Question

Sendo tg x = 4 e 0 < x <  π/2, calcule sen x e cos x

Answer 1

$\bullet \ tg \ x = \frac{sen \ x}{cos \ x} \\\\ \bullet sen^2 \ x + cos^2 \ x = 1 \\ sen^2 \ x = 1 - cos^2 \ x \\ sen \ x = \sqrt{1 - cos^2 \ x} \\\\ ---------------------$

$\bullet \ tg \ x = \frac{\sqrt{1 - cos^2 \ x}}{cos \ x} \\\\ \bullet \ tg \ x = 4 \\\\ \bullet \ sen \ x = \sqrt{1 - cos^2 \ x} \\\\ 4 = \frac{\sqrt{1 - cos^2 \ x}}{cos \ x} \\\\ 4 \ cos \ x = \sqrt{1 - cos^2 \ x} \\\\ (4 \ cos \ x)^2 = 1 - cos^2 \ x \\\\ 16 \ cos^2 x = 1 - cos^2 \ x \\\\ 17 \ cos^2 \ x = 1 \\\\ cos^2 \ x = \frac{1}{17}$

$cos \ x = \sqrt{\frac{1}{17}} \\\\ cos \ x = \frac{1}{\sqrt{17}} \rightarrow \cdot \frac{\sqrt{17}}{\sqrt{17}} \rightarrow \boxed{\frac{\sqrt{17}}{17}} \\\\ ---------------------$

$\bullet \ tg \ x = 4 \\\\ \bullet tg \ x = \frac{sen \ x}{cos \ x} \\\\ \bullet cos \ x = \frac{\sqrt{17}}{17} \\\\ 4 = \frac{sen \ x}{\frac{\sqrt{17}}{17}} \\\\ sen \ x = \boxed{\frac{4 \ \sqrt{17}}{17}}$