Question

A = $\frac{7 cos (5pi-x) - 3 cos (3pi+x)}{8 sen( \frac{pi}{2-x}) }$
RESPOSTA: 2A+1 = 0

Answer 1

Propriedades utilizadas:

$\begin{array}{rclc} \cos\left(-\alpha \right )&=&\cos\left(\alpha \right)&\;\;\;\text{(i)}\\ \\ \cos\left(x\pm \pi \right )&=&-\cos\left(\alpha \right )&\;\;\;\text{(ii)}\\ \\ \cos\left(\alpha+k\cdot2\pi \right )&=&\cos\left(\alpha \right )&\;\;\;\text{(iii)}\\ \\ \mathrm{sen}\left(\dfrac{\pi}{2}-\alpha \right)&=&\cos\left(\alpha \right )&\;\;\;\text{(iv)} \end{array}$

(na relaçao $\text{(iii)}$, $k$ é um número inteiro)


$A=\dfrac{7\cos \left(5\pi-x \right )-3\cos\left(\pi+x\right)}{8\mathrm{\,sen}\left(\dfrac{\pi}{2}-x \right )}\\ \\ \\ A=\dfrac{7\cos \left(x-5\pi \right )+3\cos\left(x\right)}{8\cos\left(x \right )}\\ \\ \\ A=\dfrac{7\cos \left(x-\pi-2\cdot 2\pi \right )+3\cos\left(x\right)}{8\cos\left(x \right )}\\ \\ \\ A=\dfrac{7\cos \left(x-\pi \right )+3\cos\left(x\right)}{8\cos\left(x \right )}\\ \\ \\ A=\dfrac{-7\cos \left(x\right )+3\cos\left(x\right)}{8\cos\left(x \right )}\\ \\ \\ A=\dfrac{\cos\left(x \right )\cdot \left(-7+3 \right )}{\cos\left(x \right ) \cdot 8}\\ \\ \\ A=\dfrac{-4}{8}\\ \\ \\ A=-\dfrac{1}{2}\\ \\ \\ 2A=-1\\ \\ \\ \boxed{2A+1=0}$