Question

ALGUEM ME AJUDA A RESOLVER ISSO 2.cos 5pi/6 - raiz de 3. Sen 11pi/6

Answer 1

Explicação passo-a-passo:

Temos que

$\cos( \frac{5\pi}{6} ) = \cos( \frac{5. {180}^{0} }{6} ) = \cos(5. {30}^{0} ) = \cos {150}^{0} = - \cos( {180}^{0} - {150}^{0} ) = - \cos( {30}^{0} ) = - \frac{ \sqrt{3} }{2} \: (1)$

$\sin( \frac{11\pi}{6} ) = \sin( \frac{11. {180}^{0} }{6} ) = \sin(11. {30}^{0} ) = \sin( {330}^{0} ) = - \sin( {360}^{0} - {330}^{0} ) = - \sin( {30}^{0} ) = - \frac{1}{2} \: \: (2)$

Assim,

$2. \cos( \frac{5\pi}{6} ) - \sqrt{3} . \sin( \frac{11\pi}{6} ) = > 2.( - \frac{ \sqrt{3} }{2} ) - \sqrt{3} .( - \frac{1}{2} ) = - \sqrt{3} + \frac{ \sqrt{3} }{2} = \frac{ - 2 \sqrt{3} + \sqrt{3} }{2} = - \frac{ \sqrt{3} }{2}$