I/D#3: Unit Q Concept 1: Pythagorean Identities

INQUIRY ACTIVITY SUMMARY

1. The Pythagorean Theorem is an identity because it is a proven fact that is always true. The Pythagoreom Theorem is a^2+b^2=c^2, but in the Unit Circle, these letters are mentioned as x, y and r, making a slight shift to the original theorem. With these letters, the Pythagorean Theorem is now x^2+y^2=r^2. In order to make the Pythagorean Theorem equal to one, we must divide both sides of this equation by r^2, as seen in the picture below.

 

Then, going back to the Unit Circle,  the ratio for cosine was x/r and the ratio for sine was y/r. These ratios can then be plugged into (x/r)^2+(y/r)^2=1 in order to get sin^2x+cos^2x=1. This is referred to as a Pythagorean Identity because it is just simply the Pythagorean Theorem rearranged in a different way. To show that this is identity is true, we can use one of the "Magic 3" ordered pairs from the Unit Circle. If we have a 45 degree angle, we know that the ordered pair is (radical2/2, radical2/2). When this is plugged into the equation, it will be radical2/2^2+radical2/2^=1. This is true because when radical2/2 is squared, it results to being 1/2 and then when it is added to the other 1/2, it is one. Therefore, the identity is true.

2.To derive the identity with Secant and Tangent, cos^2x must be divided by both sides in the equation of sin^2x+cos^2x=1. You will get:

 

In the unit circle, sine has a ratio of y/r and cosine has a ratio of x/r. When the ratios are divided,with sine being the numerator and cosine being the denominator, you get y/x, which is the ratio for tangent. The two cosines will then cancel and result in being 1. 1/(x/r)^2 (the ratio of cosine) is equal to sec^2. This is how you get tan^2x+1=sec^2x.

To derive the identity with Cosecant and Cotangent, we must now divide sin^2x+cos^2x=1 by sin^2x.

 

The sines will cancel, leaving us with 1. The ratio of x/y equals cotangent (refer to picture for further explanation). 1/sin^2x equals csc^2x because of reciprocal identities. This ends up being 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 they all somehow connect to the Unit Circle and also that they somehow involve triangles as well.

 

2. "If I had to describe trigonometry in three words, they would be..." complicated, intricate, and time-consuming. 

WPP #13 & 14: Unit P Concept 6 & 7

Please see my WPP 13-14, made in collaboration with Leslie E,  by visiting their blog here. Also be sure to check out the other awesome posts on their blog.

BQ#1: Unit P Concepts 1 and 4: Law of Sines and Area of an Oblique Triangle

 

1. Law of Sines
We need the Law of Sines to help us find sides or angles that are not necessarily right triangles. Since the Pythagorean Theorem can only be used for right triangles, the Law of Sines helps us find the sides for triangles that are not right triangles.

 

http://etc.usf.edu/clipart/36700/36740/tri19_36740_lg.gif

To derive the Law of Sines, we first need to draw an imaginary line down angle B.




http://etc.usf.edu/clipart/36700/36738/tri17_36738_lg.gif



This imaginary line can be labeled h, as the picture above shows. This allows for two right triangles to be formed. We can now use SOH CAH TOA to help us figure out the rest (just SOH in this case)

Sin A=h/c

Sin C=h/a

Since both of these have h as a common variable, we can simplify to get c sin A=a sinC
Then, divide by ac and then you will get sinA/a=sin C=c.
(this applies to angle B too, if a perpendicular line was drawn from either angle A or C.)

You finally get:

 

 http://www.clarku.edu/~djoyce/trig/lawofsines.jpg



4. Area Formulas
The area of an oblique triangle is derived from the area formula which is A=1/2bh

http://www.compuhigh.com/demo/lesson07_files/oblique.gif

 

To find the height, a verical line is drawn and labeled h. The trig functions, which are sinA=h/c, sinB=h/b and sinC=h/a can be simplified by their denominators in order to make them equal to h. Then, this is plugged into the area of a triangle. H is then replaced with either asinC or csinA. Finally, you end up with the area of an oblique triangle, which is:

 

https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgZrTRmOJrogy3QUkgKj4h6OGuoWqoGscJ0J0Vx9qiKdr2IOoRiD67ngtI6YPhH06z0jxjEKdDKrUYCPBh8lVllMYlDLUvRHJ3G3KjA0Mm_OFJp-z-5Au4VdDeRht56fsINaXJ00Sijqu4/s400/hi.bmp

 

The area of an oblique triangle related to the area I am familiar with because it is what was used to derive the area of an oblique triangle. The area of a triangle is essential in finding the area of an oblique triangle because you need to plug the height into the familiar area equation in order to derive the area of an oblique triangle. 

 

 

 

WPP#12: Unit O Concept 10: Solving Angles of Elevation and Depression Word Problems

http://image.yaymicro.com/rz_1210x1210/0/9d9/baby-boy-climbing-on-bookcase-9d9831.jpg

 

The Problem

A.) Sasha is walking to a shelf to get the flour she needs to bake a recipe. She is 124 ft. away from the shelf and the angle of elevation is 24 degrees. She wants to know the height of the shelf so that she will know how high she has to reach for the flour. What is the height of the shelf?

B.) Sasha needs to get on top of the shelf in order to get her flour. When she is on top, she realizes that she had to climb 55.2 ft. to get there. She spots a flour sack on the floor and notices that she did not have to go through the trouble of climbing to get a flour sack. She then decides that she wants to jump on the flour sack that is on the ground to avoid getting hurt by the floor. The angle of depression to the flour sack is 42 degrees. How long will her fall be?

The Solution

 

A.)

 

 

B.)

 

 



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

INQUIRY ACTIVITY SUMMARY

 

In this activity, we were given a square that told us to derive the pattern for 45-45-90 triangles. We were  only given the side length, which was 1. We were also given an equilateral triangle and we were asked to derive the pattern for 30-60-90 triangles. The information we were given was that the side of the equilateral triangle was 1. This activity is meant for us to understand why special right triangles have these patterns.

 

30-60-90 Triangle

To derive a 30-60-90 Triangle, we first need to label the given information, which is that the side lengths of the equilateral triangle are 1. Then, we need to split the equilateral triangle in half. This will create two right triangles. Since the side lengths of the equilateral triangle are one, the two right triangles now have a side that is 1/2. We then need to solve the missing side using the Pythagorean Theorem. We know that one side is 1/2 and the hypotenuse is 1. We then need to square both sides. 1/2 squared is 1/4 and 1 squared is 1. Then, we need to subtract 1/4 from both sides. We will get 3/4, and finally we need to take the square root of it. The top stays as radical 3, but the bottom is 2. The final side ends up being radical 3/2. Lastly, we need to multiply the sides by two to get rid of the fractions. We will end up with radical 3, 1, and 2. A 30-60-90 Triangle has"n" in its pattern simply because "n" is a variable that represents any number. If any number were placed in to replace "n", the pattern would still work. (refer to pictures below for steps)

 

 

 

 



45-45-90 Triangle

To derive a 45-45-90 Triangle, we first need to label the given information, which is that the side lengths of the square are 1. Then, we need to split the square into two right triangles by cutting the square diagonally. We know that the lengths of both legs of the triangle are 1, so we need to use the Pythagorean Theorem to solve for the hypotenuse. We need to square 1 twice, and then add it. Since 1 squared is simply 1, 1+1 equals 2. Then, we need to take the square root of two, so the hypotenuse ends up being radical 2. The side lengths are now 1,1,radical 2. If 'n' is multiplied to this, it ends up being n, n, n radical 2. The relationship between the sides is the same, so n applies to any number because it is just a constant. " n" basically just represents a number that can be substituted in (refer to pictures below for steps).

 

 

 

 

 

 

 

 

 

 



 

 

INQUIRY ACTIVITY REFLECTION

 

1. "Something I never noticed before about special right triangles is" that the Pythagorean Theorem helps derive the patterns for special right triangles.
 

2. "Being able to derive these problems myself aids in my learning because" now I know why these patterns are there, so I do not have to memorize things anymore and can use this logical reasoning instead to help me solve problems.



 

 





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

INQUIRY ACTIVITY SUMMARY

1. 30º Triangle:

 

 

The activity that we did during class shows us how this 30 degree triangle relates to the unit circle. Since this is a special right triangle, we have to label it according to the rules of Special Right Triangles. The hypotenuse is 2x, the horizontal value is x radical 3 and the vertical value is x. The hypotenuse must equal 1, so we need to divide it by 2x. This is done to all the sides. Then, the hypotenuse must be labeled r, the horizontal value x, and the vertical value y. Then, you need to draw a coordinate plane and finally find the ordered pairs. The ordered pairs are found by simplifying what you divided by 2x and then by thinking of this as a graph. The ordered pairs are (0,0) , (radical 3/2,0), and (radical 3/2, 1/2) (as shown in the picture above)

 

 

2. 45º Triangle

 

 

 

 

For the 45 degree triangle,we must first find the rules for special right tringles and label it according to that. The hypotenuse is x radical 2, the horizontal value is x, and the vertical value is also x. Everything must then be divided by x radical 2 because it was divided so that it would equal 1. When it is simplified, you will get 1/ radical 2, but you must remember to rationalize it because there cannot be a radical on the bottom of a fraction. The rationalized answer will be radical 2/2. Then, you must label the sides r, x, and y, the same was that the previous example was labeled. After, draw a coordinate plane and imagine it as if it were a graph. This is how you will get your points. The points will be (0,0), (radical2/2,0), and (radical 2/2, radical 2/2), as shown in the picture above.

 

3. 60º Triangle

 

 

 

 

 

The 60 degree triangle has the same rules that a 30 degree triangle has. So, the work is practically done for this. The only thing is to switch the x and y values. The ordered pairs would then be (0,0), (1/2,0), and (1/2,radical 3/2), as shown in the picture above.

 

4.This activity helps us derive the unit circle because we now know where the ordered pairs came from in the unit circle. The ordered pairs are achieved when you divide what you divided to get the hypotenuse to equal 1 (explained above) Also, When the coordinate plane is drawn, we can see that it is separated into the four quadrants that the unit circle has.

 

5. The trianges drawn all lie on the first quadrant. The triangles are simply reflected into all of the other quadrants and the x or y values change, depending on the quadrant.

 

30º Triangle:

 

The 30 degree triangle is reflected into all of the other quadrants. The x value becomes negative in the second quadrant. Both x and y values become negative in the third quadrant. The y value becomes negative in the fourth quadrant, as shown in the picture below. (changes are highlighted)

 

 

 

 

 

45º Triangle

 

The 45 degree triangle is reflected on all of the quadrants. The x value becomes negative in the first quadrant. The x and y values become negative in the third quadrant. The y value becomes negative in the fourth quadrant. (refer to picture below)

 

 

 

 

60º Triangle

 

The 60 degree triangle is reflected on all of the quadrants. The x value becomes negative in the first quadrant. The x and y values become negative in the third quadrant. The y value becomes negative in the fourth quadrant. (refer to picture below)

 

 

 

 

 

 

INQUIRY ACTIVITY REFLECTION

 

1. ''The coolest thing I learned from this activity was'' that the unit circle consists of special right triangles that make you not have to memorize the whole unit circle because of the patterns that it has. There is a meaning to the unit circle; it is not simply just a bunch of numbers.

 

2. "This activity will help me in this unit because" it will help me fill out the unit circle a lot faster and it will most likely increase my chances of filling out the unit circle accurately.

 

3. "Something I never realized before about special right triangles and the unit circle is" that by just knowing the first quadrant, you can fill out the others as well.

 



RWA #1: Unit M Concepts 4-6 - Conic Sections in real life (parabola)

1. Mathematical Definition of a Parabola- "The set of all points equidistant from a given point known as the focus and a given line known as the directrix." (http://www.lessonpaths.com/learn/i/unit-m-conic-section-applets/parabola-drawn-from-definition-geogebra-dynamic-worksheet)

2. Algebraically: The equation for a vertical hyperbola is (x-h)^2=4p(y-k).
                           The equation for a horizontal hyperbola is (y-k)^2=4p(x-h).
When the vertex of a parabola is at the origin,  you must see if the graph is y^2 or x^2. Then, you must put in the right value for p based on if you are given the focus or directrix. The standard form would then be (x-h)^2= or (y-k)^2=, which is the equation. In the equation, h and k represent the vertex, or center of the graph. P tells us what way the graph goes. The term that is squared tells us the direction of the parabola.

This link explains the parts that parabolas have and it shows diagrams as well for reference. It is a great reference to use while learning about parabolas.
http://www.purplemath.com/modules/parabola.htm

Graphically:

This picture shows where the parts of a parabola are when it is graphed.
(http://www.teacherschoice.com.au/images/parabola_types.gif)

                  

This picture shows what way the parabola will face according to the equation.
(http://home.windstream.net/okrebs/Ch6-35.gif)

The shape of a parabola is a U, but it has many components to go along with it. The vertex of a parabola is (h,k). It is important to rememer that x always goes with h and y anways goes with k. P is the direction and distance that the vertex is from the focus. P also determines if the graph goes up, down, left, or right, depending on whether is is x^2 or y^2 and positive or negative. The axis of symmetry cuts the parabola in half. It is also perpendicular to the directrix. The directrix is found outside the parabola. It is p units away from the vertex, just as the focus is. The distance away from the vertex to the focus can also determine how wide or narrow the parabola is. The farther the focus is from the vertex, the wider it gets. The variable that is squared in a parabola deterimines the direction.
               
3. Real World Application

This is a Parabolic Heater. A parabola is what makes this heater function. (http://content.costco.com/Images/Content/Product/284457.jpg)

            

This video explains how parabolas are used in everyday things. Here it explains how the Parabolic Heater works. (http://www.youtube.com/watch?v=fV9YuF__fM4)

 

A Parabolic Heater is an example of something that uses a parabola to work. The heat source is located at the focus. It then bounces off the back to be re-directed back to the person. It bounces off in parallel lines. "The circular shape of the heater provides more energy efficiency than other electric space heaters. The parabolic design converts nearly 80 percent of electric energy into radiant heat."

( http://www.ehow.com/facts_7163048_parabolic-heater_.html )

Parabola's are found everywhere. They are found in things like architecture and even nature. The shape of the parabola is what makes some things work, like the Parabolic Heater. Certain things that use parabolas to work must be constructed precisely and designed accurately in order for them to work. It is amazing how parabolas make certain things work smoothly.

4. References

• http://www.teacherschoice.com.au/images/parabola_types.gif
• http://home.windstream.net/okrebs/Ch6-35.gif
• http://www.youtube.com/watch?v=fV9YuF__fM4
• http://content.costco.com/Images/Content/Product/284457.jpg
• http://www.ehow.com/facts_7163048_parabolic-heater_.html
• http://www.purplemath.com/modules/parabola.htm
• http://www.lessonpaths.com/learn/i/unit-m-conic-section-applets/parabola-drawn-from-definition-geogebra-dynamic-worksheet