Question

considerando sen a + cos a=1/5. calcule o valor de sen a e cos a:​

Answer 1

Relação fundamental da trigonometria :

$\text{sen}^2(\text a)+\text{cos}^2(\text a) = 1$

Temos :

$\displaystyle \text{sen(a)}+\text{cos(a)}=\frac{1}{5}$

elevando ao quadrado dos dois lados :

$\displaystyle \text {sen}^2(\text a)+\text{cos}^2(\text a)+2.\text{sen(a).cos(a)}= \frac{1}{25}$

$\displaystyle 1+2.\text{sen(a).cos(a)}= \frac{1}{25} \\\\\\\ 2.\text{sen(a).cos(a)}= \frac{1}{25} - 1 \\\\\\ 2.\text{sen(a).cos(a)}= \frac{1-25}{25} \\\\\\ \text{sen(a).cos(a)}= \frac{-24}{25.2} \\\\\\ \text{sen(a).cos(a)}= \frac{-12}{25}$

pela relação fundamental da trigonometria sabemos que :

$\text{cos(a)}=\sqrt{1-\text{sen}^2(\text a)}$

substituindo :

$\displaystyle \text{sen(a).cos(a)}=\frac{-12}{25}$

$\displaystyle \text{sen(a)}.\sqrt{1-\text{sen}^2(\text a)} = \frac{-12}{25}$

elevando ao quadrado dos dois lados :

$\displaystyle \text{sen}^2(\text a).[\ 1-\text {sen}^2 (\text a)\ ] = \frac{144}{625}$

$\displaystyle \text{sen}^2(\text a)-\text {sen}^4 (\text a)= \frac{144}{625}$

$\displaystyle 625.\text{sen}^2(\text a)-625.\text{sen}^4(\text a) = 144 \\\\ 625.\text{sen}^4(\text a)-625.\text{sen}^2(\text a) + 144 = 0 \\\\\\ \text{sen}^2(\text a)= \frac{-(-625)\pm\sqrt{(-625)^2-4.625.144}}{2.625 } \\\\\ \text{sen}^2(\text a)= \frac{625\pm\sqrt{390625-360000}}{1250}$

$\displaystyle \text{sen}^2(\text a) = \frac{625\pm\sqrt{30625}}{1250} \\\\\\ \text{sen}^2(\text a)= \frac{625\pm 175}{1250} \\\\\\ \text{sen}^2(\text a) =\frac{625+175}{1250} \to \text{sen}^2(\text a) = \frac{800}{1250} \to \text{sen}^2(\text a) = \frac{16}{25} \\\\\\ \text{sen}(\text a) =\pm \sqrt{\frac{16}{25}} \\\\\\\ \boxed{\text{sen}(\text a)= \frac{4}{5}\ \ \text{ou}\ \ \text{sen}(\text a)= \frac{-4}{5}}$

Achando o valor do cos a :  

$\displaystyle \text{Se} \ \ \text{sen(a)} = \frac{4}{5} \\\\\ \text{cos}(\text a)=\sqrt{1-\text{sen}^2(\text a)} \\\\ \text{cos(a)}=\sqrt{1-\frac{16}{25}} \\\\\\ \text{cos(a)} = \sqrt{\frac{25-16}{25}} \to \text{cos(a)} =\pm\sqrt{\frac{9}{25}} \\\\\\\ \boxed{\text{cos(a)}= \frac{3}{5} \ \ \text{ou} \ \ \text{cos(a)} = \frac{-3}{5}\ }$

Temos aí 4 valores pra testar.

Se tivermos

$\displaystyle \text{sen(a)}= \frac{4}{5} \ \ \text e \ \ \text{cos(a)} = \frac{-3}{5}$

então :

$\displaystyle \text{sen(a)}+\text{cos(a)} = \frac{4}{5}+\frac{-3}{5} \\\\ {\text{sen(a)}+\text{cos(a)} = \frac{1}{5} } \to \text{(FUNCIONOU)}$

Portanto :

$\huge\boxed{\ \text{sen(a)} = \frac{4}{5} \ ; \ \text{cos(a)} = \frac{-3}{5}\ }\checkmark$