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Unit Circle

  1. May 25, 2008 #1
    [SOLVED] Unit Circle

    How can I use the unit circle to get the right answer. I understand the 30 60 90 and 45 45 90 triangles, but come to a problem when using this method with cosine. For example cosine(3pi/2) is -[2^(1/2)]/2. Using my method I get the positive. Please explain the methods of using the unit circle.
  2. jcsd
  3. May 25, 2008 #2
    Have you learned about reference angles?

    Let's use this Unit Circle as our guide: http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf go to page 3.

    Ok let's start in Quadrant 1. We know the Trigonometric values for 30-60-90 are for Cosine & Sine are ... refer to page 3.

    Well if we look at Quadrant 2, and use our knowledge about reference angles, we can see that the trigonometric values for 30 & 150 are the same, except that cosine is negative.
    Last edited: May 25, 2008
  4. May 25, 2008 #3
    Not sure what exactly are reference angles, but I do know how to convert from radians to degrees.
  5. May 25, 2008 #4
    Yes they are the same, but negative why exactly?
  6. May 25, 2008 #5
    Are you in the positive or negative x direction? If you go around the unit circle, in quadrant 2, only the x-values became negative.

    In quadrant 3, both x and y are negative, but in quadrant 4, only the y-values are negative.

    Quadrant 1: positive x and y
    Quadrant 2: negative x and positive y


    30 degrees and 150 degrees are reference angles bc 180 - 150 = 30. Same goes for 135, 180 - 135 = 45.

    If you start on the negative x-axis, and draw an angle towards the positive y-axis in a clock-wise direction, isn't that basically the same as drawing a 30 degree angle in the positive x-axis going in a counter-clockwise direction except in a different quadrant?

    Both of these have hypotenuse 1, originating from the Origin. So now draw a 150 degree angle coming from the positive x-axis, you're triangle isn't going to have a 150 degree angle, that's not how we draw it. It's actually the remaining 30 degrees left that will form the angle for your triangle.
    Last edited: May 25, 2008
  7. May 25, 2008 #6
    Okay so if I was to use cosine which is X/R. I would receive a negative answer because of the negative x. Tell me if I'm wrong.
  8. May 25, 2008 #7
    It depends what quadrant your in.
  9. May 25, 2008 #8
    Sorry a Quadrant 2.
  10. May 25, 2008 #9
    Thx for the help rocomath I think I can handle it myself from here.
  11. May 26, 2008 #10


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    I notice this has been marked "solved" but no one has pointed out to anon413 that " For example cosine(3pi/2) is -[2^(1/2)]/2" is wrong.

    Draw a unit circle on an xy- coordinate system. Angle "0" starts on the right, on the x-axis, at the point (1, 0). A circle of radius one has circumference 2pi, so going around the circle (counter-clockwise) a distance of pi/2 (one right angle) takes you 1/4 of the way around the circle or to the y- axis, (0, 1). Another pi/2 takes you to the negative x-axis, (-1, 0). The third right angle, to 3pi/2, takes you to the negative y-axis, (0, -1). Since "cosine" can be defined as the x-coordinate of such points, cos(3pi/2)= 0, not -sqrt(2)/2. (And sin(3pi/2)= -1.)

    Perhaps you mean cos(3pi/4). pi/4 is half of a right angle (and 1/8 of the entire circle) so pi/4 takes you to the point on the circle halfway between (1, 0) and (0, 1): where the line y= x intersects the circle. 2pi/4 is exactly a right angle and so takes you to (1, 0) as before. 3pi/4 takes you halfway between (1, 0) and (-1, 0), where the line y= -x crosses the circle, in the second quadrant. That point satisfies both x2+ y2= 1 and y= -x: setting y= -x in the quadratic, x2+ (-x)2= 2x2= 1 so x= +/- sqrt(1/2)= +/- sqrt(2)/2. Since we are in the second quadrant, x< 0 so x= -sqrt(2)/2: that is, cos(3pi/4)= -sqrt(2)/2. (And since y= -x, sin(3pi/4)= sqrt(2)/2.)
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