B Unit Circle and Other Trig Questions

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In constructing a right triangle within a unit circle, the leg from the terminal side can be drawn to either the x-axis or y-axis, but drawing it to the x-axis is standard for angles between 0 and 180 degrees. For an angle of 135 degrees, the triangle's angle at the origin can be calculated based on which axis the leg is drawn to, affecting the cosine and sine values. The unit circle has a radius of one, meaning any coordinates greater than one indicate it's not a unit circle. Angles like pi/2, pi, 2pi, or 3pi/2 do not form triangles, as they represent points on the circle rather than vertices of a triangle. Understanding the relationship between the angle and the coordinates (x, y) as cos(theta) and sin(theta) is crucial for solving trigonometric problems.
onemic
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When creating a right triangle in a unit circle how do you know where to place the leg from the terminal side? My textbook and Khan academy don't really explain this and it's just sort of assumed that I'd know. For example, If theta is equal to 135 degrees, where does the leg to complete the right triangle get placed? The y-axis or the x-axis?
 
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Most people draw the leg to the x-axis, but there is no reason one way would be better than another. In this case ##\frac \pi 2 < \theta < \pi##, so if you drew the leg to the x-axis, your triangle will have an angle at the origin measuring ##\pi - \theta##. If you drew the leg to the y-axis, you would have an angle at the origin measuring ##\theta - \frac \pi 2##. Once you have dealt with the right triangle, be sure to account for your signs; in this quadrant, cosine is negative and sine is positive.
 
onemic said:
When creating a right triangle in a unit circle how do you know where to place the leg from the terminal side? My textbook and Khan academy don't really explain this and it's just sort of assumed that I'd know. For example, If theta is equal to 135 degrees, where does the leg to complete the right triangle get placed? The y-axis or the x-axis?
The initial side is usually placed on the positive x-axis. The angle is usually measured counter-clockwise from the pos. x-axis. An angle of 135 degrees would have its terminal side halfway into the upper left quadrant of the unit circle.
 
RUber said:
Most people draw the leg to the x-axis, but there is no reason one way would be better than another. In this case ##\frac \pi 2 < \theta < \pi##, so if you drew the leg to the x-axis, your triangle will have an angle at the origin measuring ##\pi - \theta##. If you drew the leg to the y-axis, you would have an angle at the origin measuring ##\theta - \frac \pi 2##. Once you have dealt with the right triangle, be sure to account for your signs; in this quadrant, cosine is negative and sine is positive.

Ah, so I can always draw the leg to the x axis? I was getting confused because sometimes the leg from the terminal side would be drawn to the x-axis and sometimes the y axis

Mark44 said:
The initial side is usually placed on the positive x-axis. The angle is usually measured counter-clockwise from the pos. x-axis. An angle of 135 degrees would have its terminal side halfway into the upper left quadrant of the unit circle.

That I know, but I'm specifically talking about the leg drawn from the terminal side to complete the right triangle.
 
onemic said:
Ah, so I can always draw the leg to the x axis? I was getting confused because sometimes the leg from the terminal side would be drawn to the x-axis and sometimes the y axis
That I know, but I'm specifically talking about the leg drawn from the terminal side to complete the right triangle.
It should be drawn down to the x-axis (for angles between 0 and 180 degrees, or up to the x-axis for angles between 180 and 360 degrees.
 
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If you want the cosine of your right triangle to be the cosine of your angle, you should draw to the x-axis.
Try it out.
##\cos \frac{2\pi}{3} = -1/2##
If you draw the leg to the x-axis, you will have an angle at the origin of ##\pi/3## which has cosine of 1/2.
If you draw the leg to the y-axis, you will have an angle at the origin of ##\pi/6## which has cosine of ##\frac{\sqrt{3}}{2}##.

In any case, you should use whichever method makes the problem simplest, and always, always draw a picture.
 
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Thanks! This helped clear up a lot.
 
Another question. There are times where I see a triangle constructed in the unit circle where the terminal side intersects the circle at coordinates greater than 1. I thought that the unit circle has a radius of one and thus any x or y coordinate that intersects the circle can never be greater than 1?

Also when theta is equal to pi/2, pi, 2pi or 3pi/2 what would the opposite and adjacent sides be equal to, since you can't create a right triangle?
 
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onemic said:
Another question. There are times where I see a triangle constructed in the unit circle where the terminal side intersects the circle at coordinates greater than 1. I thought that the unit circle has a radius of one and thus any x or y coordinate that intersects the circle can never be greater than 1?
Correct. If there are coordinates on the circle that are greater than 1 or less than -1, then it's not a unit circle.
onemic said:
Also when theta is equal to pi/2, pi, 2pi or 3pi/2 what would the opposite and adjacent sides be equal to, since you can't create a right triangle?
You don't get a triangle with those angles. A better way of thinking of things is that for any point on the unit circle, with coordinates of (x, y), we have ##x = \cos(\theta)## and ##y = \sin(\theta)##. Here ##\theta## is the angle measured from the positive x-axis to the ray that extends from (0, 0) to the point (x, y) on the unit circle. See https://en.wikipedia.org/wiki/Unit_circle.
 
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Again, you can draw any picture you like as long as it helps you understand the problem.
There might be times where you would draw the right angle at the outside of the circle (radius to tangent line), forcing the radius to be one of the legs rather than the hypotenuse. This method is not generally helpful for determining the (x,y)=(cos theta, sin theta), but might be helpful for other geometric problems.
 
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Thanks, that cleared everything up
 

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