Trigno-sides

For $135^\circ < x < 180^\circ$ , we have $\cot x < -1 < \cos x, \tan x < 0 < \sin x$ and since all the points lie on the graph of $y=x^2,$ which is a convex function, the line $\overline{QR}$ must be parallel with the line $\overline{PS}.$ That is, the slopes must equal: $\cot x + \sin x = \frac{\cot^2x - \sin^2x}{\cot x - \sin x} = \frac{\tan^2x - \cos^2x}{\tan x - \cos x} = \tan x + \cos x.$ Multiplying the equation by $\sin x \cos x$ to clear denominators and write everything in terms of sines and cosines, we have $cos^2x + \sin^2x\cos x = \sin^2x + \sin x\cos^2x \\ \implies (\cos x + \sin x)(\cos x - \sin x) = \cos^2x-\sin^2x = \sin x\cos^2x - \sin^2x\cos x = \sin x\cos x(\cos x - \sin x) \\ \implies \cos x + \sin x = \sin x\cos x = \frac{1}{2}\sin(2x)$ where the last implication is due to the fact $\cos x \neq \sin x$ for the domain given in the problem. Then by squaring both sides, $1 + \sin(2x) = \cos^2x + 2\sin x\cos x + \sin^2x = \frac{1}{4}\sin^2(2x) \\ \implies \sin^2(2x) - 4\sin(2x) - 4 = 0 \\ \implies \sin(2x) = 2 \pm 2\sqrt{2}$ Now just note that $270^\circ < 2x < 360^\circ$ implies $\sin(2x) < 0$ , so $\sin(2x) = \boxed{2 - 2\sqrt{2}}$