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

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sin-1(-1/2)


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