Question

Considerando que cos (x) = √2/2 e x é um arco do 3º quadrante, então o valor de sen 3x é:

Answer 1

Resposta:

**sen(a+b)=sen(a)*cos(b)+sen(b)*cos(b)

***cos(a+b)=cos(a)*cos(b)-sen(a)*sen(b)

***cos(x)=-√2/2

x é do 3ª Quadrante ==> cos(x) < 0  e  sen(x) <0

sen²(x)+cos²(x)=1

sen²(x)+[-√2/2]² =1

sen²(x)= 1-2/4   ==>sen²(x) = 1/2  ==> sen(x)=±√(1/2) =√2/2 , como é do 3ª Quadrante ==> sen(x)=-√2/2

sen (x+2x) =sen(x)*cos(2x)+sen(2x)*cos(x)

sen(3x)=sen(x)*(cos²(x)-sen²(x)) +cos(x)*2*sen(x)*cos(x)

sen(3x)=sen(x)*(cos²(x)-sen²(x)) +cos(x)*2*sen(x)*cos(x)

sen(3x)=sen(x)*(2cos²(x) -1) +2*sen(x)*cos²(x)

sen(3x)=-√2/2 *(2*(-√2/2)²-1) +2*(-√2/2)*(-√2/2)²

sen(3x)=-√2/2 *(2*2/4-1) +2*(-√2/2)*2/4

sen(3x)=-√2/2 *(1-1) +2*(-√2/2)*2/4

sen(3x)=-√2/2 * 0 +2*(-√2/2)*2/4 =-√2/2

Letra D

Answer 2

Resposta:

$\boxed{\mathtt{D}}$

Explicação passo-a-passo:

Sabemos que:

$\displaystyle \boxed{\mathtt{\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}}}$

Porém, $\displaystyle \mathtt{x \in \left ] \pi, \frac{3\pi}{2} \right [}$.

Daí,

$\\ \displaystyle \mathsf{x = \frac{\pi}{4} + \pi} \\\\ \boxed{\mathsf{x = \frac{5\pi}{4}}}$

Por fim,

$\\ \displaystyle \mathsf{\sin (3x) = \sin \left ( 3 \cdot \frac{5\pi}{4} \right )} \\\\\\ \mathsf{\qquad \, \quad = \sin \frac{15\pi}{4}} \\\\\\ \mathsf{\qquad \ \quad = \sin \left ( \frac{8\pi}{4} + \frac{7\pi}{4} \right )} \\\\\\ \mathsf{\qquad \, \quad = \sin \frac{7\pi}{4}} \\\\\\ \mathsf{\qquad \, \quad = \sin \left ( \frac{4\pi}{4} + \frac{3\pi}{4} \right )}$

$\\ \displaystyle \mathsf{\qquad \, \quad = - \sin \frac{3\pi}{4}} \\\\\\ \boxed{\boxed{\mathsf{\sin (3x) = - \frac{\sqrt{2}}{2}}}}$

Resolução II:

Sabemos que:

$\displaystyle \boxed{\mathtt{\sin x = - \frac{\sqrt{2}}{2}}}$

Então,

$\\ \displaystyle \mathsf{\sin (3x) = \sin (2x + x)} \\\\ \mathsf{\qquad \, \quad = \sin (2x) \cdot \cos x + \sin x \cdot \cos (2x)} \\\\ \mathsf{\qquad \, \quad = \left ( 2 \cdot \sin x \cdot \cos x \right ) \cdot \cos x + \sin x \cdot \left ( \cos^2 x - \sin^2 x \right )} \\\\\\ \mathsf{\qquad \, \quad = 2 \cdot \sin x \cdot \cos^2 x + \sin x \cdot \left ( \cos x + \sin x \right ) \cdot \left (\cos x - \sin x \right )} \\\\\\ \mathsf{\qquad \, \quad = 2 \cdot \left ( - \frac{\sqrt{2}}{2} \right ) \cdot \left ( - \frac{\sqrt{2}}{2} \right )^2 + \left ( - \frac{\sqrt{2}}{2} \right ) \cdot \left ( - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} \right ) \cdot \underbrace{\mathsf{\left ( - \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \right )}}_{zero}}$

$\\ \displaystyle \mathsf{\qquad \, \quad = 2 \cdot \frac{- \sqrt{2}}{2} \cdot \frac{2}{4} + 0} \\\\\\ \mathsf{\qquad \, \quad = - \frac{4\sqrt{2}}{8}} \\\\ \boxed{\boxed{\mathsf{\sin (3x) = - \frac{\sqrt{2}}{2}}}}$