Friday, February 21, 2014

I/D #1: Unit N Concept 7: How do SRTs and the UC relate?

Inquiry Activity Summary:

This activity was meant to help us review on the rules of special right triangles and understand how they relate to the unit circle, a circle with a radius of 1. Special right triangles are special because they can yield exact trigonometric values. There are three types of special right triangles: 30º, 45º, and 60º.

(http://www.biology.arizona.edu/biomath/tutorials/trigonometric/specialangles.html)
1. The rules of a 30º triangle:
-the shortest side (opposite of 30º) has the length of x
-the hypotenuse is 2 times the length of the x  (the shortest side)
-the side adjacent to the 30º angle has a length of x times radical 3








(http://www.biology.arizona.edu/biomath/tutorials/trigonometric/specialangles.html)


 2. The rules of a 45º triangle:
-the sides making the right angle both have the length of x (which makes sense because they are both opposite from a 45º angle, making it an isosceles triangle)
-the hypotenuse has length of x times radical 2



(http://www.sparknotes.com/math/geometry2/specialtriangles/section4.rhtml)






3. The rules of a 60º triangle:(basically the same as the 30º triangle; orientation of triangle just changed)
-the shortest side has the length of x
-the hypotenuse has the length of 2x
-the side opposite the 60º angle has the length of x times radical 3







4. This activity with special right triangles helps us derive the unit circle because we now understand why the UC has certain ordered pairs around it. Say we make the hypotenuses of all the SRTs equal 1, by dividing the hypotenuses by themselves. Now what we did to one one, we must do to the other sides, so we divide the other sides of the SRTs by the respective length of their hypotenuse.

(http://www.education.com/study-help/article/unit-circle/?page=2)
For example, we would divide all the sides of the 30º triangle by 2x to get a horizontal side length of radical (3) over 2, a vertical side length of 1/2, and a hypotenuse of 1.







(http://www.education.com/study-help/article/unit-circle/?page=2)
For the 45º triangle, we divided all the sides by x times radical 2, leaving us with a hypotenuse of 1 and the other two sides of radical (2) over 2.











(http://www.education.com/study-help/article/unit-circle/?page=2)


For the 60º triangle, we divided all sides by 2x, as we did in the 30º triangle. However, since the orientation is different, the horizontal side length will be 1/2 and the vertical side length will be radical (3) over 2. The hypotenuse will still equal 1.








Now if we put the new triangles we just made onto the 1st quadrant of a coordinate plane with angle at the origin, we see why the ordered pairs of certain angles in the UC are what they are. By making the hypotenuse equal 1, it is now equal to the radius of the unit making it r. And because we made the hypotenuse equal the radius, the other sides now change. The horizontal side length of the triangles become the x-value in the ordered pair of the vertex not on the x-axis while the the vertical side length becomes the y-value in that ordered pair. That is why in the UC, a 30º angle has the ordered pair of (radical 3 / 2, 1/2), a 45º angle has (radical 2 /2, radical 2 /2), and a 60º angle has (1/2, radical 3 / 2). Also, since the UC has a radius of 1, the point at 0º will obviously be (1, 0) and the point at 90º will be (0, 1).

5. So far we have only drawn the triangles in the first quadrant (1/4 of the UC) with the focus angle at the origin. Now, to explain the rest of the UC, we must put them in the other three quadrants, but with the focus angle still at the origin (basically just reflecting them across the x and y axis).

(http://www.education.com/study-help/article/unit-circle/?page=2)
For instance, if we move the the 30º triangle to the 3rd quadrant, we see that now the horizontal length and vertical length are still the same numbers, but are now negative since they are on the negative sides of of the x and y axis respectively. So now the ordered pair of the vertex not on the x-axis would be (-radical 3 / 2, -1/2).






(http://www.education.com/study-help/article/unit-circle/?page=2)

If we move the the 60º triangle to the 2nd quadrant, we see that now the horizontal length is still the same number, but just negative since it is on the left side of the x-axis. So now the ordered pair of the vertex not on the x-axis would be (-1/2, radical 3 / 2).









(http://www.education.com/study-help/article/unit-circle/?page=2)

If we move to the 45º triangle to the 4th quadrant (reflecting it across the x-axis), the horizontal length would still be the same. The vertical length, though, would only become negative, since it is towards the bottom of the y-axis. That is why the ordered pair is now (radical 2 /2, -radical 2 /2).



Inquiry Activity Reflection:

1. “The coolest thing I learned from this activity was" that special right triangles, which we learned in Geometry, can help us derive the unit circle.

2. “This activity will help me in this unit because…” now I don't need to ignorantly memorize all of the unit circle. I just need to remember the ordered pairs of the first quadrant where it has the 0º, 30º, 45º, 60º, and 90º angles and then make any necessary changes if they are in different quadrants. And if I forget anything, I can always resort to using the SRTs to find the numbers again.

3. “Something I never realized before about special right triangles and the unit circle is…” that just by knowing the SRTs, one can quickly figure out the unit circle without memorizing it completely.

References:
-http://www.education.com/study-help/article/unit-circle/?page=2
-http://www.biology.arizona.edu/biomath/tutorials/trigonometric/specialangles.html
-http://www.sparknotes.com/math/geometry2/specialtriangles/section4.rhtml

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