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Trig. Question

  1. Sep 11, 2008 #1

    Can someone explain how to do this problem by hand. Thanks.
  2. jcsd
  3. Sep 11, 2008 #2


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    Welcome to PF!

    Hi kg90! Welcome to PF! :smile:

    May I check … do you know how to do sin-1(1/2)?
  4. Sep 11, 2008 #3


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    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.
  5. Sep 11, 2008 #4
    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..
  6. Sep 11, 2008 #5
    Think about which quadrant you are working in and where sine, cosine, etc. are positive/negative?
  7. Sep 11, 2008 #6
    Yeah, but aren't there two quadrants where that would be true?
  8. Sep 11, 2008 #7
    Yes, and therefore on the interval [0, 2pi), there will be 2 answers.
  9. Sep 11, 2008 #8
    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).
  10. Sep 11, 2008 #9


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    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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