Question

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

Answer 1

`cos^(-1)(-1/2)=theta=120^0`

Which is the same as: `2/3 pi" radians"`

Explanation:

`color(brown)("Consider the vertex A as being at the origin of an x y graph plane")`
`color(brown)("In which case the length of triangle side AB is always positive.")`
`color(brown)("Also the only way a trig ratio of the triangles vertex A")`
`color(brown)("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)(1/2)=60^0`

But we have `cos(theta)= ("adjacent")/("hypotenuse")=x/("hypotenuse")=-1/2->(-1)/2`

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

So `cos(120^0)=(-x)/c=-cos(180-120)=-cos(60)=-1/2`

Thus `color(blue)(theta= 120^0)`

so `cos^(-1)(-1/2)=theta=120^0`

For radian measure`-> 120/180xxpi = 2/3 pi" radians"`

Answer 2

`theta = 120° or240°`

Explanation:

We are asked to find which angle has a cos value of `-1/2`
`cos theta = -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 `2nd and 3rd` 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°`