Question

Sabendo que f(x) = 3tg 2x, qual é o valor de f(0)?

Answer 1

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⠀⠀⠀☞ f(0) = 0. ✅  

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⠀⠀⠀☔⠀Oi, Damon ✌. Vamos inicialmente substituir o valor de x da função por 0:

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$\LARGE\blue{\sf f(0) = 3 \cdot tg(2 \cdot 0)}$

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$\LARGE\blue{\sf f(0) = 3 \cdot tg(0)}$

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⠀⠀⠀➡️⠀Quanto vale a tangente de 0º? Lembrando que tangente (β) = seno (β) / cosseno (β) temos então que tan(0) = sen(0)/cos(0) = 0/1 = 0. Sendo assim:

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$\LARGE\blue{\sf f(0) = 3 \cdot 0}$  

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                                         $\qquad\quad\huge\green{\boxed{\rm~~~\gray{f(0)}~\pink{=}~\blue{ 0 }~~~}}$ ✅

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⠀⠀☔⠀Observe o seguinte triângulo retângulo:

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            $\setlength{\unitlength}{0.95cm}\begin{picture}(6,5)\thicklines\put(0,0){\line(1,0){9.7}}\put(0,0){\line(2,3){3}}\put(3,4.5){\line(3,-2){6.7}}\bezier(0.45,0.65)(0.8,0.6)(0.9,0)\put(0.4,0.2){ \alpha }\bezier(8.7,0.65)(8.5,0.7)(8.45,0)\put(8.7,0.2){ \beta }\put(2.75,4.1){\line(3,-2){0.4}}\put(3.15,3.85){\line(2,3){0.25}}\put(3.5,4.5){C}\put(-0.6,0){A}\put(9.9,0){B}\put(3.1,4.15){\circle*{0.1}}\put(-0.5,3.5){\Large \sf cateto }\put(-1,3){ \sf (adjacente~\grave{a}~\alpha )}\put(-0.8,2.5){ \sf (oposto~\grave{a}~\beta )}\put(6.5,3.5){\Large \sf cateto }\put(6,3){ \sf (adjacente~\grave{a}~\beta )}\put(6.2,2.5){ \sf (oposto~\grave{a}~\alpha )}\put(3,-0.8){\LARGE \sf hipotenusa }\end{picture}$

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                            $\Large\red{\boxed{\begin{array}{rcl}&\green{\underline{\footnotesize\sf Esta~imagem~n\tilde{a}o~\acute{e}~visualiz\acute{a}vel~pelo~App~Brainly.}}&\\&\green{\footnotesize\sf \bullet~Experimente~compartilhar\rightarrow copiar~e~acessar}&\\&\green{\footnotesize\sf o~link~copiado~pelo~seu~navegador~ou~Browser.}&\\\end{array}}}$

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⠀⠀☔⠀Dele extraímos as três relações entre seus 3 lados: seno, cosseno e tangente:

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                           $\red{\boxed{\pink{\boxed{\begin{array}{lcr}\green{\star}&&\green{\star}\\&\!\!\orange{\bf seno(\Theta) = \dfrac{cat.~oposto(\Theta)}{hipotenusa}}\!\!&\\\\&\!\!\orange{\bf cosseno(\Theta) = \dfrac{cat.~adjacente(\Theta)}{hipotenusa}}\!\!&\\\\&\!\!\orange{\bf tangente(\Theta) = \dfrac{cat.~oposto(\Theta)}{cat.~adjacente(\Theta)}}\!\!&\\\\&\!\!\orange{\bf \Downarrow\qquad\qquad\Downarrow\qquad\qquad\Downarrow}\!\!&\\\\&\!\!\orange{\bf tan(\Theta) = \dfrac{sen(\Theta)}{cos(\Theta)}}\!\!&\\\green{\star}&&\green{\star}\\\end{array}}}}}$

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                             $\bf\large\red{\underline{\quad\quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}$☁

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⠀⠀⠀☀️ L͎̙͖͉̥̳͖̭̟͊̀̏͒͑̓͊͗̋̈́ͅeia mais sobre relações trigonométricas:

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                                     https://brainly.com.br/tarefa/36098398 ✈  

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                                     $\huge\blue{\text{\bf\quad Bons~estudos.}}$ ☕

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                                          $\quad\qquad(\orange{D\acute{u}vidas\ nos\ coment\acute{a}rios})$ ☄

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                             $\bf\large\red{\underline{\qquad \qquad \qquad \qquad \qquad \qquad \quad }\LaTeX}$✍

                                $\sf(\purple{+}~\red{cores}~\blue{com}~\pink{o}~\orange{App}~\green{Brainly}$ ☘☀❄☃☂☻)

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                                                          $\Huge\green{\text{ \underline{\red{\mathbb{S}}\blue{\mathfrak{oli}}~}~\underline{\red{\mathbb{D}}\blue{\mathfrak{eo}}~}~\underline{\red{\mathbb{G}}\blue{\mathfrak{loria}}~} }}$ ✞

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