Saturday, February 22, 2014

I/D #1: Unit N: Concept 7: Unit Circle & Its Relation to Special Rights Triangle



We first had to label the triangle according to the rules of Special Right Triangle: the hypotenuse (r) would be 2x, the vertical side (y) would be x, and the horizontal side (x) would be x radical 3. Then in order to get the values of each side we have the simply the sides where the hypotenuse would equal 1. So we divide all the side by 2x. 2x/2x simplifies to 1, the hypotenuse. x/2x, the x's cancel out and leave 1/2. x radical three/2x, the x's cancel out, leaving radical 3/2. Next we drew the coordinate plane with the triangle lying in quadrant 1. Now the origin will be (0,0). The coordinates for 90* angle will be (radical 3/2, 0) since the distance between the origin and the 90* is radical 3/2. The outermost angle (60*) will be (radical 3/2, 1/2) because now we move up y which equals 1/2. 

We follow the same steps we did before but this time it's 45* which means two side will be the same since the other angle, besides 90*, will also be 45*. The hypotenuse (r) would be x radical 2, the vertical side (y) would equal x as well the horizontal side (x). Again we want to simplify the sides so the hypotenuse would equal 1. Now in order to do this, we divide the hypotenuse by itself to get 1, and we divide x by x radical 2. But by doing this, we leave a radical in the denominator and this is a major no no. So we rationalize by multiply radical 2 on both sides and it simplifies to radical 2/2. Again the origin will be (0,0). The 90* angle will be (radical 2/2, 0) since x's distance is radical 2/2. And the remaining angle will be (radical 2/2, radical 2/2 since y's distance is also radical 2/2. 

Last angle! The hypotenuse (r) is 2x - like the 30* triangle. The vertical side (y) is x radical 3 and the horizontal side (x) is x. Now again we must simplify the sides in order to have the hypotenuse equal 1 so we divide all sides by 2x. (y)'s x radical 3/2x will simplify to radical 2/2 while (x)'s x will simplify to be 1/2, the opposite of 30*. Once again the origin will be (0,0). 90* angle will be (1/2, 0) since x's distance is 1/2 and the last angle will be (1/2, radical 3/2) since y's distance from x is radical 3/2.

This activity was able to help me derive the Unit Circle by providing the angles for the first quadrant of the circle and establishing a foundation for the Unit Circle. Note that the coordinates for the 30*, 45*, 60*, in the first quadrant are the same as the ones I had mentioned previously, and they are also all positive! 

As I had mentioned, the triangles lied in the first quadrant, where both x and y are positive (+) The values change if drawn in quadrant 2, 3, or 4 by sign changes meaning the value being positive or negative. 

This is the 60* triangle drawn in the 2nd quadrant. Everything is still the same as it was in the 1st quadrant, only now the x value is negative due to the quadrant's location and the knowledge of positive and negative components on a graph. Because x is negative so is the x-coordinate in the 2nd quadrant.

x: negative (-) | y: positive (+)

This is the 45* triangle drawn in the third quadrant. It is still the same as the 45* triangle drawn in the 1st quadrant only flipped and now both x and y are negative (again this is due to basic knowledge of the positive and negative components of a graph) 

BOTH (-)

x: negative | y: negative

And lastly, this is the 30* triangle. Once again, it is made of the same components as the triangle in the 1st quadrant only the value of positive and negative changed. This time y is negative due to being located in quadrant 3, meaning that the y coordinate will be -1/2. X will remain positive. 

x: positive (+) | y: negative (-)


The coolest thing I learned from this activity was that you really only need to know the first three angles (knowing the quadrant angles would also be helpful) but if you have learned the magic three, it will be incorporated into the other quadrants and it's only a matter of logically knowing the positive and negative values of each quadrant. 

 This activity will help me in this unit because from here I can derive the unit circle if my mind draws a sudden blank. The outermost coordinates of the triangle gives us the coordinates to the angles in the circle and these three triangles are repeated four times throughout the circle and only certain coordinate values will change depending on their placement on the graph, more specifically it depends on the quadrant the angles are located in. Knowing this and the special right triangles coordinates will significantly help with the unit circle and the future concepts such as in finding the exact values of all six trig. functions. 

Something I never realized about special right triangles and the unit circle is that the special right triangles when layered over each other with the all three origins at (0,0) create a circle, it blew my mind that I couldn't make this connection but now that I see the special right triangles is intertwined with unit circles in creating the circle.

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