# How do you find the exact value of cos^-1 (-1/2)?

Dec 30, 2016

Use unit circle and special triangles. See below.

#### Explanation:

We have:

${\cos}^{-} 1 \left(- \frac{1}{2}\right) = x$

Rearrange:

$- \frac{1}{2} = \cos \left(x\right)$

Now we ask, which values of $x$ would produce $- \frac{1}{2}$ by the cosine function? This is possible because $- \frac{1}{2}$ is a value we encounter regularly with the cosine and sine functions. We know that a right triangle with angle ${60}^{o}$ adjacent to the horizontal side will give us a cosine of $\frac{1}{2}$. In the correct quadrant, namely quadrant $I I$, we can get $- \frac{1}{2}$. Where $a = 1$.

The cosine in this case is given by $\frac{x}{r}$ or the ratio of the adjacent side over the hypotenuse. This gives $\cos \left({60}^{o}\right) = \frac{1}{2}$.

In the second quadrant, $x < 0$ and $y > 0$, giving $\cos \left({120}^{o}\right) = - \frac{1}{2}$ (we use ${120}^{o}$ because it gives us an angle of ${60}^{o}$ above the negative $x$-axis). The radian equivalent of ${120}^{o}$ is $\frac{2 \pi}{3}$. Because the cosine function repeats every $2 \pi$ units (i.e. has period of $2 \pi$), you may also write $\arccos \left(- \frac{1}{2}\right) = \frac{2 \pi}{3} \pm 2 \pi$.