I am very VERY confused when it comes to these sort of inverse trigonometric problems...
I know that there are certain rules to follow to stay within the guidelines of the inverse sine tirg functions but I am not sure how to work with them.
Would someone be able to start me off with the function and explain the rules regarding the inverse functions. I can solve the rest on my own. If it helps, my professor said the answer was (4pi/3)...I think that was what he said...now I am not so sure. Please help please????
2 Answers

No, it is not 4π/3. That value is not on the range of the arcsine function. There is a misconception that arcsin(x) is the inverse function of sin(x). It is not. In fact, sin(x), not being injective, does not even have an inverse. However, arcsin(x) is the inverse of this function:
sin(x), for π/2 ≤ x ≤ π/2
The domain declaration leaves the function injective, so arcsin(x) is its inverse. Now, to answer your question, we must find a value in the domain having the same sine as 4π/3.
sin(4π/3)
= sin(π  4π/3)
= sin(π/3) ... and π/3 is on the domain.
arcsin[sin(4π/3)] = arcsin[sin(π/3)] = π/3

arcsin[sin(4pi/3)] = 4pi/3 ANSWER