Question

trigonometria ajuda gente

Answer 1

$\frac{sen(x).cos(x)}{tg(x).cos(\pi + x)} <==> \frac{a.b}{\frac{a}{b}.[cos(\pi).cos(x) - sen(\pi).sen(x)]} \\ \\ pois \\ cos(a + b) = cos(a).cos(b) - sen(b).sen(a) \\ cos(\pi) = -1 \\ sen(\pi) = 0 \\ \\ \frac{a.b}{\frac{a}{b}.[(-1).cos(x) - (0).sen(x)]} <==> \frac{a.b}{\frac{a}{b}.[-cos(x)]} <==> \frac{a.b}{\frac{a}{b}.(-b)} \\ \\ -b$

Answer 2

Resposta:

Explicação passo-a-passo:

cos(π + x) = -cos(π + x - π) = -cosx

$y=\frac{senx.cosx}{tgx.cos(\pi-x)}=\frac{senx.cosx}{\frac{senx}{cosx}*(-cosx)}=\frac{senx.cosx}{-senx}=-cosx=-b$