Question

sabendo que x+y = pi/4 e sen y = 1/2, calcular sen x, com 0< y< pi/2

Answer 1

$x + y = \frac{\pi}{4} \\\\ sin(y) = \frac{1}{2} \\\\ x = \frac{\pi}{4} - y \\\\ \mathbf{Equação \: Fundamental \: da \: Trigonometria: } \\\\ sin^2(y)+ cos^2(y) = 1 \\ cos^2(y) = 1 - \frac{1}{4}\\ cos(y) = \frac{\sqrt{3}}{2} \\\\ sin(x) = sin \left (\frac{\pi}{4} - y \right) = \\ sin\left (\frac{\pi}{4} \right) \times cos(y) - cos \left(\frac{\pi}{4}\right ) \times sin(y)$

*sen (π/4) = cos (π/4) = √2/2

sen(x) =

$\left( \frac{ \sqrt{2} }{2} \right ) \times \left ( \frac{ \sqrt{3} }{2} \right ) - \left ( \frac{1}{2} \right) \times \left(\frac{ \sqrt{2} }{2} \right) \\ \\ \frac{ \sqrt{6} }{4 } - \frac{ \sqrt{2} }{4} = \frac{ \sqrt{6} - \sqrt{2} }{ 4} \\\\ \boxed{sin(x) = \frac{\sqrt{6} - \sqrt{2}}{4}}$