{"id":216686,"date":"2024-05-05T04:28:51","date_gmt":"2024-05-05T07:28:51","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=216686"},"modified":"2024-05-07T02:32:39","modified_gmt":"2024-05-07T05:32:39","slug":"problemao-uma-soma-fascinante","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problemao-uma-soma-fascinante\/","title":{"rendered":".Desafio: Uma soma fascinante!"},"content":{"rendered":"<p class=\"g2\" ><strong><a href=\"http:\/\/clubes.obmep.org.br\/blog\/?p=217729\" rel=\"noopener\" target=\"_blank\">&#9999; <span style=\"color:#A0522D;font-size:14px;\">Link do problema para dispositivos da <em>Apple<\/em>.<\/span><\/a><\/strong><\/p>\n<div class=\"g1n\" style=\"line-height: 29px; padding-bottom: 15px;\">\n<p class=\"tit\" style=\"line-height: 25px; margin: 10px;\">Problema<br \/>\n<span style=\"font-size: 12px; color: #800000;\"> (Indicado a partir do 2\u00ba ano do E. M.)<\/span><\/p>\n<hr style=\"height: 1px; width: 100%; background-color: #800000; margin-top: 10px; margin-bottom: 20px; border: 0 none;\" \/>\nCalcule o valor da soma<br \/>\n[tex]\\qquad 2 \\  \\mathrm{sen}\\, 2^\\circ+ 4 \\  \\mathrm{sen} \\,4^\\circ+ 6 \\  \\mathrm{sen}\\, 6^\\circ+\\cdots +178 \\  \\mathrm{sen} \\,178^\\circ[\/tex].<br \/>\nSe necess\u00e1rio, use [tex] \\mathrm{cotg} \\  1^\\circ=57,29[\/tex].<\/p>\n<p style=\"font-size: 12px; text-align: right; line-height: 1px;\">Extra\u00eddo de <strong> XXVI OPM &#8211; N\u00edvel Gama<\/strong>.<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"cb36 g1\" style=\"margin: 30px; line-height: 25px; width: 80%; margin-left: auto; margin-right: auto;\">\n<img loading=\"lazy\" decoding=\"async\" class=\"alignleft size-full wp-image-1001\" style=\"border: none;\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2013\/01\/menino.jpg\" alt=\"explicador_p\" width=\"80\" height=\"96\" \/><\/p>\n<p class=\"c3 ti3\" style=\"line-height: 20px; margin: 10px;\">Lembretes<\/p>\n<p><span class=\"c3\"> <strong>(I)<\/strong> <\/span> Dados dois \u00e2ngulos [tex]a[\/tex] e [tex]b[\/tex], com [tex]0^{\\circ}\\lt a\\lt 180^{\\circ}[\/tex] e [tex]0^{\\circ}\\lt b\\lt 180^{\\circ}[\/tex], vale a rela\u00e7\u00e3o [tex]2 \\ \\mathrm{sen} \\, a \\ \\mathrm{sen} \\, b=\\cos (a-b)-\\cos(a+b)[\/tex].<br \/>\n<span class=\"c3\"> <strong>(II)<\/strong> <\/span> Dado um \u00e2ngulo [tex]a[\/tex], com [tex]0^{\\circ}\\lt a\\lt 180^{\\circ}[\/tex], ent\u00e3o [tex]\\cos(180^\\circ-a)=-\\cos a[\/tex].\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"g1n\" style=\"line-height: 28px;\">\n<p class=\"tit\" style=\"line-height: 28px; margin: 10px;\">Solu\u00e7\u00e3o<\/p>\n<hr style=\"height: 1px; width: 100%; background-color: #800000; margin-top: 1px; margin-bottom: 20px; border: 0 none;\" \/>\nVamos denotar a soma a ser calculada por [tex]S[\/tex].<br \/>\nAssim, multiplicando [tex]S[\/tex] por [tex]\\mathrm{sen} \\, 1^\\circ[\/tex], temos<br \/>\n[tex]\\quad S \\cdot \\mathrm{sen}  \\,1^\\circ=2 \\  \\mathrm{sen} \\, 2^\\circ \\cdot \\mathrm{sen} \\, 1^\\circ+ 4 \\  \\mathrm{sen} \\, 4^\\circ \\cdot \\mathrm{sen} \\, 1^\\circ+ 6 \\  \\mathrm{sen} \\, 6^\\circ \\cdot \\mathrm{sen} \\, 1^\\circ+\\cdots \\\\<br \/>\n\\qquad \\qquad \\cdots+178 \\  \\mathrm{sen} \\, 178^\\circ \\cdot \\mathrm{sen} \\, 1^\\circ.[\/tex]<\/p>\n<p style=\"margin:10px\">\n<p>Utilizando o fato de que [tex]2 \\ \\mathrm{sen} \\, a \\ \\cdot \\mathrm{sen} \\, b=\\cos (a-b)-\\cos(a+b)[\/tex], obtemos<br \/>\n[tex]\\quad  S \\cdot \\mathrm{sen}  \\,1^\\circ=\\cos1^\\circ-\\cos3^\\circ+2(\\cos3^\\circ-\\cos5^\\circ)+3(\\cos5^\\circ-\\cos7^\\circ)+\\dots\\\\<br \/>\n\\qquad \\qquad \\cdots+89(\\cos177^\\circ-\\cos179^\\circ)[\/tex]<br \/>\n[tex]\\quad  S \\cdot \\mathrm{sen}  \\,1^\\circ=\\cos1^\\circ-\\cos3^\\circ+2 \\cos3^\\circ-2 \\cos5^\\circ+3 \\cos5^\\circ-3 \\cos7^\\circ+\\dots\\\\<br \/>\n\\qquad \\qquad \\cdots+89 \\cos177^\\circ-89 \\cos179^\\circ[\/tex]<br \/>\n[tex]\\quad S \\cdot \\mathrm{sen}  \\,1^\\circ=\\cos1^\\circ+\\cos3^\\circ+\\cos5^\\circ+\\cos7^\\circ+\\dots\\\\<br \/>\n\\qquad \\qquad \\cdots+\\cos177^\\circ-89\\cos179^\\circ.[\/tex]<\/p>\n<p>Com a identidade trigonom\u00e9trica [tex]~\\cos(180^\\circ-a)=-\\cos a~[\/tex] podemos obter<\/p>\n<ul>\n<li>[tex]\\cos 91^\\circ=-\\cos 89^\\circ[\/tex]<\/li>\n<li>[tex]\\cos 93^\\circ=-\\cos 87^\\circ[\/tex]<\/li>\n<p>[tex]\\qquad ~~~~~~\\vdots[\/tex]<\/p>\n<li>[tex]\\cos 177^\\circ=-\\cos 3^\\circ[\/tex]<\/li>\n<li>[tex]\\cos 179^\\circ=-\\cos 1^\\circ[\/tex]<\/li>\n<\/ul>\n<p>e, ent\u00e3o<br \/>\n[tex]\\quad S \\cdot \\mathrm{sen}  \\,1^\\circ=\\cos1^\\circ+\\cos3^\\circ+\\cos5^\\circ+\\dots+\\cos89^\\circ-\\cos89^\\circ-\\dots\\\\<br \/>\n\\qquad \\qquad \\cdots-\\cos5^\\circ-\\cos 3^\\circ-89\\cos179^\\circ.[\/tex]<\/p>\n<p style=\"margin:10px\">\n<p>Logo, cancelando os opostos, segue que:<br \/>\n[tex]\\quad S \\cdot \\mathrm{sen}  \\,1^\\circ=\\cos1^\\circ\\cancel{+\\cos3^\\circ}\\cancel{+\\cos5^\\circ}+\\dots\\cancel{+\\cos89^\\circ}\\cancel{-\\cos89^\\circ}-\\dots\\\\<br \/>\n\\qquad \\qquad \\cdots\\cancel{-\\cos5^\\circ}\\cancel{-\\cos 3^\\circ}-89\\cos179^\\circ[\/tex]<br \/>\n[tex]\\quad S \\cdot \\mathrm{sen}  \\,1^\\circ=\\cos1^\\circ-89\\cos179^\\circ=\\cos1^\\circ+89\\cos1^\\circ\\\\<br \/>\n\\quad S \\cdot \\mathrm{sen}  \\,1^\\circ=90\\cos1^\\circ.[\/tex]<\/p>\n<p style=\"margin:10px\">\n<p>Portanto,<br \/>\n[tex]\\quad S=90~\\dfrac{\\cos1^\\circ}{\\mathrm{sen} \\, 1^\\circ}=90  \\ \\mathrm{cotg} \\  1^\\circ.[\/tex]<\/p>\n<p style=\"margin:10px\">\n<p>Usando o valor [tex]\\boxed{ \\mathrm{cotg} \\  1^\\circ=57,29}[\/tex], obtemos<br \/>\n[tex]\\quad S=90\\times 57,29=5\\,156,1[\/tex]<br \/>\ne, assim,<br \/>\n[tex]\\quad 2 \\  \\mathrm{sen}\\, 2^\\circ+ 4 \\  \\mathrm{sen} \\,4^\\circ+ 6 \\  \\mathrm{sen}\\, 6^\\circ+\\cdots +178 \\  \\mathrm{sen} \\,178^\\circ=5\\,156,1\\,.[\/tex]<\/p>\n<hr style=\"height: 1px; width: 100%; background-color: #800000; margin-top: 10px; margin-bottom: 15px; border: 0 none;\" \/>\n<p class=\"g2\">Solu\u00e7\u00e3o elaborada pelos <strong>Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n<p style=\"margin:10px\">\n","protected":false},"excerpt":{"rendered":"<p>&#9999; Link do problema para dispositivos da Apple. Problema (Indicado a partir do 2\u00ba ano do E. M.) Calcule o valor da soma [tex]\\qquad 2 \\ \\mathrm{sen}\\, 2^\\circ+ 4 \\ \\mathrm{sen} \\,4^\\circ+ 6 \\ \\mathrm{sen}\\, 6^\\circ+\\cdots +178 \\ \\mathrm{sen} \\,178^\\circ[\/tex]. Se necess\u00e1rio, use [tex] \\mathrm{cotg} \\ 1^\\circ=57,29[\/tex]. Extra\u00eddo de XXVI OPM &#8211; N\u00edvel Gama. &nbsp; &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problemao-uma-soma-fascinante\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":254,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-216686","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/216686","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/users\/254"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=216686"}],"version-history":[{"count":9,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/216686\/revisions"}],"predecessor-version":[{"id":217790,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/216686\/revisions\/217790"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=216686"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}