I know the cosine function is even, but I don't understand why it works the way it does. Since this function is on the unit circle, and the rotation is reversed, wouldn't $\cos(\frac{-2\pi}{3})$ be on the $4^{th}$ quarter of the unit circle? Since $\cos(\theta)$ is equal to $\frac{\text{opposite}}{\text{hypotenuse}}$ or $\frac{x}{R}$ and $R$ on the unit circle is always equal to $1$, why isn't the answer positive?
Why is $\cos(-2\pi/3) =-1/2$
algebra-precalculustrigonometry
Best Answer
Note that $-2\pi /3 $ is in the third quadrant because you have to start at $0$ and go clockwise, $120$ degrees and you will end up in the third quadrant.
Thus the cosine is negative.
The other way to look at it is that the cosine function is even and $\cos ( -2\pi /3) = \cos ( 2\pi /3)= -1/2.$