How do I prove that $2\cos (2\theta + {\pi \over 3}) \equiv - 2\sin(2\theta - {\pi \over 6})$ Using the identity $\cos (\theta  + {\pi  \over 2}) \equiv  - \sin\theta $

Using the identity $\quad\cos\left(\theta + \frac{\pi}{2}\right)\equiv−\sin\theta$, 
Let $\;\color{green}{\bf x = 2\theta - \dfrac{\pi}{6}} \implies\left(\implies \color{blue}{\bf x + \dfrac{\pi}{2}}\; = \;\left(2\theta - \dfrac{\pi}{6}\right) + \dfrac{\pi}{2} \;= \;\color{blue}{\bf 2\theta + \dfrac{\pi}{3}}\right)$
$$2\cos \left(\color{blue}{\bf 2\theta + {\pi \over 3}}\right)\; = \;2\cos\left(\color{blue}{\bf x + \dfrac{\pi}{2}}\right) \;\equiv\; - 2\sin(\color{green}{\bf x})\; = \;-2\sin\left(\color{green}{\bf 2\theta - \dfrac{\pi}{6}}\right)$$
