[OC] \u00e9 o raio perpendicular a [AB], M \u00e9 um ponto do arco AC. Designa-se por $\\theta $ a medida em radianos do \u00e2ngulo AOM $\\left( {0 \\leqslant \\theta\u00a0 \\leqslant \\frac{\\pi }{2}} \\right)$.<\/p>\n H \u00e9 a proje\u00e7\u00e3o ortogonal de M sobre OC.<\/p>\n Existir\u00e1 um ponto M tal que $\\overline {AM}\u00a0 = \\overline {MH} $?<\/p>\n Sugest\u00e3o<\/strong>:<\/p>\n Por outro lado, como $$g(0) \\times g(\\frac{\\pi }{2}) = \\left( {2\\operatorname{sen} 0 – \\cos 0} \\right) \\times \\left( {2\\operatorname{sen} \\frac{\\pi }{4} – \\cos \\frac{\\pi }{2}} \\right) =\u00a0 – 1 \\times \\sqrt 2\u00a0 < 0$$ a fun\u00e7\u00e3o $g$ tem um s\u00f3 zero (pois \u00e9 estritamente crescente) no intervalo $\\left[ {0,\\frac{\\pi }{2}} \\right]$.<\/p>\n Assim, existe esse ponto M, pois $$g(\\theta ) = 0 \\Leftrightarrow 2\\operatorname{sen} \\frac{\\theta }{2} – \\cos \\theta\u00a0 = 0 \\Leftrightarrow \\overline {AM}\u00a0 = \\overline {MH} $$ visto $r > 0$.![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12pag129-13.jpg\" "\\"Semicircunfer\u00eancia\\"")
<\/a>$C$ \u00e9 uma semicircunfer\u00eancia de di\u00e2metro [AB], de centro O e de raio $r$.<\/p>\n\n
Resolu\u00e7\u00e3o >><\/a><\/span><\/div><\/div>\n\n\nResolu\u00e7\u00e3o<\/b><\/span><\/p>\n\n
<\/a>Como $$\\operatorname{sen} M\\widehat OH = \\frac{{\\overline {MH} }}{{\\overline {MO} }}$$ e $$\\cos M\\widehat OH = \\frac{{\\overline {OH} }}{{\\overline {MO} }}$$ vem: $$\\operatorname{sen} \\left( {\\frac{\\pi }{2} – \\theta } \\right) = \\frac{{\\overline {MH} }}{r} \\Leftrightarrow \\overline {MH}\u00a0 = r\\cos \\theta $$ e $$\\cos \\left( {\\frac{\\pi }{2} – \\theta } \\right) = \\frac{{\\overline {OH} }}{r} \\Leftrightarrow \\overline {OH}\u00a0 = r\\operatorname{sen} \\theta $$O tri\u00e2ngulo [AOM] \u00e9 is\u00f3sceles, pois $\\overline {OA}\u00a0 = \\overline {OM}\u00a0 = r$.
\nConsequentemente, $$O\\widehat AM = O\\widehat MA = \\frac{{\\pi\u00a0 – \\theta }}{2} = \\frac{\\pi }{2} – \\frac{\\theta }{2}$$Assim, e como $$\\operatorname{sen} O\\widehat AM = \\frac{{\\overline {MM’} }}{{\\overline {AM} }}$$ (M’ \u00e9 a proje\u00e7\u00e3o ortogonal de M sobre AB), vem: $$\\operatorname{sen} \\left( {\\frac{\\pi }{2} – \\frac{\\theta }{2}} \\right) = \\frac{{\\overline {MM’} }}{{\\overline {AM} }} \\Leftrightarrow \\overline {AM}\u00a0 = \\frac{{\\overline {MM’} }}{{\\cos \\frac{\\theta }{2}}}$$Como $$\\overline {MM’}\u00a0 = \\overline {OH}\u00a0 = r\\operatorname{sen} \\theta $$ obt\u00e9m-se: $$\\overline {AM}\u00a0 = \\frac{{\\overline {MM’} }}{{\\cos \\frac{\\theta }{2}}} = \\frac{{r\\operatorname{sen} \\theta }}{{\\cos \\frac{\\theta }{2}}} = \\frac{{r\\operatorname{sen} \\left( {2 \\times \\frac{\\theta }{2}} \\right)}}{{\\cos \\frac{\\theta }{2}}} = \\frac{{2r\\operatorname{sen} \\frac{\\theta }{2}\\cos \\frac{\\theta }{2}}}{{\\cos \\frac{\\theta }{2}}} = 2r\\operatorname{sen} \\frac{\\theta }{2}$$
\n\u00ad<\/li>\n
\n{g’\\left( \\theta\u00a0 \\right)}& = &{{{\\left( {2\\operatorname{sen} \\frac{\\theta }{2} – \\cos \\theta } \\right)}^\\prime }} \\\\
\n{}& = &{2 \\times \\frac{1}{2}\\cos \\frac{\\theta }{2} + \\operatorname{sen} \\theta } \\\\
\n{}& = &{\\cos \\frac{\\theta }{2} + \\operatorname{sen} \\theta }
\n\\end{array}\\]
\nComo $\\cos \\frac{\\theta }{2} > 0,\\forall \\theta\u00a0 \\in \\left[ {0,\\frac{\\pi }{2}} \\right]$ e $\\operatorname{sen} \\theta\u00a0 \\geqslant 0,\\forall \\theta\u00a0 \\in \\left[ {0,\\frac{\\pi }{2}} \\right]$, ent\u00e3o $g'(\\theta ) > 0,\\forall \\theta\u00a0 \\in \\left[ {0,\\frac{\\pi }{2}} \\right]$.
\nLogo, a fun\u00e7\u00e3o $g$ \u00e9 estritamente crescente em $\\left[ {0,\\frac{\\pi }{2}} \\right]$.<\/p>\n
\n\u00ad<\/li>\n