Answer
$\cos 2$ is negative
Work Step by Step
The angles (in radians) representing borders between quadrants are
$[0]$ ... I ... $[\displaystyle \frac{\pi}{2}\approx 1.57]$
$[\displaystyle \frac{\pi}{2}\approx 1.57]$ ... II ... $[\pi\approx 3.14]$
$[\pi\approx 3.14]$ ... III ...$[\displaystyle \frac{3\pi}{2}\approx 4.71]$
$[\displaystyle \frac{3\pi}{2}\approx 4.71]$... IV... $2\pi\approx 6.28$
( when angles are positive, that is,
counted counterclockwise from the point (1,0) )
Negative (clockwise from the point (1,0)):
$[0]$ ... IV ... $[-\displaystyle \frac{\pi}{2}\approx-1.57]$
$[-\displaystyle \frac{\pi}{2}\approx-1.57]$ ... III ... $[-\pi\approx-3.14]$
$[-\pi\approx-3.14]$ ... II ...$[-\displaystyle \frac{3\pi}{2}\approx-4.71]$
$[-\displaystyle \frac{3\pi}{2}\approx 4.71]$... I... $[-2\pi\approx-6.28]$
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For any real number $s$ represented by a directed arc on the unit circle,
$\sin s=y\quad \cos s=x \quad \displaystyle \tan s=\frac{y}{x} (x\neq 0)$
$\displaystyle \csc s=\frac{1}{y} (y\displaystyle \neq 0)\\\sec s=\frac{1}{x} (x\neq 0) \displaystyle \\\cot s=\frac{x}{y} (y\neq 0)$.
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A radian measure of 2
represents an angle in quadrant II,
$1.57 < 2 < 3.14$
where the x coordinates are negative,
so
$\cos 2$ is negative.
