Question

Cos^-1(sqrt3/2) = ? cos^-1(sqrt2/2)= ? cos^-1(-1/2) = ?

Evaluate the following expressions. Your answer must be an exact angle in radians and in the interval [0,π]. Example: Enter pi/6 for π6.
cos^-1(sqrt3/2) =
cos^-1(sqrt2/2)=
cos^-1(-1/2) =

Can someone please help me understand how to do these questions. I've been trying to figure them out for a while now.

Answer 1

`cos^-1(sqrt3/2)=pi/6`
`cos^-1(sqrt2/2)=pi/4`
`cos^1(-1/2)=(2pi)/3`

Explanation:

Whenever we have `sin^-1` of something or `cos^-1` of something, we solve this using the inverse, and it's read as `"arcsin,"` `arccos"` or `"arctan"`. It's important to have our unit circle memorized for this, especially the radian values.

There are also intervals for which each inverse can only be solved for that interval on the unit circle:

`sin^-1:` Valid only from `pi/2` to `-pi/2`

`cos^-1:` Valid only from `pi` to `0`

`tan^-1:` Valid only from `pi/2` to `-pi/2`

It's difficult to explain it in words, so let's visualize it:

Since our problems is `"arccos"`, let's look at its unit circle interval:

Notice how we can only include values from `pi` to `0`, as mentioned above (if you want to know why, it's because these are the values we get when restricting the domain of the `y=cosx` graph, which is an entirely other lesson in itself. You don't exactly need to know why, as memorizing the intervals is very simple).

`Cos^-1(sqrt(3)/2)`

To start with our first problem, it can be read in English as "what radian has a `"cos"` value of `sqrt(3)/2`?"

Normally, when we first learn the unit circle in high school, the first problems we get ask `cos(pi/6)`, in which the answer would be `sqrt(3)/2`, so now we are just asking the opposite, while keeping our intervals in mind.

Since the only value on our interval that has a `cos` (or `x`) value of `sqrt(3)/2` is `pi/6`, the answer is `color(red)(pi/6)`

Now that we know how to do these, the next two should be fairly simple:

`cos^-1(sqrt(2)/2)`

The only value on the interval `pi` to `0` that has a `cos` (or `x`) value of `sqrt(2)/2` is `pi/4`, so the answer is `color(red)(pi/4)`

`cos^-1(-1/2)`

The only value on the interal `pi` to `0` that has a `cos` (or `x`) value of `-1/2` is `2pi/3`, so the answer is `color(red)((2pi)/3)`

This type of lesson is much better explained visually, as you'll see why these intervals exist after looking at the functions of our trigonometric functions and where these intervals exist.

This website does a much better job of showing you visually why these problems are solved using these intervals, and is definitely worth a look.