Suppose that $\cos a=-\frac{8}{17}$ for $\pi < a < 3\pi/2$ find $\sin 2a$ Suppose that $\cos A= -\frac{8}{17}$ for $\pi < A < 3\pi/2$ find $\sin 2A$. how is this question solved and in what steps ?

Tools
$$ 
\sin 2\theta = 2\cos\theta \sin \theta
\tag{1}
$$
$$
 \sin \theta = \pm \sqrt{ 1 - \cos^{2} \theta }
$$

Data
Let the angle
$$
 \theta = \arccos \left( -\frac{8}{17} \right)
$$
then the sine is
$$
\sin \theta = -\frac{15}{17}
$$
Solution
$$
 \sin 2\theta = 2 \cos\theta \sin \theta = 
2 
\left( -\frac{8}{17}\right)
\left( -\frac{15}{17}\right)
=
\frac{240}{289}
$$

