Evaluate cos 240 degrees exactly.

Step-by-step solution

1. 1

   Locate the angle and fix the sign. $240^\circ$ lies between $180^\circ$ and $270^\circ$, so it is in the third quadrant. There both $x$ and $y$ coordinates on the unit circle are negative, and since cosine is the $x$-coordinate, $\cos240^\circ$ must be negative. Getting the sign from the quadrant before computing any magnitude prevents the commonest error.

2. 2

   Find the reference angle. Measuring from the nearest horizontal axis, $180^\circ$:

   $240^\circ-180^\circ=60^\circ$

   So the reference angle is $60^\circ$, and the magnitude of $\cos240^\circ$ equals that of $\cos60^\circ$.

3. 3

   Use the third-quadrant identity. For an angle written as $180^\circ+\theta$,

   $\cos\!\left(180^\circ+\theta\right)=-\cos\theta$

   which for $\theta=60^\circ$ gives

   $\cos240^\circ=-\cos60^\circ$

4. 4

   Substitute the exact special-angle value. From the $30$–$60$–$90$ triangle, $\cos60^\circ=\tfrac12$, so

   $\cos240^\circ=-\frac{1}{2}$

5. 5

   Verify and cross-check. Numerically $\cos240^\circ=-0.5$ to within $10^{-15}$ ✓. Consistency check with the Pythagorean identity: $\sin240^\circ=-\tfrac{\sqrt3}{2}$, and $\left(-\tfrac12\right)^{2}+\left(-\tfrac{\sqrt3}{2}\right)^{2}=\tfrac14+\tfrac34=1$ ✓. Note $240^\circ$ corresponds to $\tfrac{4\pi}{3}$ radians, giving the unit-circle point $\left(-\tfrac12,-\tfrac{\sqrt3}{2}\right)$.