Question

Se tg x =1/2 com ''a'' um angulo no primeiro quadrante, então cos a é

Answer 1

$tga=\frac{sena}{cosa}\\\\\frac{1}{2}=\frac{sena}{cosa}\\cosa=2.sena\\sen^2+cos^2a=1\\sen^2a=1-cos^2a\\sena=\sqrt{1-cos^2a}\\\\cosa=2.\sqrt{1-cos^2a}\\\\(cosa)^2=(2.\sqrt{1-cos^2a})^2\\cos^2a=4(1-cos^2a)\\cos^2a=4-4cos^2a\\5cos^2a=4\\cos^2a=\frac{4}{5}$

$\\cos^2a=\frac{4}{5}\\\\\boxed{cosa=\frac{\sqrt{4}}{\sqrt{5}}=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}}$