{"id":56603,"date":"2016-02-08T19:18:51","date_gmt":"2016-02-08T21:18:51","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=56603"},"modified":"2023-03-31T20:22:17","modified_gmt":"2023-03-31T23:22:17","slug":"problema-de-gincana-letras-em-fila","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-de-gincana-letras-em-fila\/","title":{"rendered":".Problema de Gincana:   Um seno de [tex]3x[\/tex]"},"content":{"rendered":"<p style=\"margin:10px\">\n<div class=\"g1n\">\n<p class=\"tit\" style=\"line-height:30px;margin:10px\">Problema<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\n<p style=\"line-height: 50px;\">Se [tex]x[\/tex] \u00e9 um arco do segundo quadrante tal que  [tex]sen(x) = \\dfrac{1}{5}[\/tex], ent\u00e3o [tex]sen(3x)[\/tex] \u00e9<br \/>\n[tex](a)\\,\\dfrac{6}{25}.\\,\\,\\,\\,\\,\\,[\/tex]<br \/>\n[tex](b)\\, \\dfrac{13}{25}.\\,\\,\\,\\,\\,\\,[\/tex]<br \/>\n[tex](c)\\, \\dfrac{71}{125}.\\,\\,\\,\\,\\,\\,[\/tex]<br \/>\n[tex](d)\\,\\dfrac{119}{125}.\\,\\,\\,\\,\\,\\,[\/tex]<br \/>\n[tex](e)\\, N.R.A.[\/tex]<\/p>\n<p style=\"margin:10px\">\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g1n\">\n<p class=\"tit\" style=\"line-height:30px;margin:10px\">Solu\u00e7\u00e3o<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nSendo [tex]sen(x)=\\dfrac{1}{5}[\/tex], pela rela\u00e7\u00e3o fundamental da trigonometria, [tex]sen^2(x)+cos^2(x)=1[\/tex], temos [tex]cos^2(x) = \\dfrac{24}{25}[\/tex].<br \/>\nAl\u00e9m disso,<br \/>\n[tex]\\,\\,sen(3x) = sen(2x + x) =[\/tex]<br \/>\n[tex]\\qquad\\qquad =  sen(2x) \\cdot cos(x) + sen(x) \\cdot cos(2x) \\\\<br \/>\n\\qquad\\qquad = 2 \\cdot sen(x) \\cdot cos(x) \\cdot cos(x) + sen(x) \\cdot (2 cos^2(x) &#8211; 1)\\\\<br \/>\n\\qquad\\qquad = 2 \\cdot \\dfrac{1}{5} \\cdot \\dfrac{24}{25} + \\dfrac{1}{5} \\cdot \\left(2 \\cdot \\dfrac{24}{25} &#8211; 1\\right)\\\\<br \/>\n\\qquad\\qquad = \\dfrac{48}{125} + \\dfrac{1}{5} \\cdot \\left(\\dfrac{48}{25} &#8211; 1\\right)\\\\<br \/>\n\\qquad\\qquad = \\dfrac{48}{125} + \\dfrac{23}{125}\\\\<br \/>\n\\qquad\\qquad = \\boxed{\\dfrac{71}{125}}\\,.[\/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\" style=\"line-height:20px;\">\nSolu\u00e7\u00e3o elaborada pelos <strong>Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"gg\">\nSegunda Gincana de 2015 &#8211; Clubes de Matem\u00e1tica da OBMEP<br \/>\nN\u00edvel C &#8211; Quest\u00e3o F\u00e1cil<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Problema Se [tex]x[\/tex] \u00e9 um arco do segundo quadrante tal que [tex]sen(x) = \\dfrac{1}{5}[\/tex], ent\u00e3o [tex]sen(3x)[\/tex] \u00e9 [tex](a)\\,\\dfrac{6}{25}.\\,\\,\\,\\,\\,\\,[\/tex] [tex](b)\\, \\dfrac{13}{25}.\\,\\,\\,\\,\\,\\,[\/tex] [tex](c)\\, \\dfrac{71}{125}.\\,\\,\\,\\,\\,\\,[\/tex] [tex](d)\\,\\dfrac{119}{125}.\\,\\,\\,\\,\\,\\,[\/tex] [tex](e)\\, N.R.A.[\/tex] Solu\u00e7\u00e3o Sendo [tex]sen(x)=\\dfrac{1}{5}[\/tex], pela rela\u00e7\u00e3o fundamental da trigonometria, [tex]sen^2(x)+cos^2(x)=1[\/tex], temos [tex]cos^2(x) = \\dfrac{24}{25}[\/tex]. Al\u00e9m disso, [tex]\\,\\,sen(3x) = sen(2x + x) =[\/tex] [tex]\\qquad\\qquad = sen(2x) \\cdot cos(x) + sen(x) \\cdot cos(2x) &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problema-de-gincana-letras-em-fila\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-56603","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/56603","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\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=56603"}],"version-history":[{"count":22,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/56603\/revisions"}],"predecessor-version":[{"id":185943,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/56603\/revisions\/185943"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=56603"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}