Recall that [tex]\sin2x=2\sin x\cos x[/tex], so that
[tex]\sin\left(2\arccos\left(-\dfrac8{17}\right)\right)=2\sin\left(\arccos\left(-\dfrac8{17}\right)\right)\cos\left(\arccos\left(-\dfrac8{17}\right)\right)[/tex]
Dealing with the cosine terms is simple, since [tex]\cos(\arccos x)=x[/tex] (so long as [tex]0\le \arccos x\le\pi[/tex], which is true here). So
[tex]\cos\left(\arccos\left(-\dfrac8{17}\right)\right)=-\dfrac8{17}[/tex]
To find the sine, we can make a substitution of [tex]\theta=\arccos\left(-\dfrac8{17}\right)[/tex]. In other words, we suppose that [tex]\theta[/tex] is some angle that satisfies [tex]\cos\theta=-\dfrac8{17}[/tex].
If this is the case, then we have
[tex]\sin^2\theta=1-\cos^2\theta=\dfrac{225}{289}\implies\sin\theta=\pm\sqrt{\dfrac{225}{289}}=\pm\dfrac{15}{17}[/tex]
But which value do we choose? We know that [tex]\cos\theta[/tex] is negative, which means [tex]\theta[/tex] must be between [tex]\dfrac\pi2[/tex] and [tex]\pi[/tex]. The sine of any angle [tex]\theta[/tex] in this interval will always be positive, so in fact
[tex]\sin\theta=\sin\left(\arccos\left(-\dfrac8{17}\right)\right)=\dfrac{15}{17}[/tex]
Putting everything together, we get
[tex]\sin\left(2\arccos\left(-\dfrac8{17}\right)\right)=2\left(\dfrac{15}{17}\right)\left(-\dfrac8{17}\right)=-\dfrac{240}{289}[/tex]
