# How do you evaluate cos^-1(-1/2)?

Jun 10, 2016

${\cos}^{- 1} \left(- \frac{1}{2}\right) = \theta = {120}^{0}$

Which is the same as: $\frac{2}{3} \pi \text{ radians}$

#### Explanation:

$\textcolor{b r o w n}{\text{Consider the vertex A as being at the origin of an x y graph plane}}$
$\textcolor{b r o w n}{\text{In which case the length of triangle side AB is always positive.}}$
$\textcolor{b r o w n}{\text{Also the only way a trig ratio of the triangles vertex A}}$
$\textcolor{b r o w n}{\text{can be negative is for either x or y to by negative.}}$

Let the unknow angle be $\theta$

cos(/_A) =cos(60^0)= x/("hypotenuse")=x/c = 1/2" "

So if this was the condition (it is not!) then ${\cos}^{- 1} \left(\frac{1}{2}\right) = {60}^{0}$

But we have $\cos \left(\theta\right) = \left(\text{adjacent")/("hypotenuse")=x/("hypotenuse}\right) = - \frac{1}{2} \to \frac{- 1}{2}$

As the hypotenuse is positive then $x$ must be negative

So $\cos \left({120}^{0}\right) = \frac{- x}{c} = - \cos \left(180 - 120\right) = - \cos \left(60\right) = - \frac{1}{2}$

Thus $\textcolor{b l u e}{\theta = {120}^{0}}$

so ${\cos}^{- 1} \left(- \frac{1}{2}\right) = \theta = {120}^{0}$

For radian measure$\to \frac{120}{180} \times \pi = \frac{2}{3} \pi \text{ radians}$

Aug 15, 2017

theta = 120° or240°

#### Explanation:

We are asked to find which angle has a cos value of $- \frac{1}{2}$
$\cos \theta = - \frac{1}{2}$

In a rotation of 360° there are two such angles.

The first step is to establish the quadrants in which Cos has negative values,

Cos is negative in the $2 n d \mathmr{and} 3 r d$ quadrants
(between 90° and 270°)

To find the root angle we use cos^-1(1/2) = 60°

In the second quadrant:

theta = 180°-60° = 120°

In the third quadrant:

theta = 180° +60° = 240°