Great Wall Of Cosine (The answer's Over 9000)!

This is a dirt easy question when you know what it is actually asking.

When $|x|<1$ $\lim _{ n\rightarrow \infty }{ { x }^{ n } } =0$ . When $x=1$ , $\lim _{ n\rightarrow \infty }{ { x }^{ n } } =1$

Applying that to $y=\lim _{ n\rightarrow \infty }{ { \cos(x) }^{ n } }$ and graphing it, you'll get a beautiful horizontal line with $y=0$ all the way except at points $x=2a\pi$ where $y=-1$ and at points $x=\pi(1+2a)$ where $y=1$ , where $a$ is an integer.This is assuming $n$ is an odd number even as it approaches $\infty$ , which is what the question wants.

This makes the question a whole lot easier. It becomes obvious that $A=1$ and this occurs when the value of $x$ chosen is $2a\pi \le x < \pi(1+2a)$ . Now, I'll convert the use of radians into degrees.

Since $0° \le x \le 360°$ , the possible values that $x$ can be for $f(x)=1$ is $0° \le x < 180°$ and $x=360°$ . And since $x$ is an integer in degrees, $B=\left( \sum _{ k=0 }^{ 179 }{ k } \right) +360=16110+360=16470$ Therefore $100\left\lfloor A+B \right\rfloor = 100\left\lfloor 16470 + 1 \right\rfloor = \boxed{1647100}$