{"id":3300,"date":"2025-02-27T14:37:19","date_gmt":"2025-02-27T13:37:19","guid":{"rendered":"https:\/\/matematicastro.es\/?p=3300"},"modified":"2025-02-27T21:55:55","modified_gmt":"2025-02-27T20:55:55","slug":"expresiones-identidades-y-ecuaciones-trigonometricas","status":"publish","type":"post","link":"https:\/\/matematicastro.es\/?p=3300","title":{"rendered":"Expresiones, identidades y ecuaciones trigonom\u00e9tricas"},"content":{"rendered":"\n

El genio se descubre en la fortuna adversa; en la prosperidad se oculta<\/em><\/p>\n\n\n\n

Homero<\/a><\/em><\/p>\n\n\n\n

\n\n\n\n

En Matem\u00e1ticas I (1\u00ba de Bachillerato) se trabaja mucho la demostraci\u00f3n de identidades trigonom\u00e9tricas, la simplificaci\u00f3n de expresiones en las que aparecen razones trigonom\u00e9tricas, la resoluci\u00f3n de ecuaciones trigonom\u00e9tricas y de sistemas de ecuaciones trigonom\u00e9tricas. Veamos unos ejemplos.<\/p>\n\n\n\n

Identidades trigonom\u00e9tricas<\/h3>\n\n\n\n

Para demostrar identidades trigonom\u00e9tricas, una t\u00e9cnica es comenzar por el lado de la igualdad que aparentemente sea m\u00e1s \u00ablargo\u00bb o \u00abcomplicado\u00bb y, a trav\u00e9s de sucesivos pasos usando las f\u00f3rmulas trigonom\u00e9tricas<\/a> que se saben (se pueden llevar al examen), intentar llegar a la otra parte de la igualdad. Veamos un par de ejemplos.<\/p>\n\n\n\n

Demostrar las siguientes identidades trigonom\u00e9tricas:<\/p>\n\n\n\n

a)<\/strong> $\\displaystyle\\frac{\\cos x+\\text{sen}\\,x}{\\cos x-\\text{sen}\\,x}-\\frac{\\cos x-\\text{sen}\\,x}{\\cos x+\\text{sen}\\,x}=2\\text{tg}\\,x$<\/p>\n\n\n\n

Soluci\u00f3n<\/summary>\n

$\\displaystyle\\frac{\\cos x+\\text{sen}\\,x}{\\cos x-\\text{sen}\\,x}-\\frac{\\cos x-\\text{sen}\\,x}{\\cos x+\\text{sen}\\,x}=$<\/p>\n\n\n\n

$=\\displaystyle\\frac{\\left( \\cos x+\\text{sen}\\,x \\right)\\left( \\cos x+\\text{sen}\\,x \\right)}{\\left( \\cos x-\\text{sen}\\,x \\right)\\left( \\cos x+\\text{sen}\\,x \\right)}-\\frac{\\left( \\cos x-\\text{sen}\\,x \\right)\\left( \\cos x-\\text{sen}\\,x \\right)}{\\left( \\cos x-\\text{sen}\\,x \\right)\\left( \\cos x+\\text{sen}\\,x \\right)}=$<\/p>\n\n\n\n

$=\\displaystyle\\frac{{{\\cos }^{2}}x+2\\cos x\\,\\text{sen}\\,x+\\text{se}{{\\text{n}}^{2}}x}{\\left( \\cos x-\\text{sen}\\,x \\right)\\left( \\cos x+\\text{sen}\\,x \\right)}-\\frac{{{\\cos }^{2}}x-2\\cos x\\,\\text{sen}\\,x+\\text{se}{{\\text{n}}^{2}}x}{\\left( \\cos x-\\text{sen}\\,x \\right)\\left( \\cos x+\\text{sen}\\,x \\right)}=$<\/p>\n\n\n\n

$=\\displaystyle\\frac{2\\cos x\\,\\text{sen}\\,x+2\\cos x\\,\\text{sen}\\,x}{\\left( \\cos x-\\text{sen}\\,x \\right)\\left( \\cos x+\\text{sen}\\,x \\right)}=\\frac{\\text{sen}\\,2x+\\text{sen}\\,2x}{{{\\cos }^{2}}x-\\text{se}{{\\text{n}}^{2}}x}=\\frac{2\\,\\text{sen}\\,2x}{\\cos 2x}=2\\,\\text{tg}\\,2x$<\/p>\n\n\n\n

\n<\/details>\n\n\n\n

b)<\/strong> $\\displaystyle\\frac{\\text{tg}\\,x}{\\cos^2x}=\\frac{1+\\text{tg}^2x}{\\text{cotg}^2x}$<\/p>\n\n\n\n

Soluci\u00f3n<\/summary>\n

$\\displaystyle\\frac{1+\\text{t}{{\\text{g}}^{2}}x}{\\text{cotg}\\,x}=\\frac{1+\\displaystyle\\frac{\\text{se}{{\\text{n}}^{2}}\\,x}{{{\\cos }^{2}}x}}{\\displaystyle\\frac{\\cos x}{\\text{sen}\\,x}}=\\frac{\\displaystyle\\frac{{{\\cos }^{2}}x+\\text{se}{{\\text{n}}^{2}}\\,x}{{{\\cos }^{2}}x}}{\\displaystyle\\frac{\\cos x}{\\text{sen}\\,x}}=\\frac{\\displaystyle\\frac{1}{{{\\cos }^{2}}x}}{\\displaystyle\\frac{\\cos x}{\\text{sen}\\,x}}=$<\/p>\n\n\n\n

$=\\displaystyle\\frac{\\text{sen}\\,x}{\\cos x\\cdot {{\\cos }^{2}}x}=\\frac{\\text{sen}\\,x}{\\cos x}\\cdot \\frac{1}{\\cos {{x}^{2}}}=\\frac{\\text{tg}\\,x}{{{\\cos }^{2}}x}$<\/p>\n\n\n\n

\n<\/details>\n\n\n\n

Expresiones trigonom\u00e9tricas<\/h3>\n\n\n\n

Simplificar las siguientes expresiones trigonom\u00e9tricas:<\/p>\n\n\n\n

a)<\/strong> $\\displaystyle\\frac{\\text{sen}\\,\\alpha+\\text{cotg}\\,\\alpha}{\\text{tg}\\,\\alpha+\\text{cosec}\\,\\alpha}$<\/p>\n\n\n\n

Soluci\u00f3n<\/summary>\n

$\\displaystyle\\frac{\\text{sen}\\,\\alpha +\\text{cotg}\\,\\alpha }{\\text{tg}\\,\\alpha +\\text{cosec}\\,\\alpha }=\\frac{\\text{sen}\\,\\alpha +\\displaystyle\\frac{\\cos \\alpha }{\\text{sen}\\,\\alpha }}{\\displaystyle\\frac{\\text{sen}\\,\\alpha }{\\cos \\alpha }+\\frac{1}{\\text{sen}\\,\\alpha }}=\\frac{\\displaystyle\\frac{\\text{se}{{\\text{n}}^{2}}\\,\\alpha +\\cos \\alpha }{\\text{sen}\\,\\alpha }}{\\displaystyle\\frac{\\text{se}{{\\text{n}}^{2}}\\,\\alpha +\\cos \\alpha }{\\cos \\alpha \\,\\text{sen}\\,\\alpha }}=$<\/p>\n\n\n\n

$=\\displaystyle\\frac{\\left( \\text{se}{{\\text{n}}^{2}}\\,\\alpha +\\cos \\alpha \\right)\\cos \\alpha \\,\\text{sen}\\,\\alpha }{\\left( \\text{se}{{\\text{n}}^{2}}\\,\\alpha +\\cos \\alpha \\right)\\,\\text{sen}\\,\\alpha }=\\cos \\alpha$<\/p>\n\n\n\n

\n<\/details>\n\n\n\n

b)<\/strong> $\\displaystyle2\\text{tg}\\,\\alpha\\cdot\\cos^2\\frac{\\alpha}{2}-\\text{sen}\\,\\alpha$<\/p>\n\n\n\n

Soluci\u00f3n<\/summary>\n

$\\displaystyle2\\,\\text{tg}\\,\\alpha \\cdot {{\\cos }^{2}}\\frac{\\alpha }{2}-\\text{sen}\\,\\alpha =2\\frac{\\text{sen}\\,\\alpha }{\\cos \\alpha }\\cdot \\frac{1+\\cos \\alpha }{2}-\\text{sen}\\,\\alpha =\\frac{\\text{sen}\\,\\alpha \\left( 1+\\cos \\alpha \\right)}{\\cos \\alpha }-\\text{sen}\\,\\alpha =$<\/p>\n\n\n\n

$=\\displaystyle\\frac{\\text{sen}\\,\\alpha +\\text{sen}\\,\\alpha \\cos \\alpha }{\\cos \\alpha }-\\frac{\\text{sen}\\,\\alpha \\cos \\alpha }{\\cos \\alpha }=\\frac{\\text{sen}\\,\\alpha }{\\cos \\alpha }=\\text{tg}\\,\\alpha$<\/p>\n\n\n\n

\n<\/details>\n\n\n\n

Ecuaciones trigonom\u00e9tricas<\/h3>\n\n\n\n

Resolver las siguientes ecuaciones trigonom\u00e9tricas y dar las soluciones dentro del intervalo \\(\\left[ 0{}^\\text{o}\\,,\\,360{}^\\text{o} \\right)\\) (primera vuelta):<\/p>\n\n\n\n

a)<\/strong> $\\displaystyle\\text{tg}\\,x+2\\text{sen}\\,x=0$<\/p>\n\n\n\n

Soluci\u00f3n<\/summary>\n

$\\displaystyle\\text{tg}\\,x+2\\text{sen}\\,x=0\\Rightarrow \\frac{\\text{sen}\\,x}{\\cos x}+2\\text{sen}\\,x=0\\Rightarrow \\text{sen}\\,x+2\\text{sen}\\,x\\cos x=0\\Rightarrow$ <\/p>\n\n\n\n

$\\displaystyle\\Rightarrow \\text{sen}\\,x\\left( 1+2\\cos x \\right)=0\\Rightarrow \\begin{cases} \\text{sen}\\,x=0\\Rightarrow x=0{}^\\text{o}\\,\\,;\\,\\,x=180{}^\\text{o}\\\\ \\cos x=-\\frac{1}{2}\\Rightarrow x=120{}^\\text{o}\\,\\,;\\,\\,x=240{}^\\text{o} \\end{cases}$<\/p>\n\n\n\n

\n<\/details>\n\n\n\n

b)<\/strong> $\\displaystyle\\text{sen}\\,x\\cdot\\text{tg}\\,x=\\frac{\\sqrt{3}}{2}$<\/p>\n\n\n\n

Soluci\u00f3n<\/summary>\n

$\\displaystyle\\text{sen }x\\cdot \\text{tg}\\,x=\\frac{\\sqrt{3}}{6}\\Rightarrow \\text{sen}\\,x\\cdot \\frac{\\text{sen}\\,x}{\\cos x}=\\frac{\\sqrt{3}}{6}\\Rightarrow$<\/p>\n\n\n\n

$\\displaystyle \\Rightarrow 6\\,\\text{se}{{\\text{n}}^{2}}\\,x=\\sqrt{3}\\cos x\\Rightarrow 6\\left( 1-{{\\cos }^{2}}x \\right)=\\sqrt{3}\\cos x\\Rightarrow$<\/p>\n\n\n\n

$\\displaystyle\\Rightarrow 6-6{{\\cos }^{2}}x=\\sqrt{3}\\cos x\\Rightarrow 6{{\\cos }^{2}}x+\\sqrt{3}\\cos x-6=0$<\/p>\n\n\n\n

El discriminante de la ecuaci\u00f3n anterior es $\\displaystyle\\Delta=\\sqrt{3}^2-4\\cdot6\\cdot(-6)=3+144=147$, y $\\sqrt{147}=\\sqrt{7^2\\cdot3}=7\\sqrt{3}$. Por tanto:<\/p>\n\n\n\n

$\\displaystyle\\cos x=\\dfrac{-\\sqrt{3}\\pm7\\sqrt{3}}{2\\cdot6}=\\dfrac{-\\sqrt{3}(-1\\pm7)}{12}=\\begin{cases}\\frac{6\\sqrt{3}}{12}=\\frac{\\sqrt{3}}{2}\\\\ \\frac{-8\\sqrt{3}}{12}=\\frac{-2\\sqrt{3}}{3}\\end{cases}$<\/p>\n\n\n\n

Si $\\cos x=\\dfrac{\\sqrt{3}}{2}$, entonces $x=30{}^\\text{o}\\,\\,;\\,\\,x=330{}^\\text{o}$.<\/p>\n\n\n\n

Si $\\cos x=\\dfrac{-2\\sqrt{3}}{3}$, entonces no existe soluci\u00f3n para $x$ pues $\\dfrac{-2\\sqrt{3}}{3}\\cong -1,15$, y el coseno no puede ser un n\u00famero menor que $-1$.<\/p>\n\n\n\n

\n<\/details>\n\n\n\n

Sistema de ecuaciones trigonom\u00e9tricas<\/h3>\n\n\n\n

Resuelve el siguiente sistema de ecuaciones trigonom\u00e9tricas, dando las soluciones en el primer cuadrante.<\/p>\n\n\n\n

$\\displaystyle\\begin{cases}\\text{sen}\\,x\\cdot\\text{sen}\\,y=\\displaystyle\\frac{\\sqrt{2}}{4}\\\\ \\displaystyle\\cos x\\cdot\\text{sen}\\,y=\\frac{\\sqrt{6}}{4}\\end{cases}$<\/p>\n\n\n\n

Soluci\u00f3n<\/summary>\n

Dividiendo ambas ecuaciones tenemos:<\/p>\n\n\n\n

$\\displaystyle\\frac{\\text{sen}\\,x}{\\cos x}=\\frac{\\displaystyle\\frac{\\sqrt{2}}{4}}{\\displaystyle\\frac{\\sqrt{6}}{4}}\\Rightarrow \\text{tg}\\,x=\\frac{4\\sqrt{2}}{4\\sqrt{6}}\\Rightarrow \\text{tg}\\,x=\\sqrt{\\frac{2}{6}}\\Rightarrow$<\/p>\n\n\n\n

$\\displaystyle\\Rightarrow \\text{tg}\\,x=\\sqrt{\\frac{1}{3}}\\Rightarrow \\text{tg}\\,x=\\frac{\\sqrt{3}}{3}\\Rightarrow x=30{}^\\text{o}$<\/p>\n\n\n\n

Sustituyendo en la primera ecuaci\u00f3n:<\/p>\n\n\n\n

$\\displaystyle\\text{sen}\\,30{}^\\text{o}\\cdot \\text{sen}\\,y=\\frac{\\sqrt{2}}{4}\\Rightarrow \\frac{1}{2}\\text{sen}\\,y=\\frac{\\sqrt{2}}{4}\\Rightarrow \\text{sen}\\,y=\\frac{2\\sqrt{2}}{4}\\Rightarrow \\text{sen}\\,y=\\frac{\\sqrt{2}}{2}\\Rightarrow y=45{}^\\text{o}$<\/p>\n\n\n\n

\n<\/details>\n\n\n\n

<\/p>\n","protected":false},"excerpt":{"rendered":"

El genio se descubre en la fortuna adversa; en la prosperidad se oculta Homero En Matem\u00e1ticas I (1\u00ba de Bachillerato) se trabaja mucho la demostraci\u00f3n de identidades trigonom\u00e9tricas, la simplificaci\u00f3n de expresiones en las que aparecen razones trigonom\u00e9tricas, la resoluci\u00f3n de ecuaciones trigonom\u00e9tricas y de sistemas de ecuaciones trigonom\u00e9tricas. Veamos unos ejemplos. Identidades trigonom\u00e9tricas Para […]<\/p>\n","protected":false},"author":1,"featured_media":3334,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":{"0":"post-3300","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","6":"hentry","7":"category-uncategorized","9":"post-with-thumbnail","10":"post-with-thumbnail-large"},"_links":{"self":[{"href":"https:\/\/matematicastro.es\/index.php?rest_route=\/wp\/v2\/posts\/3300","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/matematicastro.es\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/matematicastro.es\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/matematicastro.es\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/matematicastro.es\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=3300"}],"version-history":[{"count":34,"href":"https:\/\/matematicastro.es\/index.php?rest_route=\/wp\/v2\/posts\/3300\/revisions"}],"predecessor-version":[{"id":3447,"href":"https:\/\/matematicastro.es\/index.php?rest_route=\/wp\/v2\/posts\/3300\/revisions\/3447"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/matematicastro.es\/index.php?rest_route=\/wp\/v2\/media\/3334"}],"wp:attachment":[{"href":"https:\/\/matematicastro.es\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=3300"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/matematicastro.es\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=3300"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/matematicastro.es\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=3300"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}