ID#3: Unit Q Concept 1: Pythagorean Identities

The pythagorean Theorem is an identity because it is a proven fact or formula that is always true. The derivation actually comes from the unit circle. In quadrant 1 of the unit circle we get a triangle with all positive values and the x-axis is x, the hypotenuse is r, which equals 1, and the other side is y. The first thing we do is set up our equation, which is x^2+y^2=r^2. Next we divide both sides by r and the r will cancel, leaving us with 1, and our equation should be (x/r)^2+(y/r)^2=1. With this information we can tell that x/r is cosine, and y/r is sine, so now our equation should be cos^2x+sin^2x=1.

One of the next two identities is derived from the first identity. The first thing we do is divide all sides by cos^2x and the equation should be (cos^2x/cos^2)+(sin^2x/cos^2x)=(1/cos^2x). Cos^2x/cos^2x cancels and becomes 1, thus leaving us with 1+(sin^2/cos^2x)=(1/cos^2x). From our identities we know that sin^2x/cos^2x is the same as tanx and that 1/cos is the same as secx and we end the final equation of 1+tan^2x=sec^2x.

For the last identity we must divide everything by sin^2x and we should end up with (sin^2x/sin^2)+(cos^2x/sin^2x)=(1/sin^2x). (sin^2x/sin^2x) cancels out and leaves us with 1, so our equation should be 1+(cos^2x+sin^2x)=(1/sin^2x). From our identities we know that cosx/sinx is the same as cotx and that 1/sinx is the same as cscx. Our final identity should end up being 1+cot^2x=csc^2x.

The connections I see between units N, O, P, and Q are that most of the information comes from the unit circle, such as the trig functions. 

If I had to describe trigonometry in THREE words, they would be tricky, cool, and cornfusing. 

I/D #2: Unit O Concepts 7-8: How can we derive the patterns for our special right triangles?

Inquiry Summary Activity

1. 30-60-90

   The given triangle is an equilateral triangle and all of its angles are 60 degrees and each side is equal to 1. To get the 30-60-90 triangle you must draw a line down the center of the triangle, leaving one angle being 90 degrees, the other one is 30 degrees, and the other angle is 60 degrees. Like the equilateral triangle these all add up to have a total of 180 degrees. Since the hypotenuse of the triangle remains as 1, we get 1/2 for one of the sides-the x value-since the line was drawn down the center of the triangle and the original number for that side was 1. With this information we use the Pythagorean theorem to find the missing side value. When you solve for the missing side value-the y value- you end up with y^2 equaling radical 3/4 but since the 4 can break down into 2x2 the answer is radical 3/2. Next you multiply each side by 2. When you multiply the angle opposite of 30 degrees you should end up with "n" after multiplying 1/2 by 2, when you multiply the hypotenuse by two you should get "2n" since the value of the hypotenuse was 1, and for the other side you multiply 2 by radical 3/2, the two's cancel out and you're left with radical 3. The "n" values are there to show that the value can keep expanding. 

2. 45-45-90

   We are given a square with all equal sides whose values are all 1. In order for us to find our 45-45-90 triangle we first draw a line that goes from one corner of the square to another. The 90 degree angle is the corner that does not have a line drawn on it and the other two angles are both 45 degrees. Since we have our x and y values for the triangle we use the Pythagorean theorem to solve for the hypotenuse. The hypotenuse ends up being radical 2 and the other two sides remain as 1. Since the value can keep expanding we use "n", thus giving us the x and y sides being "n" and the hypotenuse ends up as "n radical 2." 

Inquiry Activity Reflection

Something I never noticed before about special right triangles is that "n" can expand. 

Being able to derive these patterns myself aids in my learning because I now know why each value is what it is. 

I/D Uint N Concept 7: How Do SRT and UC relate?

Inquiry Activity Summary

   1. The 30* Triangle
The 30* triangle is one of the special right triangles and it requires simplification in order to be able to label the triangle correctly. The first step to the simplification of this triangle is to label it correctly according to the rules of special right triangles. The labeling should be that the hypotenuse is 2x, the leg opposite of the 30* is x, and the longer leg of the triangle is radical 3. To simplify that we must divide the length of each leg and the hypotenuse by 2x because it is the only way our hypotenuse will equal 1, which is our "r." The leg opposite of the 30* should end up being 1/2, our "y," and the longer leg should end up being radical 3/2, the value of our x. When we graph the triangle in the unit circle we draw the coordinate plane and this triangle should be in the first quadrant. The origin of the triangle is (0,0) , going across the horizontal line of the graph we should have (radical 3,2), and when we go upwards from there we get (radical 3/2, 1/2). 

2. The 45* Triangle 
The 45* triangle follows the same rules as the 30 60 90 triangle when it comes to the labeling of the triangle. To simplify the 45 45 90 triangle you must divide by x, except for the two other legs, those must be divided by the reciprocal of 1/x so the x can cancel out and leave you with (radical 2/2) for both legs of the triangle. When graphing it you must use the coordinate plane and it should land in quadrant one, once again, and the origin of the 45 45 90 triangle is (0,0), when you move over on the horizontal line of the graph you should end up with (radical 2/2, radical 2/2), and you should end up with the same when moving up on the graph. 

3. The 60* Triangle 
The same rules as the 30* triangle applies to this one and the only difference is the value change. In this triangle the horizontal value is x and the longer leg is radical 3/2. The labeling still remains the same, the hypotenuse is still 1, the angle that is opposite of the 60* is still y and the horizontal line is 1/2x. When plotting the graph the origin is (0,0), the horizontal points should be (1/2.0) and when you reach the top point, it is (1/2, radical 3/2). 

4. How does this activity help derive the unit circle?
This activity helped me learn that the unit circle is just a bunch of triangles and that they all make up the points in the unit circle, depending on which quadrant they lay on. The quadrant they lay on can determine whether both the x and y are going to be positive, only in quadrant one, in quadrant two only the x is negative, in quadrant 3 both the x and y are negative, and in quadrant 4 the y value is negative. 

5. How do the values change if you draw the triangles in quadrants II, III, and IV? 
The values change depending on the quadrants. If the triangle is in quadrant II then only the x will be negative, if the triangle is in quadrant III then both the x and y will be negative, and if the triangle is in quadrant IV then only the y is going to be negative. 

Inquiry Activity Reflection

The coolest thing I learned from this activity was that all the points on the unit circle are repeated in each quadrant but with different values, like negative and positive. 

This activity will help me in this unit because it helped me to understand where everything in the unit circle comes from and how to differentiate everything that's on the unit circle. With all this information I will be able to fill out the unit circle to help me on the test. 

Something I never realized before about special right triangles and the unit circle is that they are both related to each other. I didn't know that the points in the unit circle came from the special right triangles.