{"id":100454,"date":"2018-03-21T17:00:37","date_gmt":"2018-03-21T20:00:37","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=100454"},"modified":"2023-04-15T20:43:05","modified_gmt":"2023-04-15T23:43:05","slug":"problemao-desigualdade-em-trigonometria","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problemao-desigualdade-em-trigonometria\/","title":{"rendered":".Problem\u00e3o: Desigualdade em trigonometria"},"content":{"rendered":"

\n

\n

Problema
\n (Indicado a partir do 2\u00ba ano do E. M.)<\/span><\/p>\n

\n

Num tri\u00e2ngulo [tex]ABC[\/tex], indicamos por [tex]a[\/tex], [tex]b[\/tex] e [tex]c[\/tex] as medidas dos lados opostos aos \u00e2ngulos [tex]\\hat{A}[\/tex], [tex]\\hat{B}[\/tex] e [tex]\\hat{C}[\/tex], respectivamente.
\n(a)<\/strong> Prove que [tex]\\dfrac{a}{b+c} = \\dfrac{sen \\hat{A}}{sen \\hat{B} + sen \\hat{C}} \\, [\/tex].
\n(b)<\/strong> Escreva [tex]sen \\hat{A}[\/tex] em fun\u00e7\u00e3o de [tex]sen \\dfrac{\\hat{A}}{2}\\,\\,[\/tex] e [tex]\\,\\,cos \\dfrac{\\hat{A}}{2} \\, [\/tex].
\n(c)<\/strong> Transforme [tex]sen \\hat{B} + sen \\hat{C}[\/tex] num produto.
\n(d)<\/strong> Utilize os itens anteriores para provar que [tex]sen \\dfrac{\\hat{A}}{2} \\leq \\dfrac{a}{b+c} \\, [\/tex].<\/p>\n

\n<\/div>\n

\n

\n

Solu\u00e7\u00e3o <\/p>\n

\n

(a)<\/strong> Pela lei dos senos, temos [tex]\\dfrac{a}{sen \\hat{A}} = \\dfrac{b}{sen \\hat{B}} = \\dfrac{c}{sen \\hat{C}}[\/tex].
\nPortanto,
\n[tex]\\qquad \\begin{cases}a \\, sen \\hat{B}=b \\, sen \\hat{A}\\\\
\na \\, sen \\hat{C}=c \\, sen \\hat{A}
\n\\end{cases}.[\/tex]
\nSomando termo a termo essas duas equa\u00e7\u00f5es, segue que:
\n[tex]\\qquad a \\, sen \\hat{B}+a \\, sen \\hat{C}=b \\, sen \\hat{A}+c \\, sen \\hat{A}[\/tex]
\n[tex]\\qquad a \\left(sen \\hat{B}+sen \\hat{C}\\right)=\\left(b+c\\right) \\, sen \\hat{A}.[\/tex]
\nComo [tex]sen \\hat{B}+sen \\hat{C}\\ne 0 \\, [\/tex] e [tex] \\, b+c \\ne 0[\/tex] (Por qu\u00ea?)<\/span>, segue que:
\n[tex]\\qquad \\boxed{ \\dfrac{a}{b+c} = \\dfrac{sen \\hat{A}}{sen\\hat{B}\\, +\\, sen\\hat{C}}} \\, . \\qquad \\textcolor{#800000}{(i)}[\/tex]<\/p>\n

\n

(b)<\/strong> Pela f\u00f3rmula do seno do arco duplo, temos
\n[tex]\\qquad \\boxed{sen \\hat{A} = 2 \\cdot sen \\dfrac{\\hat{A}}{2} \\cdot cos \\dfrac{\\hat{A}}{2}} \\, . \\qquad \\textcolor{#800000}{(ii)}[\/tex]<\/p>\n

\n

(c)<\/strong> Pelas transforma\u00e7\u00f5es em produto, temos
\n[tex]\\qquad \\boxed{sen \\hat{B} + sen \\hat{C} = 2 \\cdot sen \\dfrac{\\hat{B} + \\hat{C}}{2} \\cdot cos \\dfrac{\\hat{B} – \\hat{C}}{2}} \\, . \\qquad \\textcolor{#800000}{(iii)}[\/tex]<\/p>\n

\n

(d) <\/strong>Utilizando as igualdades [tex]\\textcolor{#800000}{(i)}[\/tex], [tex]\\textcolor{#800000}{(ii)}[\/tex] e [tex]\\textcolor{#800000}{(iii)}[\/tex], segue que:
\n[tex]\\qquad \\begin{align*} \\dfrac{a}{b+c} & \\stackrel{\\textcolor{#800000}{(i)}}{=}\\dfrac{sen \\hat{A}}{sen\\hat{B}\\, +\\, sen\\hat{C}}\\\\
\n&\\stackrel{\\textcolor{#800000}{(ii)}}{=}\\dfrac{2 \\cdot sen \\dfrac{\\hat{A}}{2} \\cdot cos \\dfrac{\\hat{A}}{2}}{sen\\hat{B}\\, +\\, sen\\hat{C}}\\\\
\n&\\stackrel{\\textcolor{#800000}{(iii)}}{=}\\dfrac{2 \\cdot sen \\dfrac{\\hat{A}}{2} \\cdot cos \\dfrac{\\hat{A}}{2}}{2 \\cdot sen \\dfrac{\\hat{B} + \\hat{C}}{2} \\cdot cos \\dfrac{\\hat{B} – \\hat{C}}{2}}\\\\
\n&=\\dfrac{\\cancel{2} \\cdot sen \\dfrac{\\hat{A}}{2} \\cdot cos \\dfrac{\\hat{A}}{2}}{\\cancel{2} \\cdot sen \\left(90^{\\circ} – \\dfrac{\\hat{A}}{2}\\right) \\cdot cos \\dfrac{\\hat{B} – \\hat{C}}{2}}\\\\
\n&=\\dfrac{ sen \\dfrac{\\hat{A}}{2} \\cdot \\cancel{cos \\dfrac{\\hat{A}}{2}}}{\\cancel{sen \\left(90^{\\circ} – \\dfrac{\\hat{A}}{2}\\right)} \\cdot cos \\dfrac{\\hat{B} – \\hat{C}}{2}}\\\\
\n\\end{align*}[\/tex]
\nAssim, [tex]\\dfrac{a}{b+c} = \\dfrac{sen \\dfrac{\\hat{A}}{2}}{ cos \\dfrac{\\hat{B} – \\hat{C}}{2}} \\, . \\quad \\textcolor{#800000}{(iv)}[\/tex]
\nSendo [tex]0\\leq cos \\dfrac{\\hat{B} – \\hat{C}}{2} \\leq 1[\/tex], temos [tex]\\boxed{\\dfrac{1}{cos \\dfrac{\\hat{B} – \\hat{C}}{2}} \\geq 1} \\, . \\quad \\textcolor{#800000}{(v)}[\/tex]
\nPor outro lado, como [tex]\\dfrac{\\hat{A}}{2}[\/tex] \u00e9 um \u00e2ngulo agudo, [tex]\\boxed{sen \\dfrac{\\hat{A}}{2} \\gt 0} \\, . \\quad \\textcolor{#800000}{(vi)}[\/tex]
\nDessa forma, por [tex]\\textcolor{#800000}{(v)}[\/tex] e [tex]\\textcolor{#800000}{(vi)}[\/tex], temos que
\n[tex]\\qquad sen \\dfrac{\\hat{A}}{2} \\leq \\dfrac{sen \\dfrac{\\hat{A}}{2}}{ cos \\dfrac{\\hat{B} – \\hat{C}}{2}}[\/tex],
\ndonde, por [tex]\\textcolor{#800000}{(iv)}[\/tex], segue finalmente que [tex]\\fcolorbox{black}{#eee0e5}{$sen \\dfrac{\\hat{A}}{2} \\leq \\dfrac{a}{b+c}$} \\, .[\/tex]<\/p>\n

\n

Solu\u00e7\u00e3o elaborada pelos Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Problema (Indicado a partir do 2\u00ba ano do E. M.) Num tri\u00e2ngulo [tex]ABC[\/tex], indicamos por [tex]a[\/tex], [tex]b[\/tex] e [tex]c[\/tex] as medidas dos lados opostos aos \u00e2ngulos [tex]\\hat{A}[\/tex], [tex]\\hat{B}[\/tex] e [tex]\\hat{C}[\/tex], respectivamente. (a) Prove que [tex]\\dfrac{a}{b+c} = \\dfrac{sen \\hat{A}}{sen \\hat{B} + sen \\hat{C}} \\, [\/tex]. (b) Escreva [tex]sen \\hat{A}[\/tex] em fun\u00e7\u00e3o de [tex]sen \\dfrac{\\hat{A}}{2}\\,\\,[\/tex] e [tex]\\,\\,cos … <\/p>\n

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-100454","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/100454","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=100454"}],"version-history":[{"count":8,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/100454\/revisions"}],"predecessor-version":[{"id":187257,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/100454\/revisions\/187257"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=100454"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}