
    1 KB (175 words) - 18:30, 14 January 2020
  • ...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
  • 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
  • ...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
  • ...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
  • <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
  • ...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
  • <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
  • ...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
  • ...-\beta) = 0</math>, <math>\tan(\beta) = \frac{1}{2000}</math>, find <math>\tan(\alpha)</math>.
    4 KB (618 words) - 13:33, 21 January 2020
  • ...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
  • ...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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