Answer
$\lt \dfrac{-\sqrt 3}{2}, \dfrac{-1}{2}\gt$
Work Step by Step
Let the components of a vector $v$ be given as $v=\lt v_x,v_y\gt$
Now, $v_x=-1 \cos (\dfrac{\pi}{6})=\dfrac{-\sqrt 3}{2}$
and
$v_y=-1 \sin (\dfrac{\pi}{6})=\dfrac{-1}{2}$
Hence, $v=\lt \dfrac{-\sqrt 3}{2}, \dfrac{-1}{2}\gt$
