Question

30 PONTOS!
Questão 45 do vestibular CEDERJ 2018.2. (A questão está no anexo também)

O valor de Cos (135/180). π + Cos (45/180). π + sen (225/180) π, é igual à:

(As alternativas estão no anexo, sei que é a "D" contudo quero o calculo)

Answer 1

Explicação passo-a-passo:

I)

$\sf cos~\Big(\dfrac{135\pi}{180}\Big)=cos~\Big(\dfrac{135\pi\div45}{180\div45}\Big)$

$\sf cos~\Big(\dfrac{135\pi}{180}\Big)=cos~\Big(\dfrac{3\pi}{4}\Big)$

$\sf cos~\Big(\dfrac{135\pi}{180}\Big)=cos~\Big(\dfrac{3\cdot180^{\circ}}{4}\Big)$

$\sf cos~\Big(\dfrac{135\pi}{180}\Big)=cos~\Big(\dfrac{540^{\circ}}{4}\Big)$

$\sf cos~\Big(\dfrac{135\pi}{180}\Big)=cos~135^{\circ}$

$\sf cos~\Big(\dfrac{135\pi}{180}\Big)=-cos~(180^{\circ}-135^{\circ})$

$\sf cos~\Big(\dfrac{135\pi}{180}\Big)=-cos~45^{\circ}$

$\sf cos~\Big(\dfrac{135\pi}{180}\Big)=-\dfrac{\sqrt{2}}{2}$

II)

$\sf cos~\Big(\dfrac{45\pi}{180}\Big)=cos~\Big(\dfrac{45\pi\div45}{180\div45}\Big)$

$\sf cos~\Big(\dfrac{45\pi}{180}\Big)=cos~\Big(\dfrac{\pi}{4}\Big)$

$\sf cos~\Big(\dfrac{45\pi}{180}\Big)=cos~\Big(\dfrac{180^{\circ}}{4}\Big)$

$\sf cos~\Big(\dfrac{45\pi}{180}\Big)=cos~45^{\circ}$

$\sf cos~\Big(\dfrac{45\pi}{180}\Big)=\dfrac{\sqrt{2}}{2}$

III)

$\sf sen~\Big(\dfrac{225\pi}{180}\Big)=cos~\Big(\dfrac{225\pi\div45}{180\div45}\Big)$

$\sf sen~\Big(\dfrac{225\pi}{180}\Big)=sen~\Big(\dfrac{5\pi}{4}\Big)$

$\sf sen~\Big(\dfrac{225\pi}{180}\Big)=sen~\Big(\dfrac{5\cdot180^{\circ}}{4}\Big)$

$\sf sen~\Big(\dfrac{225\pi}{180}\Big)=sen~\Big(\dfrac{900^{\circ}}{4}\Big)$

$\sf sen~\Big(\dfrac{225\pi}{180}\Big)=sen~225^{\circ}$

$\sf sen~\Big(\dfrac{225\pi}{180}\Big)=-sen~(225^{\circ}-180^{\circ})$

$\sf sen~\Big(\dfrac{225\pi}{180}\Big)=-sen~45^{\circ}$

$\sf sen~\Big(\dfrac{225\pi}{180}\Big)=-\dfrac{\sqrt{2}}{2}$

Logo:

$\sf cos~\Big(\dfrac{135\pi}{180}\Big)+cos~\Big(\dfrac{45\pi}{180}\Big)+sen~\Big(\dfrac{225\pi}{180}\Big)$

$\sf =-\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{2}}{2}$

$\sf =0-\dfrac{\sqrt{2}}{2}$

$\sf =\red{-\dfrac{\sqrt{2}}{2}}$

Letra D