All-in Trig

Using the derivative rules for each of the trig functions, we can find the derivative of f(x): $\begin{aligned}f'(x)&=\cos x-\sin x+\sec^2x-\csc^2x+\sec x\tan x-\csc x\tan x\\ &=\cos x-\sin x+\frac{1}{\cos^2x}-\frac{1}{\sin^2x}+\frac{\sin x}{\cos^2x}-\frac{\cos x}{\sin^2x}\\ &=\cos x-\sin x+\frac{1+\sin x}{1-\sin^2x}-\frac{1+\cos x}{1-\cos^2x}\\ &=\cos x-\sin x+\frac{1}{1-\sin x}-\frac{1}{1-\cos x}\\ &=\cos x-\sin x+\frac{\sin x-\cos x}{(1-\sin x)(1-\cos x)}\end{aligned}$ Now we let f'(x)=0 to find the turning points: $\cos x-\sin x+\frac{\sin x-\cos x}{(1-\sin x)(1-\cos x)}=0\\ -(\sin x-\cos x)(1-\sin x)(1-\cos x)+(\sin x-\cos x)=0\\ (\sin x-\cos x)[(1-\sin x)(1-\cos x)-1]=0\\ (\sin x-\cos x)(\sin x\cos x-\sin x-\cos x)=0\\ (\sin x-\cos x)=0\qquad\text{and}\qquad(\sin x\cos x-\sin x-\cos x)=0$ The equation on the left is easy; solving with auxiliary angle method gives us $\sqrt2\sin(x-\frac{\pi}{4})=0$ . Since all the trig functions in the original equation have a period of either $\pi$ or 2 $\pi$ , the whole function has a period of 2 $\pi$ , so we only need to consider x values between 0 and 2 $\pi$ , and so the two x values we get are $x=\frac{\pi}{4},\frac{5\pi}{4}$ . Plugging these into our original equation to get the y values gives us the stationary points $(\frac{\pi}{4},2+3\sqrt2),(\frac{5\pi}{4},2-3\sqrt2)$ .

The equation on the right is more complicated, and requires more algebra: $\sin x\cos x-\sin x-\cos x=0\\ 2\sin x\cos x+1-1-2\sin x-2\cos x=0\\ 2\sin x\cos x+\sin^2x+\cos^2x-1-2(\sin x+\cos x)=0\\ (\sin x+\cos x)^2-2(\sin x+\cos x)-1=0\\ \sin x+\cos x=\frac{2\pm\sqrt{2^2+4}}{2}\\ \sin x+\cos x=1\pm\sqrt2$ However, since $\sin x+\cos x=\sqrt2\sin(x+\frac{\pi}{4}),\sin x+\cos x$ cannot be greater than $\sqrt2$ , so the only possible answer is $\sin x+\cos x=1-\sqrt2$ . This gives us two x values between 0 and 2 $\pi$ : $\sqrt2\sin(x+\frac{\pi}{4})=1-\sqrt2\\ x+\frac{\pi}{4}=\pi-\sin^{-1}\left(\frac{1}{\sqrt2}-1\right),\qquad2\pi+\sin^{-1}\left(\frac{1}{\sqrt2}-1\right)\\ x=\frac{3\pi}{4}-\sin^{-1}\left(\frac{1}{\sqrt2}-1\right),\qquad\frac{7\pi}{4}+\sin^{-1}\left(\frac{1}{\sqrt2}-1\right)$ This would be complicated to solve for y, but I can rearrange the original equation to make the substitution much simpler: $\begin{aligned}f(x)&=\sin x+\cos x+\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}+\frac{1}{\cos x}+\frac{1}{\sin x}\\ &=\sin x+\cos x+\frac{\sin^2x+\sin x+\cos^2x+\cos x}{\sin x\cos x}\\ &=\sin x+\cos x+\frac{2(\sin^2x+\cos^2x+\sin x+\cos x)}{2\sin x\cos x+1-1}\\ &=\sin x+\cos x+\frac{2(1+\sin x+\cos x)}{2\sin x\cos x+\sin^2x+\cos^2x-1}\\ &=\sin x+\cos x+\frac{2(1+\sin x+\cos x)}{(\sin x+\cos x)^2-1}\\ &=\sin x+\cos x+\frac{2(1+\sin x+\cos x)}{(\sin x+\cos x+1)(\sin x+\cos x-1)}\\ &=\sin x+\cos x+\frac{2}{\sin x+\cos x-1}\end{aligned}$ Now when we substitute in to get our y value, we can just put the value we got for $\sin x+\cos x$ straight into this equation. $y=1-\sqrt2+\frac{2}{1-\sqrt2+1}=1-2\sqrt2$ So the turning points for $0\leq x<2\pi$ are: $\color{#D61F06}(\frac{\pi}{4},2+3\sqrt2),\color{#3D99F6}(\frac{5\pi}{4},2-3\sqrt2),\color{#624F41}(\frac{3\pi}{4}-\sin^{-1}(\frac{1}{\sqrt2}-1),1-2\sqrt2),\color{#20A900}(\frac{7\pi}{4}+\sin^{-1}(\frac{1}{\sqrt2}-1),1-2\sqrt2)$ .

We could go through and find whether these points are local maximums, local minimums or horizontal points of inflection using the first derivative test or second derivative text, but instead I'm going to find out some other information about the graph of y=f(x) to deduce these properties. Let's see where the graph cuts the x-axis by letting x=0: $\sin x+\cos x+\frac{2}{\sin x+\cos x-1}=0\\ (\sin x+\cos x)^2-(\sin x+\cos x)+2=0\\ \sin x+\cos x=\frac{1+\sqrt{-7}}{2}$ As $\sin x+\cos x$ cannot be imaginary, we can see that this graph cannot cut the x-axis. We can also tell that the graph has an asymptote where the denominator equals zero, i.e. at the following x values: $\sin x+\cos x-1=0\\ \sin(x+\frac{\pi}{4})=\frac{1}{\sqrt2}\\ x+\frac{\pi}{4}=\frac{\pi}{4},\frac{3\pi}{4}\\ x=0, \frac{\pi}{2}$ Now we know the locations of all asymptotes and turning points, and that the graph never cuts the x-axis. We can put this all together logically in one graph, most easily by drawing on paper and avoiding creating new turning points or cutting the x-axis. It should end up looking something like this: We can see from the graph that smallest possible positive value of f(x) is at the point $\color{#D61F06}(\frac{\pi}{4},2+3\sqrt2)$ and the largest possible negative value of f(x) is at the points $\color{#624F41}(\frac{3\pi}{4}-\sin^{-1}(\frac{1}{\sqrt2}-1,1-2\sqrt2)$ and $\color{#20A900}(\frac{7\pi}{4}+\sin^{-1}(\frac{1}{\sqrt2}-1),1-2\sqrt2)$ . Therefore $P=2+3\sqrt2,N=1-2\sqrt2$ .

So the answer is $\lfloor1000(2+3\sqrt2+1-2\sqrt2)\rfloor=\lfloor1000(3+\sqrt2)\rfloor=\boxed{4414}$