Look at the picture.
[tex]r=\sqrt{x^2+y^2}-\text{from the Pythagorean theorem}\\\\\sin\theta=\dfrac{y}{r}\\\\\cos\theta=\dfrac{x}{r}\\\\\tan\theta=\dfrac{y}{x}\\\\\cot\theta=\dfrac{x}{y}[/tex]
We have the point
[tex](4,\ -8)\to x=4,\ y=-8[/tex]
Calculate r:
[tex]r=\sqrt{4^2+(-8)^2}=\sqrt{16+64}=\sqrt{80}=\sqrt{16\cdot5}=\sqrt{16}\cdot\sqrt5=4\sqrt5[/tex]
Substitute:
[tex]\cos\theta=\dfrac{4}{4\sqrt5}=\dfrac{1}{\sqrt5}\to\boxed{C)}[/tex]
