• Matéria: Matemática
  • Autor: brenoreis64
  • Perguntado 4 anos atrás

Sejam x e y dois números reais, com 0 < x < π/2 e π/2 < y < π. sabendo que sen y 4/5 e 10.sen x 5.cos y = 3 Nessas condições, determine o que está sendo pedido abaixo:
a) cos y b) tg x

Respostas

respondido por: CyberKirito
1

\large\boxed{\begin{array}{l}\sf 0&lt;x&lt;\dfrac{\pi}{2}\longrightarrow \in~1^o~quadrante\\\sf\dfrac{\pi}{2}&lt;y&lt;\pi\longrightarrow \in~2^o~quadrante\\~~\rm a)\\\sf sen(y)&gt;0~e~~cos(y)&lt;0.\\\sf sen(y)=\dfrac{4}{5}\longrightarrow sen^2(y)=\dfrac{16}{25}\\\\\sf cos^2(y)=\dfrac{25}{25}-\dfrac{16}{25}=\dfrac{9}{25}\\\sf cos(y)=-\sqrt{\dfrac{9}{25}}\\\huge\boxed{\boxed{\boxed{\boxed{\sf cos(y)=-\dfrac{3}{5}}}}}\end{array}}

\large\boxed{\begin{array}{l}\rm b)\\\sf 10\cdot sen(x)\cdot 5\cdot cos(y)=3\\\sf 10\cdot sen(x)\cdot\backslash\!\!\!5\cdot\bigg(-\dfrac{\backslash\!\!\!3}{\backslash\!\!\!5}\bigg)=\backslash\!\!\!3\\\\\sf -10 sen(x)=1\\\sf sen(x)=-\dfrac{1}{10}~absurdo\,pois\,x\in~1^o~quadrante\\\sf e\,portanto\,sen(x)&gt;0.\\\sf A\,quest\tilde ao\,n\tilde ao\,foi\,bem\,elaborada.\\\sf mas\,admitindo-se\,que\,x\,pertenc_{\!\!,}a\\\sf ao\,3^o~quadrante\,ter\acute iamos:\end{array}}

\large\boxed{\begin{array}{l}\sf sen(x)=-\dfrac{1}{10}\longrightarrow sen^2(x)=\dfrac{1}{100}\\\\\sf cos^2(x)=\dfrac{100}{100}-\dfrac{1}{100}=\dfrac{99}{100}=\dfrac{9\cdot11}{100}\\\\\sf cos(x)=-\sqrt{\dfrac{9\cdot11}{100}}=-\dfrac{3\sqrt{11}}{10}\\\sf DA\acute I\\\sf tg(x)=\dfrac{sen(x)}{cos(x)}\\\\\sf tg(x)=\dfrac{-\frac{1}{\diagdown\!\!\!\!\!\!10}}{-\frac{3\sqrt{11}}{\diagdown\!\!\!\!\!10}}\\\\\sf tg(x)=\dfrac{1}{3\sqrt{11}}\cdot\dfrac{\sqrt{11}}{\sqrt{11}}\end{array}}

\large\boxed{\begin{array}{l}\sf tg(x)=\dfrac{\sqrt{11}}{3\cdot11}\\\\\huge\boxed{\boxed{\boxed{\boxed{\sf tg(x)=\dfrac{\sqrt{11}}{33}}}}}\end{array}}

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