MHB Trigonometric of tangent and sine functions

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The discussion focuses on simplifying the expression involving tangent and sine functions: \(\left(\tan \dfrac{2\pi}{7}-4\sin \dfrac{\pi}{7}\right)\left(\tan \dfrac{3\pi}{7}-4\sin \dfrac{2\pi}{7}\right)\left(\tan \dfrac{6\pi}{7}-4\sin \dfrac{3\pi}{7}\right). A user named Dan shares a numerical approximation of the result, approximately -8.7083, and asks if they are close to the correct answer. The conversation revolves around verifying this simplification and discussing the properties of the trigonometric functions involved. The overall goal is to confirm the accuracy of the numerical result and the simplification process.
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Simplify $\left(\tan \dfrac{2\pi}{7}-4\sin \dfrac{\pi}{7}\right)\left(\tan \dfrac{3\pi}{7}-4\sin \dfrac{2\pi}{7}\right)\left(\tan \dfrac{6\pi}{7}-4\sin \dfrac{3\pi}{7}\right)$.
 
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I got about -8.7083121069873265814843145114959219999438416220762493062309701968483080116471438626714199913273209202440106287205185869613863265722279384849632901398226721432702744787955199077781973792055631357781755415456733331388124469116498805756011983384506433640247288593850820089590928253650738597742412754348439402301589178427587352708579714081417809559865987679591010005488147679340458470850085945675037017144904042287279335497160556553452963874716319026939725908000014025345344112227718546669183167834689374961602264946845122520973969612465747159538537772604016219227651262298967156522926816165972425283143645081705710618441263404832174787619393871431981552587401078586609898568054592224581007986448669288148444215128498186229896741206349315952054866256984612864167199842472866575465952447085845790874114020207148838309575537421068097653460699858226519033408205869208586639796284569971790165922274544500088737294024496093237634962135759638417523292668843921768916664652250085254395250841082981914897137469833874653795905403859395226103590166597307639852476273392879083873723649605334228340639098097814440444549718831560592044450598723461692804007733588937741590648889021935199757480637837660523461803937847078950866107512314217003731198786042996011750451712135884514444878781114357178952544416819768594073353584110967734337432198244086867356343864416552649586451520413569547092694404444532795994994438271445333716459116502792251091594744077024600155160457563503684373664211499889132746530159707148096441069149624420565803877227318131627873671715105932608979648146278844613590157353436714676787933231516958800501704045205913556958442328349254532043183475963879427381686337157339601066369199608533673687386762949721994907024739645071023865884051877078161059364899178505753714909479228031574944638783941268199880771092085126999780242627307210910698666574484989827761671955524479251591513340499083326461726929118742316319408632872884749801756534634165971359295929387392681708974873301182019170001880197471302122860499952496977...

Am I close?

-Dan
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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