Question

How do you find the exact value of `arccos(-1/sqrt(2))`?

Answer 1

`(3pi)/4`

Explanation:

`arccos(-1/sqrt2)`

First, it would be helpful to rationalize `-1/sqrt2` because unit circle values are usually rationalized.

`-1/sqrt2*sqrt2/sqrt2=-sqrt2/2`

Arccos is asking for the ANGLE with a cosine of the given value.

The range of arccos is between zero and `pi`. So if you are finding an arccos of a positive value, the answer is between zero and `pi/2`. If you are finding the arccos of a negative value, the answer is between `pi/2` and `pi`.

According to the unit circle, the angle in the second quadrant (between `pi/2` and `pi`) with a cosine of `-sqrt2/2` is `(3pi)/4`.

Answer 2

`(3pi)/4, (5pi)/4`

Explanation:

cos x = - 1/(sqrt2) = - sqrt2/2
On the trig unit circle, there are 2 arcs that have the same cos value:
`x = (3pi)/4` and `x = (5pi)/4`
Answers for `(0, 2pi)`:
`(3pi)/4, (5pi)/4`
Check with calculator:
`cos ((3pi)/4) = cos 135^@ = - 0.707`
`cos ((5pi)/4) = cos 225^@ = - 0.707`.

Answer 3

`(3pi)/4`

Explanation:

`color(blue)(arccos(-1/(sqrt2))`

First we should understand what the question is about.

It means that, we need to find an angle, when it is inside a cosine function gives `-1/sqrt2`

Let's rationalize it

`rarr-1/sqrt2*sqrt2/sqrt2`

`rArr-sqrt2/2`

Now let's find out the angle using the unit circle

The angle is `color(green)((3pi)/(4)`