{"id":122634,"date":"2024-04-16T17:51:41","date_gmt":"2024-04-16T17:51:41","guid":{"rendered":"https:\/\/www.todamateriabr.com.br\/blog\/?p=122634"},"modified":"2025-07-13T15:28:44","modified_gmt":"2025-07-13T15:28:44","slug":"qual-e-o-valor-minimo-de-senx-cosx-se-nao-souber-favor-nao-r","status":"publish","type":"post","link":"https:\/\/www.todamateriabr.com.br\/blog\/qual-e-o-valor-minimo-de-senx-cosx-se-nao-souber-favor-nao-r\/","title":{"rendered":"Qual \u00e9 o valor m\u00ednimo de sen\u2074x + cos\u2074x? Se n\u00e3o souber, favo"},"content":{"rendered":"

Pergunta:<\/h2>\n

Qual \u00e9 o valor m\u00ednimo de sen\u2074x + cos\u2074x? Se n\u00e3o souber, favor n\u00e3o responder, se n\u00e3o voc\u00ea ir\u00e1 me atrapalhar.Quero saber se fiz da maneira correta, perceba que \u00e9 elevado a 4, n\u00e3o a 2, se n\u00e3o seria a rela\u00e7\u00e3o fundamental.\u200b <\/p>\n

\n
Respostas <\/p>\n\n

<\/p>\n

Consigo pensar em 3 formas para resolver. <\/p>\n

1\u00aa forma : M\u00ednimo de uma par\u00e1bola <\/strong><\/p>\n

[tex]\\displaystyle \\sf \\text{sabemos que} : \\\\\\\\ sen^2(x)+cos^2(x) = 1 \\\\\\\\\\ cos^2(x) = 1-sen^2(x) \\\\\\\\\\\\\\ \\text{queremos o m\\’inimo de } :\\\\\\\\ M = sen^4(x)+cos^4(x) \\\\\\\\ \\text{Fa\\c ca} :\\\\\\\\ M = sen^4(x)+\\left[cos^2(x)\\right]^2 \\\\\\\\\\ M = sen^4(x)+\\left[1-sen^2(x)\\right]^2 \\\\\\\\\\ M = sen^4(x) +1-2sen^2(x)+sen^4(x) \\\\\\\\ M = 2sen^4(x)-2sen^2(x)+1\\\\\\\\ \\text{note q M \\’e uma par\\’abola na vari\\’avel }sen^2(x).\\\\\\\\ Assim,[\/tex]<\/p>\n

[tex]\\displaystyle \\sf \\text{m\\’inimo } =\\frac{-\\Delta}{4a} \\\\\\\\ \\text{m\\’inimo }=\\frac{-((-2)^2-4\\cdot 2\\cdot 1)}{4\\cdot 2}\\\\\\\\\\ \\text{m\\’inimo }=\\frac{-(4-8)}{4\\cdot 2} \\\\\\\\\\ \\text{m\\’inimo }=\\frac{\\not 4}{\\not 4\\cdot 2}=\\frac{1}{2}\\\\\\\\\\ \\large\\boxed{\\begin{matrix}\\text{Portanto o valor m\\’inimo : } \\\\\\\\ \\displaystyle \\sf sen^4(x)+cos^4(x)=\\frac{1}{2}\\end{matrix}\\ }\\checkmark[\/tex]<\/p>\n<\/p>\n

2\u00aa forma : Desigualdade das m\u00e9dias <\/strong><\/p>\n

[tex]\\displaystyle \\sf \\text{Pela desigualdade das m\\’edias sabemos que } : \\\\\\\\ \\sqrt{\\frac{(a_1)^2+(a_2)^2+..+(a_n)^2}{n}}\\geq \\frac{a_1+a_2+..+a_n}{n}\\\\\\\\\\ \\text{Fa\\c ca} : \\\\\\\\ a_1 = sen^2(x) \\ ; \\ a_2=cos^2(x)\\\\\\\\ Assim, \\\\\\\\\\ \\sqrt{\\frac{(sen^2(x))^2+(cos^2(x))^2}{2}}\\geq \\frac{\\overbrace{\\sf sen^2(x)+cos^2(x)}^{\\displaystyle 1}}{2} \\\\\\\\\\\\ \\sqrt{\\frac{sen^4(x)+cos^4(x)}{2}} \\geq \\frac{1}{2} \\\\\\\\\\ \\left(\\sqrt{\\frac{sen^4(x)+cos^4(x)}{2}} \\right)^2\\geq \\left(\\frac{1}{2} \\right)^2[\/tex]<\/p>\n

[tex]\\displaystyle \\sf \\frac{sen^4(x)+cos^4(x)}{2} \\geq \\frac{1}{4} \\\\\\\\\\\\ sen^4(x)+cos^4(x) \\geq \\frac{2}{4} \\\\\\\\\\\\\\ sen^4(x)+cos^4(x)\\geq \\frac{1}{2}\\\\\\\\\\\\ \\large\\boxed{\\begin{matrix}\\text{valor m\\’inimo} : \\\\\\\\ \\displaystyle \\sf \\ sen^4(x)+cos^4(x) = \\frac{1}{2} \\ \\end{matrix}}\\checkmark[\/tex]<\/p>\n<\/p>\n<\/p>\n

3\u00aa forma : Desigualdade de cauchy <\/strong><\/p>\n

[tex]\\displaystyle \\sf \\text{usando a desigualdade de cauchy schwarz } : \\\\\\\\ (a_1\\cdot b_1+a_2\\cdot b_2) ^2\\leq \\left[(a_1)^2+(a_2)^2\\right]\\cdot \\left[(b_1)^2+(b_2)^2\\right] \\\\\\\\\\\\ \\text{queremos} : \\\\\\\\ \\left[sen^4(x)+cos^4(x)\\right]_{\\displaystyle \\text{m\\’inimo}}=\\ ?\\\\\\\\ \\text{Fa\\c ca} : \\\\\\\\ \\underbrace{\\sf \\left(1\\cdot sen^2(x)+1\\cdot cos^2(x)\\right}_{\\displaystyle 1})^2\\leq (1^2+1^2) \\cdot \\left[(sen^2(x))^2+(cos^2(x))^2\\right][\/tex]<\/p>\n

[tex]\\displaystyle \\sf 1\\leq (2)\\cdot \\left[sen^4(x)+cos^4(x)\\right] \\\\\\\\\\\\ sen^4(x)+cos^4(x) \\geq \\frac{1}{2} \\\\\\\\\\\\\\ \\large\\boxed{\\begin{matrix}\\text{valor m\\’inimo} : \\\\\\\\ \\displaystyle \\sf \\ sen^4(x)+cos^4(x) = \\frac{1}{2} \\ \\end{matrix}}\\checkmark[\/tex]<\/p>\n<\/p>\n

<\/span> <\/p>\n

Muito obrigada ^^ <\/span> <\/p>\n","protected":false},"excerpt":{"rendered":"

Resposta para a quest\u00e3o:
\nQual \u00e9 o valor m\u00ednimo de sen\u2074x + cos\u2074x? Se n\u00e3o souber, favor n\u00e3o responder, se n\u00e3o voc\u00ea ir\u00e1 me atrapalhar.Quero saber se fiz da maneira correta, perceba que \u00e9 elevado a 4, n\u00e3o a 2, se n\u00e3o seria a rela\u00e7\u00e3o fundamental.\u200b <\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-122634","post","type-post","status-publish","format-standard","hentry","category-perguntas-e-respostas"],"_links":{"self":[{"href":"https:\/\/www.todamateriabr.com.br\/blog\/wp-json\/wp\/v2\/posts\/122634","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.todamateriabr.com.br\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.todamateriabr.com.br\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.todamateriabr.com.br\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.todamateriabr.com.br\/blog\/wp-json\/wp\/v2\/comments?post=122634"}],"version-history":[{"count":0,"href":"https:\/\/www.todamateriabr.com.br\/blog\/wp-json\/wp\/v2\/posts\/122634\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.todamateriabr.com.br\/blog\/wp-json\/wp\/v2\/media?parent=122634"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.todamateriabr.com.br\/blog\/wp-json\/wp\/v2\/categories?post=122634"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.todamateriabr.com.br\/blog\/wp-json\/wp\/v2\/tags?post=122634"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}