Question

Relações trigonométricas (50 pontos)

Answer 1

$\displaystyle \text{cos(x)}=\frac{4}{5} \ , \ 0<\text x<\frac{\pi}{2}$

Se o ângulo está no primeiro quadrante e temos o cosseno =4/5, podemos pensar no triângulo 3,4 e 5. Assim, temos :

$\displaystyle \text{cos(x)}=\frac{4}{5} \ , \ \text{sen(x)}=\frac{3}{5} \ , \ \text{tg(x)}=\frac{3}{4}$

A questão pede a relação :

$\displaystyle \text{sen}^2(\text x)-3.\text{sen(x)} \\\\ \underline{\text{substituindo o valor de sen(x)}} : } \\\\ (\frac{3}{5})^2-3.\frac{3}{5} \\\\\\ \frac{9}{25}-\frac{9}{5} \\\\\\ \frac{9}{25}-\frac{5.9}{25} \\\\\\ \frac{9-45}{25} \\\\\\ \huge\boxed{\frac{-36}{25}}\checkmark$

2ª forma de resolver :

$\displaystyle \text{cos(x)}=\frac{4}{5} \ , \ 0<\text x<\frac{\pi}{2}$

A questão pede :

$\displaystyle \text{sen}^2(\text x)-3.\text{sen(x)}$

Usando a relação fundamental da trigonometria :

$\displaystyle \text{sen}^2(\text x)+\text{cos}^2(\text x) = 1 \\\\ \text {sen}^2(\text x) = 1 - \text{cos}^2(\text x) \\\\ \text {sen(x)}=\pm\sqrt{1-\text{cos}^2(\text x)}$

no caso o intervalo de x é no primeiro quadrante, portanto :

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

Substituindo na expressão que queremos :

$\displaystyle \text{sen}^2(\text x)-3\text{sen}(\text x) \\\\\ 1-\text{cos}^2(\text x)-3\sqrt{1-\text{cos}^2(\text x)} \\\\\\ 1-(\frac{4}{5})^2 - 3\sqrt{1-(\frac{4}{5})^2} \\\\\\ 1-\frac{16}{25} - 3\sqrt{1-\frac{16}{25}} \\\\\\ \frac{25-16}{25}-3\sqrt{\frac{25-16}{25}} \\\\\\ \frac{9}{25} - 3\sqrt{\frac{9}{25}} \\\\\\ \frac{9}{25}-\frac{3.3}{5} \\\\\\ \frac{9-5.9}{25} \to \frac{9-45}{25}\\\\\\ \huge\boxed{\frac{-36}{25}}\checkmark$

Answer 2

A resposta do cara de cima está totalmente correta!...