Question

Questão de Trigonometria: 2º EM

Answer 1

$cos\frac{13\pi}{6} = cos390 = cos 30 = \frac{\sqrt{3}}{2} \\\\cos\frac{11\pi}{4} = cos495 = cos 135 = -cos45 = \frac{-\sqrt{2}}{2} \\\\\\sen\frac{-7\pi}{6} = -sen210 = sen30 = \frac{{1}}{2} \\\\tg\frac{31\pi }{3} = tg1860 = tg60 = \frac{sen60}{cos60} = \frac{\frac{\sqrt{3}}{2} }{\frac{1}{2} } = \sqrt{3} \\\\\\6(\frac{\sqrt{3} }{2} )^{2} - 4(\frac{-\sqrt{2}}{2})^{2} + \frac{1}{2} +( \sqrt{3})^{2}\\\\\frac{9}{2} - 2 + \frac{1}{2} + 3 = 6$

Resposta: 6