Why is $\cos(-2\pi/3) =-1/2$?

Question

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?

Answer 1

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 $\cos ( -2\pi /3) = \cos ( 2\pi /3)= -1/2.$

Answer 2

The convention I was taught that the quadrants start with "northeast" as 1, "northwest" as 2, "southwest" as 3, and "southeast" as 4. See the diagram on this web page.

I think of $cos$ as representing increments along the X axis, and $sin$ representing increments along the Y axis. To that end, $cos$ is positive in quadrants 1 and 4, and negative in 2 and 3. $Sin$ is positive in Q1 and 2, negative in Q3 and 4.

For your specific question, a negative angle is measured from the positive X axis; so $-2\pi/3$ starts at a positive X point, and swings clockwise around backwards into Q3.

Answer 3

$$cos(\frac{-2π}{3}) = cos(\frac{2π}{3}) = cos(π-\frac{π}{3})=-cos(\frac{π}{3})=-\frac{1}{2}$$

Hope it helps.

You are using old trigonometry $\frac{opposite}{hypotenuse}$ to explain a more generalised result (without using any sign conventions), which is impossible. You have to use angle sum identity for $cos(π-\frac{π}{3})=cos(π)cos(\frac{π}{3})+sin(π)sin(\frac{π}{3})=-\frac{1}{2}$ $$OR$$

On a cartesian plane if you draw an unit circle, the angle $-\frac{2π}{3}$ will be in 3rd quadrant, the $opposite$ side to the angle will be $negative$ and $hypotenuse$ will be $positive$ (Note: $hypotenuse=√(base^2+opposite^2)\ so\ always\ positive$) therefore $$cos(-\frac{2π}{3})=-\frac{opposite}{hypotenuse}=-\frac{1}{2}$$ Visit this link to see figure https://i.stack.imgur.com/1xsEe.jpg