Question

Determine o valor de y = $\frac{sen 30º - cos 45º}{tg 30º} + \frac{cos 60º + sen 45º}{tg 60º}$:

Answer 1

Só precisa substituir:
$\frac{0,5-0,7071}{0,5774} + \frac{0,5+0,7071}{1,7321}$ = $-\frac{0,2071}{0,5774} + \frac{1,2071}{1,7321}$
$-0,3586+0,6969=1,0555$
Espero ter ajudado =)

Answer 2

$y=\frac{sen(30)-cos(45)}{tan(30)}+\frac{cos(60)+sen(45)}{tan(60)} \\ \\ y=\frac{\frac{1}{2}-\frac{\sqrt2}{2}}{\frac{\sqrt3}{3}}+\frac{\frac{1}{2}+\frac{\sqrt2}{2}}{\sqrt3} \\ \\ y=\frac{\frac{1-\sqrt2}{2}}{\frac{\sqrt3}{3}}+\frac{\frac{1+\sqrt2}{2}}{\sqrt3} \\ \\ y=\frac{1-\sqrt2}{2}.\frac{3}{\sqrt3}+\frac{1+\sqrt2}{2}.\frac{1}{\sqrt3} \\ \\ y=\frac{3.(1-\sqrt2)+1+\sqrt2}{2\sqrt3} \\ \\ y=\frac{3-3\sqrt2+1+\sqrt2}{2\sqrt3}=\frac{4-2\sqrt2}{2\sqrt3}=\frac{2(2-\sqrt2)}{2\sqrt3}$
$y=\frac{2(2-\sqrt2)}{2\sqrt3}=\frac{2- \sqrt{2}}{\sqrt{3}}=\boxed{\frac{2\sqrt{3}-\sqrt{6}}{3}}$