Dissertativas 3 - 2ª Fase Dia 1 - ITA 2025

$\mathrm{\alpha }, \mathrm{\beta }\in \left[\frac{\mathrm{\pi }}{2}, \frac{3\mathrm{\pi }}{2}\right]\Rightarrow \cos \mathrm{\alpha }<0 \mathrm{e} \cos \mathrm{\beta }<0$

Do sistema: $\left\{\begin{array}{l}\mathrm{sen} \mathrm{\alpha }-\mathrm{sen} \mathrm{\beta }=\frac{1}{4} \left(\mathrm{i}\right)\\ \mathrm{sen} \mathrm{\alpha }-2\mathrm{sen} \mathrm{\beta }+\cos \mathrm{\beta }=\frac{3}{4} \left(\mathrm{ii}\right)\end{array}\right.$

Fazendo (i)$-$(ii): $\mathrm{sen} \mathrm{\beta }-\cos \mathrm{\beta }=-\frac{1}{2}\Rightarrow 1-2\mathrm{sen} \mathrm{\beta }·\cos \mathrm{\beta }=\frac{1}{4}\Rightarrow 2\mathrm{sen} \mathrm{\beta }·\cos \mathrm{\beta }=\frac{3}{4}$.

Ou seja, $\mathrm{sen} \mathrm{\beta }<0$.

Fazendo 2(i)$-$(ii): $\mathrm{sen} \mathrm{\alpha }-\cos \mathrm{\beta }=-\frac{1}{4}\Rightarrow \mathrm{sen} \mathrm{\alpha }<0$.

Fazendo 3(i)$-$(ii):

$2\mathrm{sen} \mathrm{\alpha }-\mathrm{sen} \mathrm{\beta }-\cos \mathrm{\beta }=0\Rightarrow 2\mathrm{sen} \mathrm{\alpha }=\mathrm{sen} \mathrm{\beta }+\cos \mathrm{\beta }\Rightarrow 4{\mathrm{sen}}^2\mathrm{\alpha }=1+2\mathrm{sen} \mathrm{\beta }·\cos \mathrm{\beta }=\frac{7}{4}$

Portanto, $\mathrm{sen} \mathrm{\alpha }=-\frac{\sqrt{7}}{4} \mathrm{e} \cos \mathrm{\alpha }=-\frac{3}{4}$.

Por fim, tem-se:

$\left\{\begin{array}{l}\mathrm{sen} \mathrm{\beta }-\cos \mathrm{\beta }=-\frac{1}{2}\\ \mathrm{sen} \mathrm{\beta }+\cos \mathrm{\beta }=-\frac{\sqrt{7}}{2}\end{array}\right. \Rightarrow \left\{\begin{array}{l}\mathrm{sen} \mathrm{\beta }=-\frac{1}{4}\left(\sqrt{7}+1\right)\\ \cos \mathrm{\beta }=\frac{1}{4}\left(-\sqrt{7}+1\right)\end{array}\right.$

Logo:

 $$\begin{gathered} \mathrm{sen}\left(\mathrm{\alpha }+\mathrm{\beta }\right)=-\left(\frac{\sqrt{7}}{4}\right)·\left(\frac{\left(-\sqrt{7}+1\right)}{4}\right)+\left(-\frac{3}{4}\right)·\left(-\frac{1}{4}\right)·\left(\sqrt{7}+1\right)= \\ =\frac{1}{16}\left(7-\sqrt{7}+3\sqrt{7}+3\right)=\frac{2\sqrt{7}+10}{16}\Rightarrow \boxed{\mathrm{sen}\left(\mathrm{\alpha }+\mathrm{\beta }\right)=\frac{\sqrt{7}+5}{8}} \end{gathered}$$