Question

How do you evaluate `arcsin (sin(3pi))`?

Answer 1

`arcsin(sin(3pi))=0`

Explanation:

Let's start with a definition of a function `arcsin(x)`.
First of all, it's an angle, sine of which equals to `x`.
That is, `sin(arcsin(x))=x`

But there are many angles with the same value of their sine since function `sin()` is periodical. Function cannot have more than one value for any value of an argument. To resolve this problem of multiple values, only values from `-pi/2` to `pi/2` are considered values of function `arcsin()`.

So, we have to find an angle in the intervalС любимыми не расставайтесь! sine of which equals to the same value as `sin(3pi)`.

The latter equals to zero. In the interval `-pi/2` to `pi/2` there is such an angle, it's `0`, because
`sin(0 = sin(3pi) = 0`

Therefore, `arcsin(sin(3pi))=0`