• Matéria: Matemática
  • Autor: andersonmonte
  • Perguntado 9 anos atrás

Dado tg x = √3 , calcule sen x e cos x
                    3


ellisactassot: VOCE ESTA USANDO QUE FORMULA PRA RESOLVER O EXERCICIO? SEN²X+COS²X=1 ?
andersonmonte: sim
ellisactassot: RESPONDE SEPARADAMENTO , TG 3=60 GRAUS = PI/3 SENO DE 60° =3/2 COSENO 60=1/2
ellisactassot: DESCULPA EU RESPONDO SO O COSENO :/

Respostas

respondido por: Lukyo
3
\large\begin{array}{l} \mathsf{tg\,x=\sqrt{3}}\\\\ \mathsf{\dfrac{sen\,x}{cos\,x}=\sqrt{3}}\\\\ \mathsf{sen\,x=\sqrt{3}\,cos\,x\qquad(i)} \end{array}


\large\begin{array}{l} \textsf{Elevando os dois lados ao quadrado, obtemos}\\\\ \mathsf{(sen\,x)^2=(\sqrt{3}\,cos\,x)^2}\\\\ \mathsf{sen^2\,x=(\sqrt{3})^2\,cos^2\,x}\\\\ \mathsf{sen^2\,x=3\,cos^2\,x\qquad}\textsf{(mas }\mathsf{cos^2\,x=1-sen^2\,x}\textsf{)}}\\\\ \mathsf{sen^2\,x=3\cdot (1-sen^2\,x)}\\\\ \mathsf{sen^2\,x=3-3\,sen^2\,x} \end{array}

\large\begin{array}{l} \mathsf{sen^2\,x=3-3\,sen^2\,x}\\\\ \mathsf{sen^2\,x+3\,sen^2\,x=3}\\\\ \mathsf{4\,sen^2\,x=3}\\\\ \mathsf{sen^2\,x=\dfrac{3}{4}}\\\\ \mathsf{sen\,x=\pm\sqrt{\dfrac{3}{4}}}\\\\ \mathsf{sen\,x=\pm\,\dfrac{\sqrt{3}}{2}} \end{array}

\large\begin{array}{l} \begin{array}{rcl} \mathsf{sen\,x=-\,\dfrac{\sqrt{3}}{2}}&~\textsf{ ou }~&\mathsf{sen\,x=\dfrac{\sqrt{3}}{2}} \end{array}\\\\\\ \textsf{O sinal do seno vai depender do quadrante a que x pertence.} \end{array}


\large\begin{array}{l} \textsf{Como a tangente \'e positiva, x pode ser do primeiro ou}\\\textsf{do terceiro quadrantes.}\\\\\\ \bullet~~\textsf{Se x for do primeiro quadrante, ent\~ao}\\\\ \mathsf{0<sen\,x<1}\\\\ \mathsf{sen\,x=\dfrac{\sqrt{3}}{2}\qquad\checkmark}\\\\\\ \textsf{e encontramos o cosseno pela equa\c{c}\~ao (i):}\\\\ \mathsf{sen\,x=\sqrt{3}\,cos\,x} \end{array}

\large\begin{array}{l} \mathsf{cos\,x=\dfrac{1}{\sqrt{3}}\cdot sen\,x}\\\\ \mathsf{cos\,x=\dfrac{1}{\;\diagup\!\!\!\!\!\!\!\! \sqrt{3}}\cdot \dfrac{\;\diagup\!\!\!\!\!\!\!\! \sqrt{3}}{2}}\\\\ \mathsf{cos\,x=\dfrac{1}{2}\qquad\checkmark} \end{array}


\large\begin{array}{l} \bullet~~\textsf{Se x for do terceiro quadrante, ent\~ao}\\\\ \mathsf{-1<sen\,x<0}\\\\ \mathsf{sen\,x=-\,\dfrac{\sqrt{3}}{2}\qquad\checkmark}\\\\\\ \textsf{e encontramos o cosseno de forma an\'aloga \`a que}\\\textsf{foi feita para x no primeiro quadrante:}\\\\ \mathsf{cos\,x=-\,\dfrac{1}{2}\qquad\checkmark} \end{array}


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\large\textsf{Bons estudos! :-)}

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