Question

Calcule o valor exato de sen(36º)



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Por favor responder de forma detalhada. Respostas com brincadeiras serão eliminadas.

Answer 1

$Para\ essa\ quest\tilde{a}o\ vamos\ recorrer\ a\ um\ decag\acute{o}no\ de\ lado\ L_1_0\\ inscrito\ a\ uma\ circunfer\hat{e}ncia\ de\ raio\ R\\\\ Sabendo\ que\ o\ \hat{a}ngulo\ central\ de\ um\ decag\acute{o}no\ regular\ pode\ ser\\ calculado\ pela\ express\tilde{a}o\ a\ seguir\ :\\\\ a_c\ =\ \frac{360^\circ}{n}\\\\ a_c\ =\ \frac{360^\circ}{10}\\\\ a_c\ =\ 36^\circ$

$Agora\ imaginemos\ o\ tri\hat{a}ngulo\ OAB\ formado\ pelo\ lado\ L_1_0\ e\\ por\ dois\ segmentos\ que\ correspondem\ ao\ raio\ R\ .\\\\ Olhar\ anexo\ (1)\\\\ Pelo\ fato\ de\ \Delta OAB\ ser\ is\acute{o}sceles\ ent\tilde{a}o\ seus\ demais\ \hat{a}ngulos\\ correspondem\ \grave{a}\ 72^\circ\\\\ No\ anexo\ (2)\ irei\ tra\c{c}ar\ a\ bissetriz\ relativa\ ao\ v\acute{e}rtice\ B\\\\ No\ anexo\ (3)\ recorrendo\ aos\ tri\hat{a}ngulos\ is\acute{o}sceles\ irei\\ reescrever\ os\ lados\ desses\ tri\hat{a}ngulos$

$Recorrendo\ ao\ Teorema\ da\ Bissetriz\ Interna\ (TBI)\ , \\\\ \frac{L_1_0}{R-L_1_0}\ =\ \frac{R}{L_1_0}\\\\ (L_1_0)^2\ =\ R^2\ -\ (L_1_0).R\\\\ (L_1_0)^2\ +\ R.(L_1_0)\ -\ R^2\ =\ 0\\\\ Assim\ temos\ uma\ equa\c{c}\tilde{a}o\ do\ segundo\ grau\ ,\\\\ L_1_0\ =\ \frac{-b\ ^+_-\ \sqrt{b^2\ -\ 4ac}}{2a} \\\\ L_1_0\ = \ \frac{-R\ ^+_-\ \sqrt{R^2+4R^2}}{2}\\\\ Desprezando\ a\ solu\c{c}\tilde{a}o\ negativa\ que\ n\tilde{a}o\ conv\acute{e}m\ , \ temos\ :$

$\boxed{\boxed{L_1_0\ =\ \frac{ \sqrt{5} -1}{2}\ .\ R}}\\\\\\ Retorne\ ao\ anexo\ (1)\\\\ Agora\ irei\ aplicar\ o\ Teorema\ dos\ Cossenos\ no\ \Delta OAB\\\\ (L_1_0)^2\ =\ R^2\ +\ R^2\ -\ 2.(R).(R).cos\ 36^\circ\\\\ Como\ calculamos\ anteriormente\ L_1_0\ em\ fun\c{c}\tilde{a}o\ de\ R\ agora\\ iremos\ substitu\acute{i}-lo\ ,\\\\ \Big[\Big( \frac{ \sqrt{5} -1}{2}\Big)\ .\ R \Big]^2\ =\ 2R^2\ -\ 2.R^2.cos\ 36^\circ\\\\ \Big( \frac{5\ -\ 2\sqrt{5}+1}{4}\Big)\ .\ R^2\ =\ R^2\Big(2\ -\ 2.cos\ 36^\circ\Big)$

$Simplificando\ os\ dois\ membros\ da\ equa\c{c}\tilde{a}o\ :\\\\ \frac{6\ -\ 2\sqrt{5}}{4}\ = \ 2\ -\ 2.cos\ 36^\circ\\\\ 6\ -\ 2\sqrt{5}\ =\ 8\ -\ 8.cos\ 36^\circ\\\\ 8.cos\ 36^\circ\ =\ 2\ +\ 2\sqrt{5}\\\\ cos\ 36^\circ\ = \ \frac{1\ +\ \sqrt{5}}{4}$

$Seja\ a\ Rela\c{c}\tilde{a}o\ Fundamental\ da\ Trigonometria\ dada\ por\ :\\\\ \boxed{\boxed{ sen^2\ a\ +\ cos^2\ a\ =\ 1}}\\\\\\ Atrav\acute{e}s\ dela\ vamos\ calcular\ o\ sen \ 36^\circ\\\\ sen^2\ 36^\circ\ +\ cos^2\ 36^\circ\ =\ 1\\\\ Substituindo\ o\ valor\ do\ cos\ 36^\circ\ achado\ anteriormente\ ,\\\\\ sen^2\ 36^\circ\ +\ \Big(\frac{1\ +\ \sqrt{5}}{4}\Big)^2\ =\ 1\\\\ sen^2\ 36^\circ\ +\ \frac{1\ +\ 2\sqrt{5}\ +\ 5}{16}\ =\ 1\\\\ 16\ .\ sen^2\ 36^\circ\ +\ 6\ +\ 2\sqrt{5}\ = 16$

$16\ .\ sen^2\ 36^\circ\ +\ 6\ +\ 2\sqrt{5}\ = 16\\\\ 16\ .\ sen^2\ 36^\circ\ = 10\ -\ 2\sqrt{5}\\\\ sen^2\ 36^\circ\ =\ \frac{5\ -\ \sqrt{5}}{8}\\\\ sen\ 36^\circ\ =\ ^+_-\ \frac{\sqrt{10\ -\ 2 \sqrt{5}}}{4}\\\\ Desprezando\ a\ solu\c{c}\tilde{a}o\ negativa\ que\ n\tilde{a}o\ conv\acute{e}m\ (\ 36^\circ\ \acute{e}\\ um\ arco\ do\ primeiro\ quadrante\ )\ ,\ temos\ que\ :\\\\ \boxed{ \boxed{sen\ 36^\circ\ = \ \frac{ \sqrt{10-2 \sqrt{5}}}{4}}}$