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

Inquiry Activity Summary


1.       The Pythagorean Identity comes from a derivation of the Pythagorean Theorem and we can basically call it an identity because it is a proven fact, which will be shown later in the work.

        From this picture we have the standard Pythagorean Theorem and we will be replacing it with trig functions. We use the values of sine and cosine in the side values and we use the hypotenuse of r which will always equal 1. We know this because we are solving with values from a unit circle. We then see that we can the whole thing by r which will basically leave it in the form known as one of the Pythagorean Identities. We then use 45° angle to prove this identity and thus fulfilling its definition, but how do we find the other two?


2.       The other two identities are found in just a few easy steps or rather one step and a few glances at the identities chart.

       As you can see we are really only dividing by sine or cosine and figuring out what each value produces in terms of different identities. So for example in the first part where we are trying to derive for the secant and tangent one, we find that tangent equals sin over cosine. 

Inquiry Activity Reflection

1. The connections that I see between Units N, O, P, and Q so far are that when we are deriving for these identities, we use values found on the unit circle from Unit N like the radius always equaling 1 which is very convenient for us. I also see a connection between P and Q because of the way we are deriving things from one to another reminds me of how we get the Law of Cosines and Law of Sines. It may seem like they are two entirely different things but they are actually connected together with a little math. 

2. If I had to describe trigonometry in three words, they would be connectable mathematical equations. 

I/D #2: Unit O: Concepts 7-8: Deriving the SRTs

Inquiry Activity Summary

1. 30-60-90

             To follow this image, view from right to left starting with what we were given first, an equilateral triangle with sides all equaling 1. Since it's an equilateral triangle, we will also have all angles equal to 60°. We cut the triangle down it's altitude because we want to have right triangles, in this case we get two of them but notice that one of the angles is now 30°, one of the sides is unknown, which happens to be the longer leg, and the shorter leg has been halved to 1/2. 

            To find the unknown side, we use the Pythagorean theorem  but instead of using 1/2 we should double it so that we can have an easier time solving for it. We can do this because the triangle would still hold proportionate to a 30-60-90 as long as we double the other known side. We find the unknown side to be √3 and we set up our values into a special ratio for our special right triangle by putting n into the values so it can fit any 30-60-90 triangle as long as it fits the ratio. 

2. 45-45-90

              Once again start with what we were given, this time a square with all sides equaling 1 and the angles each being 90°. We cut across this square to get our two right triangles, both having sides equaling 1 and having two 45° angles but now their hypotenuse has an unknown value. To find this unknown value you use the Pythagorean Theorem and you get √2. We have our final values and turn them into the special ratio for our 45-45-90 triangle by using n as a variable for any other special right triangles we may have to make sure the sides correspond to the ratio of 1, 1, and radical 2.

          

Inquiry Activity Reflection

1. Something I never noticed before about special right triangles is that they can be constructed from other shapes. To go into more detail I just thought they were weird triangles that had special rules attached to them but they actually appear in a lot of places. 

2. Being able to derive these patterns myself aids in my learning because it gives me an idea of where these special rules come from and how they come together. 

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

Inquiry Activity Summary

1. The 30­° Triangle 

(http://www.gradeamathhelp.com/image-files/30-60-90-triangle.jpg)

         
     
                Before we begin to solve the special right triangle section of the unit circle, we must first follow the special right triangle rules as shown in the first picture. For this triangle, we start from the 30° angle and move from there. So the hypotenuse will stay the same as the 2 value, but the side opposite from our angle will be the 1 value because that is what is next to the x in our rule and the side that will be horizontal will have a set radical 3 value. We will also pretend the triangle is on a coordinate grid where the horizontal side is basically the y value and the vertical side is the x while the hypotenuse will take the radius value.

              Now we can solve for this special right triangle with the parameters given. Since we are solving for a section of a unit circle, the radius always has to be one so there is one value given. To get to the value, we must divide the hypotenuse by 2 and do the same for each side. The y value becomes 1/2 and the x value becomes radical 3 over 2. The coordinates are easy to plot if you are still imagining a coordinate grid and are shown next to the corners of the triangle, with our origin starting at the 30°. 

   2. The 45° Triangle

(http://www.gradeamathhelp.com/image-files/45-45-90-triangle.jpg)

           

                 What we did previously will now be used to solve for this triangle. Now we have a 45 45 90 right triangle and we are starting from one of the 45° angles. Use the new set of special right triangle rule for this triangle and now we have to divide the hypotenuse by radical 2 to get our desired one value. Do that division for each side and now You have your x and y values as labelled in the right picture. Remember you cannot have a radical in the denominator so make sure you multiply by a radical over radical value to get rid of that radical. This time the ordered pairs are circled a midst all the work that is shown. 

  3. The 60° Triangle

(http://www.gradeamathhelp.com/image-files/30-60-90-triangle.jpg)

              Truth be told the resourceful learner would just use the values solved for in the 30° triangle and switch them accordingly. But an explanation will be provided for the slower learners in the crowd. We use the same rule for our 60° triangle as the first one because it is apart of the same 30,60,90 triangle, and we still divide the hypotenuse by 2 to get our desired 1 value and do the same thing for the other sides. But now we must switch the x and y values we had because the angle we are starting from is different. We must also pay careful attention to our coordinates because our values have been switched. The coordinates are circled again and the x and y values have been changed so make sure you read carefully.

 4. How does this help you derive the unit circle? 

        This triangle activity helped me figure out why the coordinates change as I go through a unit circle. It makes perfect sense now that I look at it, the unit circle is just composed of a bunch of triangles. The coordinate we care about is the one that is connected to the y value and the hypotenuse, that is where I figured out why we have such crazy numbers like radical 3 over 2. It also helps me remember what coordinates go where for the unit circle because now I look at the unit circle like a grid where the radius is always 1. Here is a picture I found that better illustrates the idea that these triangles fit in the unit circle. 

(http://upload.wikimedia.org/wikipedia/commons/thumb/9/94/Special_right_triangles_for_trig.svg/220px-Special_right_triangles_for_trig.svg.png)

 5. What quadrant does the triangle drawn in this activity lie in?  How do the values change if you draw the triangles in Quadrant II, III, or IV?  Re-draw the three triangles, but this time put one of the triangles in Quadrant II, one in Quadrant III, and one in Quadrant IV.  Label them as you did in the activity and describe the changes that occur.

The triangles drawn in this activity belong in quadrant I because we start from the origin and go right on the x axis and our triangle points upwards into quadrant I. The values change signs whenever we move them into other quadrants, but as long as we keep the same angle, the values will stay the same. For example the x and y value will be negative in quadrant III and only the y value will be negative in quadrant IV. Lets draw a each angle in a different spot to see these differences.

Here we have a 45° angle in quadrant II. The only thing that changes here is the sign for the x value and you can clearly see that the coordinate changes accordingly. The x and y value are labelled here to show you the difference. 

Here we have a 30° angle in quadrant III. This time both the x and the y value change signs as represented in the image, The coordinate value changes signs while in this quadrant as well. 

Here we have a 60° angle in quadrant IV and this time only the y value changes signs as seen in the labels above.

Inquiry Activity Reflection

The coolest thing I learned from this activity was how the unit circle changes values according to rules for a triangle. I have always wondered about the similarity between triangles and circle when eating a pizza but I've never really put too much thought into it.

This activity will help me in this unit because simply put, it will make filling in the unit circle much easier. Although I still like memorizing it a lot more than this.

Something I never realized before about special right triangles and the unit circle is how closely related they are to each other. Special right triangles are the reason we have such wonky numbers in our unit circle when we are filling out the coordinates.