Valor de $\cot12^{\circ} \cdot \cot24^{\circ} \cdot \cot28^{\circ}\cdot \cot32^{\circ} \cdot \cot48^{\circ}\cdot \cot88^{\circ}$

Question

Necesito una solución sin calculadora.

Intenté concertarlo en una expresión de la tangente pero eso no tenía ningún valor de $$\cot12^{\circ} \cdot \cot24^{\circ} \cdot \cot28^{\circ}\cdot \cot32^{\circ} \cdot \cot48^{\circ}\cdot \cot88^{\circ}$$ es $1$ .

Answer 1

Denote $$X = \cot12^{\circ} \cdot \cot24^{\circ} \cdot \cot28^{\circ}\cdot \cot32^{\circ} \cdot \cot48^{\circ}\cdot \cot88^{\circ}.\tag{1}$$

Utilizar la identidad $$ \tan x \cdot \tan(60^{\circ}-x) \cdot \tan(60^{\circ}+x) = \tan (3x), \tag{2} $$ (ver laboratorio bhattacharjee Las respuestas de la gente aquí o aquí ) o uno equivalente para $\cot(\bullet)$ : $$ \cot x \cdot \cot(60^{\circ}-x) \cdot \cot(60^{\circ}+x) = \cot (3x) \tag{2'} , $$ se puede reescribir $X$ como: $$X = \cot12^{\circ} \cdot \cot24^{\circ} \cdot \color{violet}{\cot28^{\circ}}\cdot \color{violet}{\cot(60^{\circ}-28^{\circ})} \cdot \cot48^{\circ}\cdot \color{violet}{\cot(60^{\circ}+28^{\circ})} \\ =\cot12^{\circ} \cdot \cot24^{\circ} \cdot \cot 48^{\circ} \cdot \underline{\cot84^{\circ}} \\ =\cot12^{\circ} \cdot \cot24^{\circ} \cdot \cot48^{\circ}\cdot \cot 84^{\circ} \cdot \color{tan}{\cot 72^{\circ}} \cdot \color{tan}{\tan 72^{\circ}}\\ =\color{green}{\cot12^{\circ}} \cdot \cot24^{\circ} \cdot \color{green}{\cot(60^{\circ} - 12^{\circ})}\cdot \cot 84^{\circ} \cdot \color{green}{\cot (60^{\circ} + 12^{\circ})} \cdot \color{tan}{\tan 72^{\circ}}\\ = \cot24^{\circ} \cdot \underline{\cot 36^{\circ}} \cdot \cot84^{\circ} \cdot \color{tan}{\tan 72^{\circ}} \\ = \color{orange}{\cot24^{\circ}} \cdot \color{orange}{\cot (60^{\circ} - 24^{\circ})} \cdot \color{orange}{\cot(60^{\circ} + 24^{\circ})} \cdot \color{tan}{\tan 72^{\circ}} \\ = \underline{\cot 72^{\circ}} \cdot \tan 72^{\circ} \\ = 1. \tag{3} $$

Answer 2

Una pista:

$\cot12^\circ\cot24^\circ\cot28^\circ\cot32^\circ\cot48^\circ\cot88^\circ$

$=\cot24^\circ\dfrac{\cos(48^\circ-12^\circ)+\cos(48^\circ+12^\circ)}{\cos(48^\circ-12^\circ)-\cos(48^\circ+12^\circ)}\dfrac{\cos(32^\circ-28^\circ)+\cos(32^\circ+28^\circ)}{\cos(32^\circ-28^\circ)-\cos(32^\circ+28^\circ)}\tan2^\circ$

$=\cot24^\circ\dfrac{\cos36^\circ+\cos60^\circ}{\cos36^\circ-\cos60^\circ}\dfrac{\cos4^\circ+\cos60^\circ}{\cos4^\circ-\cos60^\circ}\tan2^\circ$

$=\cot24^\circ\dfrac{\cos36^\circ+\dfrac{1}{2}}{\cos36^\circ-\dfrac{1}{2}}\dfrac{\cos4^\circ+\dfrac{1}{2}}{\cos4^\circ-\dfrac{1}{2}}\tan2^\circ$

$=\cot24^\circ\dfrac{2\cos36^\circ+1}{2\cos36^\circ-1}\dfrac{2\cos4^\circ+1}{2\cos4^\circ-1}\tan2^\circ$

Answer 3

$$\cot12\cot24\cot48\cot84$$

$$=\tan 6\tan42\tan66\tan78$$

Observe que $\tan5x=\tan30$ para $x=6,42,-66,78$

Utilice https://trans4mind.com/personal_development/mathematics/trigonometry/multipleAnglesTangent.htm#List_of_Formulae ,

para encontrar $\tan6,\tan42,\tan78,\tan114=-\tan66=\tan(-66),\tan150=\tan(-30)$ son las raíces de $$t^5-5\tan30\cdot t^5+\cdots+\tan30=0$$

$$\implies\prod_{r=0}^4\tan(36r+6)=\dfrac{\tan30}1$$

$r=4,\tan(36r+6)=?$