Calculating sin-1(-1/2) by Hand

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Homework Help Overview

The discussion revolves around calculating sin-1(-1/2) by hand, exploring the properties of the sine function and its inverse. Participants are examining the implications of negative values in trigonometric functions and the relevant quadrants.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the definition of sine and how it applies to negative values. There is discussion about the quadrants where sine is negative and the implications for the inverse sine function. Some participants are reflecting on their understanding of the principal value of the inverse sine function.

Discussion Status

The discussion is ongoing, with participants providing insights and asking clarifying questions. There is a recognition of multiple angles that yield the same sine value, and some guidance has been offered regarding the principal value of the inverse sine function.

Contextual Notes

There is an assumption that participants are familiar with the basic properties of sine and its inverse, but specific definitions and interpretations are being questioned. The discussion also highlights the interval for the principal value of the inverse sine function.

kg90
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sin-1(-1/2)


Can someone explain how to do this problem by hand. Thanks.
 
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Welcome to PF!

kg90 said:
sin-1(-1/2)


Can someone explain how to do this problem by hand. Thanks.

Hi kg90! Welcome to PF! :smile:

May I check … do you know how to do sin-1(1/2)?
 
What is your definition of sine? (Since a right triangle cannot have an angle whose sine is -1/2, I assume you are NOT using "opposite side divided by hypotenuse" as your definition.
 
Hey thanks for replying. I do know how to do most problems like this. I just don't know how you get the negative answer for the problem I posted above. I always end up with a positive.

HallsofIvy, I'm not sure..
 
Think about which quadrant you are working in and where sine, cosine, etc. are positive/negative?
 
NoMoreExams said:
Think about which quadrant you are working in and where sine, cosine, etc. are positive/negative?

Yeah, but aren't there two quadrants where that would be true?
 
Yes, and therefore on the interval [0, 2pi), there will be 2 answers.
 
Think about it this way, you probably know or can convince yourself (and if you can't, try to) that f(x) = sin(x) crosses the x - axis twice on the interval [0, 2pi) (which points would that be?). Now how would you do that? Well you would say f(x) = sin(x) = 0, so I need to find x such that x = sin^{-1)(0). Now in your case you have g(x) = sin(x) - 1/2 and you are trying to find those x-intercepts, so you solve g(x) = sin(x) - 1/2 = 0 or in other words x = sin^{-1}(1/2).
 
kg90 said:
Hey thanks for replying. I do know how to do most problems like this. I just don't know how you get the negative answer for the problem I posted above. I always end up with a positive.
NoMoreExams said:
Yes, and therefore on the interval [0, 2pi), there will be 2 answers.

Hi kg90! :smile:

(have a pi: π :smile:)

"sin-1" normally means the principal value of sin-1

that's the one in (-π/2,π/2] …

in other words, nearest to 0. :smile:

it may help to think of the sin as the y coordinate of the angle …
so sin-1(-1/2) is the angle where the y coordinate is -1/2 :wink:
 

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