Question

Trigonometria.
1)Se: Sen x = 1/2, x um arco do segundo quadrante.
Qual a equação?

Answer 1

Relação fundamental da trigonometria:

$\boxed{\boxed{sen^{2}(x)+cos^{2}(x)=1}}$

Ângulos do segundo quadrante:
- sen(x) > 0
- cos(x) < 0
- tan(x) < 0 (quociente do sinal do seno com o sinal do cosseno)

Consequentemente:
- cossec(x) > 0
- sec(x) < 0
- cotg(x) < 0
____________________

$sen^{2}(x)+cos^{2}(x)=1$

Como sen(x) = 1/2:

$\left(\dfrac{1}{2}\right)^{2}+cos^{2}(x)=1\\\\\\\dfrac{1}{4}+cos^{2}(x)=1\\\\\\\cos^{2}(x)=1-\dfrac{1}{4}\\\\\\cos^{2}(x)=\dfrac{4-1}{4}\\\\\\cos(x)=\pm\sqrt{\dfrac{3}{4}}=\pm\dfrac{\sqrt{3}}{\sqrt{4}}=\pm\dfrac{\sqrt{3}}{2}$

Como o cosseno é negativo no segundo quadrante:

$\boxed{\boxed{cos~x=-\dfrac{\sqrt{3}}{2}}}$
____________________________

Achando tg(x):

$tg(x)=\dfrac{sen(x)}{cos(x)}\\\\\\tg(x)=\dfrac{(\frac{1}{2})}{(-\frac{\sqrt{3}}{2})}\\\\\\tg(x)=\dfrac{1}{2}\cdot\left(-\dfrac{2}{\sqrt{3}}\right)\\\\\\tg(x)=-\dfrac{1}{\sqrt{3}}\\\\\\\boxed{\boxed{tg(x)=-\dfrac{\sqrt{3}}{3}}}$

Achando cotg(x):

$cotg(x)=\dfrac{1}{tg(x)}\\\\\\cotg(x)=\dfrac{1}{(-\frac{1}{\sqrt{3}})}\\\\\\cotg(x)=\dfrac{1}{1}\cdot\left(-\dfrac{\sqrt{3}}{1}\right)\\\\\\\boxed{\boxed{cotg(x)=-\sqrt{3}}}$

Achando cossec(x):

$cossec(x)=\dfrac{1}{sen(x)}\\\\\\cossec(x)=\dfrac{1}{(\frac{1}{2})}\\\\\\cossec(x)=\dfrac{1}{1}\cdot\dfrac{2}{1}\\\\\\\boxed{\boxed{cossec(x)=2}}$

Achando sec(x):

$sec(x)=\dfrac{1}{cos(x)}\\\\\\sec(x)=\dfrac{1}{(-\frac{\sqrt{3}}{2})}\\\\\\sec(x)=\dfrac{1}{1}\cdot\left(-\dfrac{2}{\sqrt{3}}\right)\\\\\\sec(x)=\dfrac{-2\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}\\\\\\\boxed{\boxed{ec(x)=-\dfrac{2\sqrt{3}}{3}}}$