Compute $\cos 240^\circ$.
Let $P$ be the point on the unit circle that is $240^\circ$ counterclockwise from $(1,0)$, and let $D$ be the foot of the altitude from $P$ to the $x$-axis, as shown below.

[asy]

pair A,C,P,O,D;

draw((0,-1.2)--(0,1.2),p=black+1.2bp,Arrows(0.15cm));

draw((-1.2,0)--(1.2,0),p=black+1.2bp,Arrows(0.15cm));

A = (1,0);

O= (0,0);

label("$x$",(1.2,0),SE);

label("$y$",(0,1.2),NE);

P = rotate(240)*A;

D = foot(P,A,-A);

draw(O--P--D);

draw(rightanglemark(O,D,P,2));

draw(Circle(O,1));

label("$O$",O,SE);

label("$P$",P,SW);

//label("$A$",A,SE);

label("$D$",D,N);

[/asy]

Triangle $POD$ is a 30-60-90 triangle, so $DO = \frac{1}{2}$ and $DP = \frac{\sqrt{3}}{2}$.  Therefore, the coordinates of $P$ are $\left(-\frac12,-\frac{\sqrt{3}}{2}\right)$, so $\cos 240^\circ = \boxed{-\frac{1}{2}}$.
-\frac{1}{2}
Level 3
Geometry
Problem: Compute $\cos 240^\circ$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To compute $\cos 240^\circ$, first recognize that 240 degrees is in the third quadrant of the unit circle. In this quadrant, the cosine value is negative. 

Next, determine the reference angle for 240 degrees. The reference angle is found by subtracting 180 degrees from 240 degrees, which gives us 240° - 180° = 60°. 

Now, recall the cosine value of the reference angle. We know that $\cos 60^\circ = \frac{1}{2}$. 

Since cosine is negative in the third quadrant, we take the negative of the reference angle's cosine value. Therefore, $\cos 240^\circ = -\cos 60^\circ = -\frac{1}{2}$. 

Thus, the final answer is $-\frac{1}{2}$.<|END_OF_LATENT|> The final answer is $-\frac{1}{2}$.