Question

Verifique se é verdadeira a igualdade :
1 + cotg² x = cossec² x para x = 3π/4 e x = 7π/6

Answer 1

Primeiro, sabendo que:

$cotg(x) = \dfrac{1}{tg(x)} = \dfrac{cos(x)}{sen(x)}$

e:

$cossec(x) = \dfrac{1}{sen(x)}$

Agora, substituindo:

$1 + \left(\dfrac{cos(x)}{sen(x)}\right)^2 = \left(\dfrac{1}{sen(x)}\right)^2$

$1 = \dfrac{1}{sen^2(x)} -\dfrac{cos^2(x)}{sen^2(x)}$

$1 = \dfrac{1-cos^2(x)}{sen^2(x)}$

$1 = \dfrac{1-cos^2(x)}{sen^2(x)}$

Multiplicando ambos os lados por $sen^2(x)$:

$1 \cdot sen^2(x) = \dfrac{1-cos^2(x)}{sen^2(x)} \cdot sen^2(x)$

$sen^2(x) = 1 - cos^2(x)$

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

Essa propriedade é verdadeira para qualquer ângulo x.

Apenas para confirmar:

Sabendo que: $\pi = 180\textdegree$

Assim:

$\dfrac{3 \cdot \pi}{4} = \dfrac{3 \cdot 180\textdegree}{4} = 3 \cdot 45\textdegree = 135\textdegree$

e:

$\dfrac{7 \cdot \pi}{6} = \dfrac{7 \cdot 180\textdegree}{6} = 7 \cdot 30\textdegree = 210\textdegree$

$sen^2(135\textdegree) + cos^2(135\textdegree) = sen^2(45\textdegree) + (-1)^2 \cdot cos^2(45\textdegree) = \left(\dfrac{\sqrt{2}}{2}\right)^2 + \left(\dfrac{\sqrt{2}}{2}\right)^2 = \dfrac{2}{4}+ \dfrac{2}{4} = \dfrac{4}{4} = 1$

e:

$sen^2(210\textdegree) + cos^2(210\textdegree) = (-1)^2 \cdot sen^2(30\textdegree) + (-1)^2 \cdot cos^2(30\textdegree) = \left(\dfrac{1}{2}\right)^2 + \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{1}{4}+ \dfrac{3}{4} = \dfrac{4}{4} = 1$