Question

Se sen x + cos x = 1/3 , calcule sen 2x.

Answer 1

Olá Eduarda!

Resposta:

$\boxed{\mathtt{- 8/9}}$

Explicação passo-a-passo:

$\\ \displaystyle \mathsf{\sin x + \cos x = \frac{1}{3}} \\\\ \mathsf{\left ( \sin x + \cos x \right )^2 = \left ( \frac{1}{3} \right )^2} \\\\ \mathsf{\sin^2 x + 2 \cdot \sin x \cdot \cos x + \cos^2 x = \frac{1}{9}} \\\\ \mathsf{\underbrace{\mathsf{\left ( \sin^2 x + \cos^2 x \right )}}_{1} + \sin (x + x) = \frac{1}{9}} \\\\ \mathsf{\sin \left ( 2x \right ) = \frac{1}{9} - 1} \\\\ \boxed{\boxed{\mathsf{\sin (2x) = - \frac{8}{9}}}}$

Obs.:

$\\ \displaystyle \mathtt{\bullet \qquad \sin (a + b) = \sin a \cdot \cos b + \sin b \cdot \cos a} \\\\ \mathtt{\bullet \qquad \boxed{\mathtt{\sin (a + a)}} = \sin a \cdot \cos a + \sin a \cdot \cos a = \boxed{\mathtt{2 \cdot \sin a \cdot \cos a}}}$