I/D3: Unit Q - Pythagorean Identities

INQUIRY ACTIVITY SUMMARY:

1. To understand where sin^2x+cos^2x=1 comes from, we must understand that what an identity is. An identity is a proven fact and formula that are always true. The Pythagorean Theorem is an identity because it meets these requirements. The Pythagorean Theorem is a^2+b^2=c^2. However, it can also be re-written as x^2+y^2=r^2. This is because when we look at a triangle in the unit circle, the horizontal leg is x, the vertical leg is y and the length of the hypotenuse is referred to as r. To further simplify the theorem, we want it to equal to one. To accomplish this, we will have to divide both sides of the equation by r^2, because r^2/r^2 is equal to 1. After completing this step, we are left with (x/r)^2+(y/r)^2=1. Isn't this beginning to look very similar? Because we know the unit circle, we know that cosine on it is x/r. We also know that the ratio for sine is y/r. Knowing this we can substitute Cos^2x and Sin^2x for (x/r)^2 and (y/r)^2 in the equation. We are now left with Cos^2x + Sin^2x=1. Now, we can understand that  Cos^2x + Sin^2x=1 is referred to as a Pythagorean Identity because it is both derived from the Pythagorean theorem and is also a formula that always proves true.

Let's choose 30* for on of the magic 3 ordered pairs from the unit circle to prove this identity true. We know  the ordered pair is (rad3/2, 1/2). We can plug rad3/2 in for cos^2x because cosine is x/r. We can also plug in 1/2 for sin^2x because sine is y/r. When we square both of these, we end up with 3/4+1/4 and this is equal to one, which simplifies to 1=1, proving this identity true.

2. From sin^2x+cos^2x=1, we can divide the entire equation by cos^2x. This leaves us with (sin^2x/cos^2x)+(cos^2x/cos^2x)=(1/cos^2x). Using our ratio identities, we know that (sin^2x/cos^2x) is equal to tan^2x, (cos^2x/cos^2x) is equal to 1, and (1/cos^2x) is equal to sec^2x. We can now rewrite our new equation as tan^2+1=sec^2x.

Additionally, we can divide the entire equation by sin^2x. This leaves us with (sin^2x/sin^2x)+(cos^2x/sin^2x)=(1/sin^2x). Using our knowledge or ratio identities, we know that (sin^2x/sin^2x) is equal to 1, (cos^2x/sin^2x) is equal to cot^2x, and (1/sin^2x) is equal to csc^2x.We can rewrite this new equation as 1+cot^2x=csc^2x.

INQUIRY ACTIVITY REFLECTION:

1. “The connections that I see between Units N, O, P, and Q so far are…” that all triangles are interconnected with the unit circle's right triangles and that the ratios derived can be used to solve for non-right triangles as well. 2. “If I had to describe trigonometry in THREE words, they would be…” Relationships involving triangles!

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

INQUIRY ACTIVITY SUMMARY

1. 30-60-90 triangle: We are given an equilateral triangle with a side length of 1. I drew a vertical line straight down the center of the equilateral triangle to get 2 30-60-90 triangles.

 If I look at one of the triangles, I know that the hypotenuse (the side across from the 90* angle) is 1 and the shortest side is 1/2 because we cut the side in half. To find the height , opposite of the 60* angle, I used the Pythagorean theorem, a^2+b^2=c^2.

 I plugged in 1/2 for a  and 1 for c. Squaring 1/2 gave me 1/4 and squaring 1 is 1. I wanted to solve for b^2 so i subtracted 1/4 from both sides, giving me 3/4. Next, I took the square root of both sides, leaving me with b=rad3/2. To translate this into a normal pattern( without fractions, I multiplied all the side lengths by 2. This gave me a hypotenuse of 2, the shortest leg was 1 and the height was rad3. 

Lastly, I added n to all of the side lengths. This is so that this concept can be applied to all 30-60-90 triangles. By adding n, we emphasize the relationship between the sides, so that we can still solve for sides with different values (not just one) by applying the ratio.

2.  45-45-90 triangle: When given a square with a side length of 1, we can derive the patterns for this special right triangle. First, I drew a diagonal line connecting 2 opposite angles to form 2 45-45-90 degree triangles.

 Knowing the length of 2 of the sides, which is 1, I used the Pythagorean Theorem (a^2+b^2=c^2) to solve for the hypotenuse. I plugged 1 into a and 1 into b. First I squared them giving me 1+1=c^2 -> 2=c^2. Next, I took the square root of both sides to give me the hypotenuse of rad2.

 Now that I have the length for all 3 sides, I added n to each of the lengths so that it can apply to all 45-45-90 triangles when the side length is not 1. Through using this ratio, the relationship of the 45-45-90 degree angles will remain constant with the measure of the sides.

INQUIRY ACTIVITY REFLECTION

1. Something I never noticed before about special right triangles is how the ratios came to be. I never realized that the ratios came from the Pythagorean theorem!
2. Being able to derive these triangles myself aids in my learning because I now conceptually understand the roots of the rules of special right triangles. Through this activity, I am not simply memorizing information, because I now understand where these values came from. Even if I forget the ratios, I will be able to derive these patterns.

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

INQUIRY ACTIVITY SUMMARY

1.        Describe the 30* triangle: The first thing I did was label according to the rules of the special right triangles. For the the 30* triangles, the hypotenuse is 2x, the opposite side is x, and the adjacent angle is xrad3. To simplify the 3 sides so that the hypotenuse is 1, 2x was divided by 2x to give me one. Since I divided the hypotenuse by 2x, I would have to divide the other two sides as well. The opposite side would be x divided by 2x, giving me 1/2. The adjacent side would be xrad3 divided by 2x to give me rad3/2. Next, I labeled the hypotenuse r, the horizontal value x and the vertical value y.

          I then drew a coordinate plane so that the origin is located at the labeled angle measure. The next step is to find the ordered pairs. The labeled angle is at the origin, so the ordered pair is (0,0). When looking at the 90* angle, we can see that the length of the x-side is rad3/2 and it lies on the x axis (the y -value is 0). This means the ordered pair would be (rad3/2, 0). Looking at the 60* angle, we notice that it the point is rad3/2 distance right of the origin and 1/2 distance above the origin, giving it the ordered pair (rad3/2, 1/2).

2.        Describe the 45* triangle: I labeled the triangle according to the rules of special right triangles. For the 45* triangle, the hypotenuse is xrad2, the horizontal side is x, and the vertical side is x as well. To make the hypotenuse 1, I divided xrad2 by xrad2 to give me one. Now I have to divide the two other sides by xrad2 as well. Since both are x, when I divide by xrad2, I get rad2/2. Then label the hypotenuse r, the horizontal value x and the vertical value y.

           Since the labeled angle, 45* is at the origin, the ordered pair is (0,0). The length of the x-side is rad2/2 and the point lies on the x-axis. This means the ordered pair would be (rad2/2, 0). Because the angle made from the hypotenuse and opposite side is rad2/2 right of the origin and the length or the vertical side is rad2/2,   the ordered pair would be (rad2/2, rad2/2).

3.        Describe the 60* triangle: This is very similar to the 30* triangle except the angles are flipped. The first thing I did was label according to the rules of the special right triangles. For the the 60* triangles, the hypotenuse is 2x, the opposite side is xrad3, and the adjacent angle is x. To simplify the 3 sides so that the hypotenuse is 1, 2x was divided by 2x to give me one. Since I divided the hypotenuse by 2x, I would have to divide the other two sides as well. The opposite side would be xrad3 divided by 2x, giving me rad3/2. The adjacent side would be x divided by 2x to give me 1/2. Next, I labeled the hypotenuse r, the horizontal value x and the vertical value y.

          I then drew a coordinate plane so that the origin is located at the labeled angle measure (60*). The next step is to find the ordered pairs. The labeled angle, 60*, is at the origin, so the ordered pair is (0,0). When looking at the 90* angle, we can see that the length of the x-side is 1/2 and it lies on the x axis (the y -value is 0). This means the ordered pair would be (1/2, 0). Looking at the 30* angle, we notice that it the point is 1/2 distance right of the origin and rad3/2 distance above the origin, giving it the ordered pair (1/2, rad3/2).

4.This activity helps us derive the UC by enabling us to understand the reasons to the values of certain points on the circle. The points on a unit circle is made from 30*, 45*, and 90* angles that are seen throughout all of the 4 quadrants. We now understand where the ordered pairs come from. Through simplifying the 3 sides of the triangle to make the hypotenuse equal to one, I was able to find the length of all three sides in proportion to each other. With this, I was able to find the 3 ordered pairs of the triangle. For all three triangles, the point formed from the joining of the hypotenuse and opposite angle represents a point on the unit circle. For the 30*, 45*, and 90* triangles, the ordered pairs of the previously explained point are (rad3/2, 1/2) (rad2/2, rad2,2) and (1/2, rad3/2). These three ordered pairs are repeatedly seen in all three quadrants. 
If you draw right triangles connecting the ordered pair to the origin on a unit circle, these triangles are identical to the triangles we encountered with this activity. Because of this activity, we now know where the ordered pairs come from, rather than just memorizing the graph, we are now able to explain why it is this certain way. We can also explain why the point at 0* is (1,0), the ordered pair at 90* is (0,1), at 180* is (-1,0), and the ordered pair at 270* is (0, -1) because the radius is 1.

 
5. The triangle in this activity lies in the first quadrant. The values of the the ordered pairs change as you draw it in the other 3 quadrants. For quadrant 2, all of the x values will be negative because it lies to the left of the origin. For quadrant 3, the x and y values will both become negative because the triangle lies to the left of the y axis and below the x axis. Only the y values will be negative in quadrant 4 because the triangles is below the x axis. 

This is a 30* triangle in Quadrant 2, note that the all x-values are negative.

This is a 45* triangle in Quadrant 3, note that both the x and y values are negative.

This is a 60* triangle in Quadrant 4, note that all the y-values are negative. 

INQUIRY ACTIVITY REFLECTION

1. The coolest thing I learned from this activity was how all the 4 quadrants follow the same patterns of special triangles, the only difference is that some are negative.

2. This activity will help me in this unit because I now understand where the values of the Unit Circle comes from, that the numbers are not random, and that you can find them using special right triangles.

3. Something I never realized before about special right triangles and the unit circle is how they are interconnected. I never noticed that you could be able to draw these special right triangles with in the Unit Circle and have the ordered pairs the same.