Question

Para PI < & < 3PI/2, determine sen&, cos& e sec& se tg& = 3/2.

Answer 1

Temos um ângulo $\theta,$

com $\mathsf{\pi<\theta<\dfrac{3\pi}{2}}$

($\theta$ é do 3º quadrante),


e $\mathsf{tg\,\theta=\dfrac{\,3\,}{2}}.$

_______


Daqui, tiramos que

$\mathsf{\dfrac{sen\,\theta}{cos\,\theta}=\dfrac{\,3\,}{2}}\\\\\\ \mathsf{2\,sen\,\theta=3\,cos\,\theta\qquad\quad(i)}$


Elevando os dois lados ao quadrado, ficamos com

$\mathsf{(2\,sen\,\theta)^2=(3\,cos\,\theta)^2}\\\\ \mathsf{4\,sen^2\,\theta=9\,cos^2\,\theta}\qquad\quad\textsf{(mas }\mathsf{cos^2\,\theta=1-sen^2\,\theta}\textsf{)}\\\\ \mathsf{4\,sen^2\,\theta=9\cdot (1-sen^2\,\theta)}\\\\ \mathsf{4\,sen^2\,\theta=9-9\,sen^2\,\theta}\\\\ \mathsf{4\,sen^2\,\theta+9\,sen^2\,\theta=9}\\\\ \mathsf{13\,sen^2\,\theta=9}$

$\mathsf{sen^2\,\theta=\dfrac{9}{13}}\\\\\\ \mathsf{sen\,\theta=\pm\,\sqrt{\dfrac{9}{13}}}\\\\\\ \mathsf{sen\,\theta=\pm\,\dfrac{3}{\sqrt{13}}}$


Mas como $\theta$ é do 3º quadrante, o seno é negativo. Logo,

$\boxed{\begin{array}{c}\mathsf{sen\,\theta=-\,\dfrac{3}{\sqrt{13}}} \end{array}}\qquad\quad\checkmark$


•   Achando o cosseno:

Por $\mathsf{(i)},$ temos que

$\mathsf{2\,sen\,\theta=3\,cos\,\theta}\\\\ \mathsf{cos\,\theta=\dfrac{\,2\,}{3}\,sen\,\theta}\\\\\\ \mathsf{cos\,\theta=\dfrac{\,2\,}{\diagup\!\!\!\! 3}\cdot \left(-\,\dfrac{\diagup\!\!\!\! 3}{\sqrt{13}}\right)}\\\\\\ \boxed{\begin{array}{c}\mathsf{cos\,\theta=-\,\dfrac{2}{\sqrt{13}}} \end{array}}\qquad\quad\checkmark$


•   Achando a secante:

A secante é o inverso do cosseno, portanto

$\mathsf{sec\,\theta=\dfrac{1}{cos\,\theta}}\\\\\\ \mathsf{sec\,\theta=\dfrac{1}{-\,\frac{2}{\sqrt{13}}}}\\\\\\ \boxed{\begin{array}{c}\mathsf{sec\,\theta=-\,\dfrac{\sqrt{13}}{2}} \end{array}}\qquad\quad\checkmark$


Caso tenha problemas para visualizar a resposta pelo aplicativo, experimente abrir pelo navegador: https://brainly.com.br/tarefa/7917704


Dúvidas? Comente.


Bons estudos! :-)


Tags:  razões trigonométricas sen cos sec tan tg seno cosseno secante tangente trigonometria