[tex]\sin^2\theta+\cos^2\theta=1\implies \cos\theta=\pm\sqrt{1-\sin^2\theta}[/tex]
Since [tex]\sin\theta=-\dfrac8{17}[/tex], it follows that
[tex]\cos\theta=\pm\sqrt{1-\dfrac{64}{289}}=\pm\sqrt{\dfrac{225}{289}}=\pm\dfrac{15}{17}[/tex]
where the sign depends on the location of the terminal side of the angle. Since the terminal side lies in the fourth quadrant, that means the cosine of the angle is positive, so
[tex]\cos\theta=\dfrac{15}{17}[/tex]
