Daron Rolfson
Dec 05, 2024
arcsin[sin(4pi/3)] is what?
I am quite confused when it comes to inverse trigonometric problems. I understand that there are specific rules to follow when working with inverse sine functions, but I'm not sure how to apply them in this case. Could someone please help me understand how to approach the function arcsin[sin(4π/3)] and explain the relevant rules regarding inverse functions? I believe I can handle the rest on my own. My professor mentioned that the answer was -4π/3, but now I'm uncertain about that. Any assistance would be greatly appreciated!
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
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