If I put a star beside it, that means that I am using radians, otherwise, I am using degrees. The answer is probably $\cos(4*)<\cos(2*)<\cos(8*)<\cos(6*).$ First off, $\cos(2*) = \cos(360/(\pi)), \cos(4*) = \cos(720/(\pi)), \cos(6*) = \cos(1080/(\pi))$, and $\cos(8*) = \cos(1440/(\pi)).$ Setting $\pi = 3.14$, we can now estimate that $\cos(2*) \approx \cos(114), \cos(4*) \approx \cos(229), \cos(6*) \approx \cos(345)$, and $\cos(8*) \approx \cos(99).$ Therefore, $\cos(2*),\cos(4*),\cos(8*)<0$ and $\cos(6*)>0.$ This means that $\cos(6*)$ is the greatest. Also, since 99 is just over 90, $\cos(8*)$ is a large negative. Similarily, since 114 is close to 90, $\cos(4*)$ is a large negative, but smaller than $\cos(8*).$ Lastly, since $\cos(229) = \cos(131),$ we can conclude that $\cos(4*)<\cos(2*)<\cos(8*)<\cos(6*).$ Hope this helps! Please tell me if there are any errors in my solution.

Com10atorics wrote: Arrange the numbers $\cos 2, \cos 4, \cos 6$ and $\cos 8$ from the biggest to the smallest where $2,4,6,8$ are given as radians. $\cos x$ is decreasing over $(0,\pi)$ We have $0<2\pi-6<8-2\pi<2<2\pi-4<\pi$ So $\cos (2\pi -6)>\cos(8-2\pi)>\cos(2)>\cos (2\pi-4)$ Which is $\boxed{\cos 6>\cos 8>\cos 2>\cos 4}$