Asked by Annabelle
How do I simplify arcsin (sin 6 pi) given the interval 0 ≤ theta < 2pi
Answers
Answered by
Reiny
since sin(?) and arcsin(? are inverses of each other,
arcsin( sin (6π) ) = 6π , but that is beyond your stated boundary.
so arcsin(sin 6π)
= arcsin 0
= 0 , π , 2π
arcsin( sin (6π) ) = 6π , but that is beyond your stated boundary.
so arcsin(sin 6π)
= arcsin 0
= 0 , π , 2π
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.