Question

Preciso de ajuda urgente.
Trigonometria, alguém? Por favor.

Answer 1

Primeiro as coisas básicas:

$sec(x)=\frac{1}{cos(x)}$
$cossec(x)=\frac{1}{sen(x)}$
$cotg(x)=\frac{1}{tg(x)}=\frac{cos(x)}{sen(x)}$


$cossec(30)=\frac{1}{sen(30)}=\frac{1}{\frac{1}{2}}=2$

$cossec(120)=\frac{1}{sen(120)}=\frac{1}{sen(60)}=\frac{1}{\frac{\sqrt(3)}{2}}$
$=\frac{2}{\sqrt(3)}=\frac{2\sqrt(3)}{3}$

$cossec(270)=\frac{1}{sen(270)}=\frac{1}{- sen(90)}=\frac{1}{-1}=-1$

$cossec(135)=\frac{1}{sen(135)}=\frac{1}{sen(45)}=\frac{1}{\frac{\sqrt(2)}{2}}$
$=\frac{2}{\sqrt(2)}=\sqrt(2)$

$cossec(225)=\frac{1}{sen(225)}=\frac{1}{-sen(45)}=\frac{1}{-\frac{\sqrt(2)}{2}}$
$=-\frac{2}{\sqrt(2)}=-\sqrt(2)$

$cossec(300)=\frac{1}{sen(300)}=\frac{1}{-sen(60)}=\frac{1}{-\frac{\sqrt(3)}{2}}$
$=-\frac{2}{\sqrt(3)}=-\frac{2\sqrt(3)}{3}$


$cotg(\frac{\pi}{3})=\frac{cos(\frac{\pi}{3})}{sen(\frac{\pi}{3})}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$
$=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$

$sec(\frac{\pi}{3})=\frac{1}{cos(\frac{\pi}{3})}=\frac{1}{\frac{1}{2}}=2$

$cossec(\frac{\pi}{3})=\frac{1}{sen(\frac{\pi}{3})}=\frac{1}{\frac{\sqrt{3}}{2}}$
$=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}$

$sec(\frac{9\pi}{4})=\frac{1}{cos(\frac{9\pi}{4})}=\frac{1}{cos(\frac{\pi}{4})}$
$=\frac{1}{\frac{\sqrt{2}}{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$


$cotg(-1755)=\frac{cos(-1755)}{sen(-1755)}=\frac{cos(45)}{sen(45)}$
$=\frac{\frac{\sqrt{2}{2}}{\sqrt{2}{2}}=1$


$sec(1050)=\frac{1}{cos(1050)}=\frac{1}{cos(30)}=\frac{1}{\frac{\sqrt{3}}{2}}$
$=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}$

$cotg(-1410)=\frac{cos(-1410)}{sen(-1410)}=\frac{cos(30)}{sen(30)}$
$=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}$

Logo cotg(-1410) é maior.


$x=180$

$y=\frac{5cosec(\frac{x}{2})-2sen(x)}{5sen(\frac{x}{2})}$
$y=\frac{5cosec(\frac{180}{2})-2sen(180)}{5sen(\frac{180}{2})}$
$y=\frac{5cosec(90)-2sen(180)}{5sen(90)}$
$y=\frac{5cosec(90)-0}{5}$

FALTA ACABAR AQUI, MAS TENHO QUE SAIR, AJEITO ISSO ASSIM QUE VOLTAR.
POR FAVOR NÃO DELETEM.