Question

sabendo que tg x= 2, determine $\frac{cos(2x)}{cos(2x) + sen(2x)}$

Answer 1

tg (2x) = (tg x + tg x) / ( 1 - tg ² x)

2 +2 / 1 - 2 ² =

4 / 1-4 =

-4/3

sen (2x) / cos (2x) = -4/3

sen (2x) = -4cos (2x)/3

Substituindo na expressão:

[cos (2x)] / [cos (2x) - 4cos (2x)/3] =

3 cos (2x) / - cos (2x) =

3 / -1 =

-3

Answer 2

Resposta:

tg(x+x)=(tg(x)+tg(x))/(1-tg(x).tg(x))

tg(2x)=2tg(x)/(1-tg²(x))

tg(x)=2

tg(2x)=2.2/(1-2²)=4/(1-4)= -4/3

$\displaystyle\frac{cos(2x)}{cos(2x)+sen(2x)} =\\\\\\\frac{\frac{cos(2x)}{cos(2x)}}{\frac{cos(2x)}{cos(2x)}+\frac{sen(2x)}{cos(2x)}} =\\\\\\\frac{1}{1+tg(2x)} =\frac{1}{1-\frac{4}{3}}=\frac{1}{\frac{3-4}{3}} =-3$