`arccos(1/2)` Evaluate the expression without using a calculator

`arccos(1/2)`

Let this expression be equal to y.

`y = arccos(1/2)`

Rewriting this in terms of cosine function the equation becomes:

`cos(y) =1/2`

Base on the Unit Circle Chart, cosine is 1/2 at angles pi/3 and (5pi)/3.

`y=pi/3, (5pi)/3`

Then, consider the original equation again.

`y = arccos(1/2)`

Take note that the range of arccosine is `0lt=ylt=pi` . Between `pi/3` and `(5pi)/3` , it is only `pi/3` that belongs to this interval. So the solution to the original equation is:

`y = arccos(1/2)`

`y=pi/3`

Therefore, `arccos(1/2) = pi/3` .