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How do you use the unit circle to evaluate inverse functions

  1. Feb 17, 2008 #1

    a sample problem: arcsin(-1/2)

    2. The attempt at a solution

    do i look at the unit circle and find the y-coordinate or x-coordinate that has -1/2?

    i did ASTC, and figure that it'd be in either quad 3 or quad 4; to tell you the truth i dont understand how to use the unit circle to figure this out.

    i can enter it in the calculator but my calculator doesn't change the answer into a function, for instance 9pi/5 or 5pi/6 or 5pi/7
  2. jcsd
  3. Feb 17, 2008 #2


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    Just DRAW IT. Identify the points on the unit circle which have sine of -(1/2). There are two of these points. One is in Q3 and the other is in Q4. You read arcsine(-1/2) as "The angle whose sine is negative one-half". You are interested in identifying this angle.
  4. Feb 17, 2008 #3
    i dont understand what you are trying to say mr.

    i have it drawn, i am looking at it. i see two quadrants that have the points but i do not see the points themselves.

    can you elaborate on how to find the two points. what you said is was not very helpful.

    which points do i look at? the x and y values? x coordinates? or y coordinates? be more specific because i dont comprehend you.
    Last edited: Feb 17, 2008
  5. Feb 17, 2008 #4
    How do you define the unit circle what is x-coordinate and what is the y-coordinate they correspond to some trigonometry functions.... You are trying to find the coordinate that corresponds to an angle that will give -1/2 when you take the sin of it... like arcsin (.5) = 30
  6. Feb 17, 2008 #5
    im lost. what?
  7. Feb 17, 2008 #6
    Any point on the unit circle corresponds to what? Tell me please
  8. Feb 17, 2008 #7
    the radius?
  9. Feb 17, 2008 #8
    There is a coordinate axis drawn with the unit circle or not? What does a point on the unit circle mean with reference to that coordinate axis? The name UNIT circle embodies the radius in it
  10. Feb 17, 2008 #9
    does it correspond to the angle in radians?
  11. Feb 17, 2008 #10
  12. Feb 17, 2008 #11
    yeah it looks similar to what i have drawn, that point is pi/3 or better known as the coordinate (1/2, (square root of 3) / 2 )
  13. Feb 17, 2008 #12


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    Biest has asked twice about a point and you respond with a number. I think your confusion goes back to what you said in your first post:
    In order to use the unit circle to give you sine or cosine or their inverse functions you have to know that: cos(x) is the x coordinate and sin(x) is the y coordinate of a point on the unit circle. In other words each point is (cos(x), sin(x)). x is the angle (in radians it is the same as the distance around the circle's circumference from (1, 0) to (cos(x), sin(x)).

    Draw a unit circle on a coordinate and then draw the horizontal line y= -1/2. It will cross the circle in two places. Since the y coordinate is negative those two points are in the third and fourth quadrants.
  14. Feb 17, 2008 #13
    Read the post before this one and draw a line intersecting the x-axis through the point you mention here.... that should give you one of the answers
  15. Feb 17, 2008 #14
    one of the answers is 5pi/3?
  16. Feb 17, 2008 #15
    30 degrees off
  17. Feb 17, 2008 #16
    well you see, the line intersecting the pi/3 also intersects 5pi/3...
  18. Feb 17, 2008 #17
    Yes, but you are looking at the x-coordinate not the y-coordinate... i wanted you to look at the lower quadrants. You still have not understood what the POINTS on the unit circle are.... that is key here.
  19. Feb 17, 2008 #18
    ok can we start with the points then
  20. Feb 17, 2008 #19
    there is a link to a diagram.... look at it carefully. it defines what the points are
  21. Feb 17, 2008 #20
    where at

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