Sabemos que o seno do ângulo x em questão é igual a [tex]-\frac{3}{5}[/tex], pois ângulos pertencentes ao 4º quadrante possuem sen < 0, cos > 0 e tg < 0:
[tex]\bullet \ sen \ x \\\\\ \circ sen \ x = -\frac{3}{5}, \text{pois se} \ x \in 4^\circ \text{quadrante,} \ sen \ x < 0 \\\\\\ \bullet cos \ x \\\\ \circ cos \ x = \frac{cateto \ adjacente}{hipotenusa} \\\\ \circ sen \ x = \frac{cateto \ oposto}{hipotenusa} \rightarrow \text{se} \ sen \ x = -\frac{3}{5}, \ cateto \ oposto = 3, hipotenusa = 5 \\\\ \circ hipotenusa^2 = (cateto \ oposto)^2 + (cateto \ adjacente)^2 \\\\ 5^2 = 3^2 + (c.a.)^2 \\\\ (c.a.)^2 = 25 - 9 \\\\ (c.a) = 16 \\\\ c.a. = 4[/tex]
[tex]\text{Assim:} \\\\ \circ cos \ x = \frac{cateto \ adjacente \ \rightarrow \ 4}{hipotenusa \ \rightarrow \ 5} \\\\ \boxed{cos \ x = \frac{4}{5}}[/tex]
[tex]\bullet \ tg \ x \\\\ \circ tg \ x = \frac{cateto \ oposto}{cateto \ adjacente} \\\\ tg \ x = \frac{3}{4} \\\\ \text{Sendo} \ x \in 4^\circ \text{quadrante}, tg \ x < 0 \\\\ \boxed{tg \ x = -\frac{3}{4}} \\\\\\\\ \bullet cotg \ x = (tg \ x)^{-1} \\\\ cotg \ x = (-\frac{3}{4})^{-1} \\\\ \boxed{cotg \ x = -\frac{4}{3}}[/tex]
