Question

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

Answer 1

`(7pi)/6 + 2pik and (11pi)/6+ 2pik` where `k` is all integers.

Explanation:

`arcsin(-1/2) = sin^-1(-1/2)`

This is asking:

if `sin(theta) = -1/2` then what is `theta`?

If we look at the unit circle:

sin = `-1/2` at `(7pi)/6 and (11pi)/6`

since the range is not limited the exact answer would be:

`(7pi)/6 + 2pik and (11pi)/6+ 2pik` where `k` is all integers.