Question

Questão está na imagem!

Answer 1

Resposta:

2

Explicação passo-a-passo:

$lembrando\ que\ sin(a+2k\pi)=sin(a) \\\frac{7\pi}{2} = \frac{3\pi}{2} + \frac{4\pi}{2} = \frac{3\pi}{2} + 2\pi\\sin(\frac{7\pi}{2}) = sin(\frac{3\pi}{2} ) = -1\\\\lembrando\ que\ tg(a+\pi) = tg(a)\\ \frac{5\pi}{4} = \frac{\pi}{4} + \frac{4\pi}{4} =\frac{\pi}{4} + \pi\\tg(\frac{5\pi}{4}) = tg(\frac{\pi}{4}) = 1$

$lembrando\ que\ sin(a+\pi)=-sin(a)\ e\ que\ sin(-a)=-sin(a)\\\frac{5\pi}{6} = \frac{6\pi}{6} - \frac{\pi}{6} = \pi - \frac{\pi}{6}\\sin(\pi - \frac{\pi}{6}) = -sin( - \frac{\pi}{6}) = sin(\frac{\pi}{6}) = \frac{1}{2} \\\\lembrando\ que\ cos(a+\pi) =-cos(a)\ e\ que\ cos(-a)=cos(a)\\\frac{4\pi}{3} = \frac{3\pi}{3} + \frac{\pi}{3} = \pi + \frac{\pi}{3}\\cos( \pi + \frac{\pi}{3}) = -cos(\frac{\pi}{3}) = -\frac{1}{2} \\\\Juntando\ tudo\ temos:\\\frac{(-1)^2 + 1}{\frac{1}{2} - (-\frac{1}{2}) } = 2$