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Homework Help: Find all solutions of inverse sine (-1/2)

  1. Nov 26, 2011 #1

    DryRun

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    The problem statement, all variables and given/known data
    Find all solutions of inverse sine (-1/2). Angles lie between 0 and 2∏.

    The attempt at a solution
    I know it sounds very simple, but i must be having some of the basics wrong, as i can't figure out the second solution.

    I used the calculator (in radian mode) to find inverse of sin (-1/2) which gives -∏/6
    Since the value of inverse sin (-1/2) must be between 0 and 2∏, i went with the automatic reflex of adding 2∏ to that answer which gives 11∏/6 but then i checked the answer and there is a second solution! And i have no idea how to get that second solution. I need to know a simple, yet consistent method that i can use across all problems of that type, if that's possible.
     
  2. jcsd
  3. Nov 26, 2011 #2

    I like Serena

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    Keep adding or subtracting 2pi.
     
  4. Nov 26, 2011 #3

    DryRun

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    This doesn't make any sense. The boundaries are 0 and 2∏. If i keep adding or subtracting 2∏, the value will go overboard.

    Example: 11∏/6 + 2∏ = more than 2∏ or -∏/6 - 2∏ = less than zero.
     
  5. Nov 26, 2011 #4

    I like Serena

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    My bad.
    Within 0 and 2pi there is one other solution.
    You should use the trig identity sin(pi - x)=sin(x)
     
  6. Nov 26, 2011 #5

    ehild

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    Your teacher certainly has shown the definition of sine and cosine in the unit circle. Remember: four quadrants, sine and cosine ++ in the first, +- in the second, -- in the third and -+ in the fourth. You have found the solution in the fourth quadrant, still having an other one in the third.

    ehild
     
  7. Nov 26, 2011 #6

    DryRun

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    I've been scribbling and got stuck, as i keep evaluating:

    sin (pi - (-1/2)) = sin (-1/2), which are both equivalent, obviously.

    I have no idea how to use that formula.

    @ehild I know about using CAST on the quadrant. But i'm still stuck as i don't know how to use it to solve for this particular problem, since i only know about adding or subtracting 2pi to get the values but i got only one of the answers using that method. The restriction is that the answers must lie between 0 and 2pi.
     
    Last edited: Nov 26, 2011
  8. Nov 26, 2011 #7

    I like Serena

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    You're mixing up the sine with the inverse-sine (aka arcsin).

    Try this:

    sin( pi - (-pi/6) ) = sin( -pi/6 )
     
  9. Nov 26, 2011 #8

    DryRun

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    OK, i got the second answer using that formula. 7∏/6
    But is that the only way to do it? I would prefer using the quadrant. I plotted the line in the quadrant based on sin (-1/2), so the opposite is -1 and the hypotenuse is 2. The angle is situated in the 4th quadrant. I drew a circle anti-clockwise showing the 2∏ addition that i already used to get the first answer. However, to derive the second answer (7∏/6) from the quadrant still eludes me.

    Also, if i had not checked the answers, how would i have known that there are 2 answers? Is there a quick method to know the number of solutions between defined boundaries?
     
    Last edited: Nov 26, 2011
  10. Nov 26, 2011 #9

    I like Serena

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    Perhaps you should take a look at the "Unit Circle".
    See: http://en.wikipedia.org/wiki/Unit_circle

    In this case you are looking for points on the unit circle where the y-coordinate is (-1/2).
    There are 2 of them.
     
  11. Nov 26, 2011 #10

    DryRun

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    OK, other than the first point in the fourth quadrant, i see another point in the second quadrant. But the argument of the second point gives me: pi+ (-pi/6) = 5pi/6 which is not the same answer as that which i obtained using the formula above.
     
  12. Nov 26, 2011 #11

    I like Serena

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    For starters 5pi/6 is in the 2nd quadrant.
    But then, you used pi + x, instead of pi - x.

    And also you have apparently looked at a point with the same x-coordinate, instead of a point with the same y-coordinate.
    That would give you an angle with the same cosine, instead of the same sine.
     
  13. Nov 26, 2011 #12

    ehild

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    I have not seen a unit circle in this thread yet...
    Unit circle means radius 1. Take a point along the circle and connect it with the centre. The length of that radius is 1 and its projection on the x axis is x=cosα and the projection on the y axis is sinα. The radius together with the lines parallel to both axes make a triangle (painted yellow) with hypotenuse equal to 1 and sides of length |sinα| and |cosα|. You see that the angle of this yellow triangle (60° in your case) subtracts from 180° in the second quadrant, adds to 180° in the third one and subtracts from 360° in the fourth quadrant to make the angle in question (in light blue).
    To solve a problem when the sin of an angle is given, draw a horizontal line y=sin(α). It intersects the unit circle in two points, giving two angles (the light blue and the pink ones in the figure).

    Looking at the figures you can notice that

    sin(180°-α)=sin(α)
    sin(180°+α)=-sin(α)

    and

    cos(180°-α)=-cos(α)
    cos(180°+α)=-cos(α)

    Keep the in mind.

    ehild
     

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    Last edited: Nov 26, 2011
  14. Nov 27, 2011 #13

    DryRun

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    I finally understand where i made the mistake in this problem. Thank you for your help and patience, ehild and "I Like Serena".

    Can i use this method as a general way to find all the solutions for a given sine, cosine or tan of any given angle? That is, i will first find arcsin, arccos or arctan, and then plot the line and where it intersects the unit circle, i will make a horizontal line that cuts the unit circle at another point, and then i can deduce the other solutions. Correct?
     
  15. Nov 27, 2011 #14

    I like Serena

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    Correct! :smile:
     
  16. Nov 27, 2011 #15

    DryRun

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    Thanks again. :smile:
     
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