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

Inquiry Activity Summary:
1. Okay so we know that we are dealing with identities because the title is a given, yet what are identities? Well, an identity is basically any proven fact or formula of which brings the Pythagorean Theorem to mind considering that it is a formula that works and is always correct when put to use. Okay so let's start off with the Pythagorean Theorem. If we were given the terms x, y, and r then we would write out the formula like this: x^2+y^2=r^2, but we would tweak it a bit in order to have it equal to 1. We would divide by r^2, do we agree? Yes! Because that is pretty much the only way to equal an equation to 1, by dividing by whatever number the equation is equaled to. Then we would have (x/r)^2+(y/r)^2=1.
 
     Now, if we go back to the unit circle, oh yeah I forgot to mention that the unit circle does come back to haunt haha, we know that cosine is x/r or in other words, adjacent over hypotenuse and sine is y/r or opposite over hypotenuse, so ..do we see anything? Hmm.. Yea! It matches to much of the Pythagorean Theorem! We would just have to square the two variables on the left, x and y, and then equal them to 1. In the end we could have cos^2theta + sin^2theta= 1. Remember that cosine is x and sin is y so in reality we just do some switching around and have fun with it and therefore sin^2x+cos^2x=1 is referred to as a Pythagorean Identity. To prove the theory with some examples and so you can see it visually look at the images below if let's say we did the magic 45 degree.

2. Now, to derive the identity with Secant and Tangent we would start off with the original sin^2x+cos^2x=1. The first thing we would do is divide everything by cos^2x and we would get sin^2x/cos^2x + cos^2x/cos^2x= 1/cps^2x and we can still simplify this further! We know that sin^2x/cos^2x is equal to tan^2x because tangent is opposite over adjacent (y/x). For the easy one, we know that anything over itself is 1 so cos^2x/cos^2x is 1 and then lastly we know that 1/cos^2x is the same thing as saying sec^2x. We simply are plugging in until it simplifies nicely. To conclude we have tan^2x+1=sec^2x. Yay! 

    To derive the identity with Co-secant and Cotangent we would start with the same original one, but divide by sin^2x. We know that after doing so the sines will equal 1 and cos^2x over sin^2x is the same thing as saying cot^2x so we would have 1+cot^2x= ...csc^2x since 1/sin^2x is csc^2x. 

    In order to see what is going on the picture below will hopefully help. 

 

Inquiry Activity Reflection: 

1. "The connection that I see between units N, O, P, and Q so far is...that everything in some sort of way comes together to conclude and involve the unit circle and triangles are involved as well. 

2. "If I had to describe trigonometry in 3 words, they would be...ratios, substitutions, and strangely comprehensible. 

I/D #2: Unit O Concept 7-8: Deriving the patterns for the 45-45-90 and 30-60-90 Triangles

Inquiry Activity Summary:

Today in  class we were given a square and an equilateral triangle both having a side length of 1 and we were to explain the thought process step by step in order to be able and literally know how and why the special right triangles have their constants and basically where they come from. Now, for the 45-45-90 triangle we were given the square that had a length of 1 while for the 30-60-90 triangle we were given the equilateral triangle with a length of 1 and from there we were to completely derive it and see how you get n, n, n radical 2 for the 45 degree one and n, n radical 3, and 2n. Okay so although it seems like we know nothing and we feel like we cannot figure out..we can step by step.

1. Okay so beginning with the square that we were given for the 45 degree special triangle, we can automatically label all four sides with a 1 since we know that a square is all equal in sides and the directions did say that the square contained a side length of 1. Now, we can bisect the square diagonally of course because if we did it vertically it will not create a triangle, but a rectangle. Then, it only gets easier and this is why, we know that two sides are equaled to 1 and  so in order to find the third side we can use the wonderful ...PYTHAGOREAN THEOREM! After solving it by using a^2+b^2=c^2 we should get c= radical 2. Okay now I know what you are thinking well then where does the n come from and well truth is n is just used as a constant it really isn't anything else but a constant/variable like x could be. So we could label all three sides of the right triangle as n like any variable and then we find that the sides conclude to being n, n, and n radical 2. The reason we use n is to that we can see the relationship with all sides and so we can see the possible numbers that could be interchangeable. There is a visual below so that it can be easier in understanding and seeing a picture may clear confusions.

 

2. Okay so now for the 30-60-90 triangle we are given an equilateral triangle of which we all know has three angles that are 60 degrees. We can begin by labeling the three sides with one since the directions as well told us that there was a side length equal to 1. In order to have it become a right triangle we simply bisect it vertically and the reason is considering that we will get a 90 degree, a 30, and a 60 degree like shown below. By cutting it so, it basically creates out special triangle with the wanted angles. Now, by just focusing on the triangle we want we can see that the bottom becomes 1/2 after bisecting it, while the hypotenuse still is 1, and now what about the height? Well, we can use the Pythagorean Theorem again and after plugging in and solving, we should get b= radical 3/2. Okay so now we have 1, radical 3/2, and 1/2, yet we are not done. After, we can do some more solving by multiplying by 2. This is so because well 1/2 times 2 gives us one and that is what we originally had, yet now we have to do it to all sides. Next, once we multiply we should get 2 for the hypotenuse, 1 for the bottom horizontal line and lastly radical 3 for the height. Now, almost finishing, where the heck does the n mean and why does it fit in? Well, n is just a constant of which we use to compare the sides and it basically represents the different number values. Now, after we label all sides with the n as simply a variable we end up getting n, n radical 3, and 2n. The figures below help much visually when it comes to understanding.

Inquiry Activity Reflection:

1. "Something I never noticed before about special right triangles is that they can appear and simply pop out of no where and especially out of other shapes."

2. "Being able to derive these patterns myself aids in my learning because I can actually see what I am learning and exactly where this comes from that this didn't just appear form thin air." 

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

Inquiry Activity Summary:

1. 30 degrees 

Several different instructions were given in order to complete this special right triangle of 30 degrees and much labeling in order to understand. To start with, the three sides of the triangle had to be labeled for their identification. So, one thing we know for sure and that is that our hypotenuse is equaled to 1 considering that these triangles all come from within a unit circle and what is a unit circle? Well a unit circle is one of which has a radius of one and therefore, we can see how the longer side (hypotenuse) is that radius of the circle. Now, according to the special rules of right triangles we know that the side adjacent to the 30 degrees (form the picture it is the horizontal side labeled x) will be n radical 3. Also, we find out that the side opposite from the 30 degrees (due to the picture it is the vertical side labeled y) will be n and finally we get to the hypotenuse (r) of which is said to be 2n. Okay now that that is over with, we can actually find real number values and that is done by using two trig functions: Sine and Cosine. Sine of 30 degrees is done by using SOH and therefore will be opposite over hypotenuse; we can write it out as y/r or x/2x of which equals 1/2 once it is simplified. Doing the same thing with cosine but this time using CAH, adjacent over hypotenuse, it would be x/r and with the variables it would be radical 3/2 after the simplifying like shown in the image above. If we were to treat the triangle on a coordinate plane with the origin of the 30 degrees being (0,0) then the corner as you move horizontally will be (radical 3/2,0) and lastly, the point at the highest tip will be (radical 3/2, 1/2).

2. 45 degrees

Now the same types of instructions apply to this 45 degree triangle, yet the numbers will be different. Okay so to start we can label the adjacent side, the horizontal side of the given angle on the left hand corner x. The opposite side of the angle or the vertical side will be labeled as y and lastly for now the hypotenuse will be r=1. To label according to STR, both x and y will be n, and as for the hypotenuse is will be n radical 2 after simplifying of course like shown in the image. The trig functions world the same as last time with sine and cosine. Sine and cosine are actually the same number values and they both include the same type of simplification since for both the x and y it is n/n radical 2 of which reduces to radical 2 over 2. From my perspective this special triangle is the easiest and includes less work. Picturing the image on a coordinate plane, the origin will be (0,0) and as you move horizontally across then the next point will be (radical 2/2, 0) and then as for the last vertical tip point it will be (radical 2/2, radical 2/2).  

3. 60 degrees

Okay so we are finally looking at the last type of triangle! Yay! We shall label the horizontal and vertical and hypotenuse like the same as the others, with x, y, and r=1. X is the adjacent and y is the opposite one. Considering the special rules, x will be n and y will be n radical 3. To find sine of 60 degrees then it will be y/r and then to find actual values it will turn into radical 3 over 2. Also, for cosine it will be x/r and then 1/2 after the simplification as shown. Treating the triangle as if on a coordinate plane starting as the origin (0,0) and then as you keep going across horizontally the point will become (1/2,0). Lastly, the last point will be (1/2,radical 3/2). 

4. This activity helps derive the unit circle by being able to see special right triangles from within the circle. We can see how the SOH CAH TOA can develop and we can see the different types of degrees within it.

5.

Depending on which quadrant the SRT is found, that will be the determination on whether it will be positive or negative. If it will be found on quadrant one then both x and y will be positive; if it is then in quadrant 2 then only the x value will be negative while the y is positive. Also, the third quadrant both x and y are negative. Lastly, for quadrant 4 then only the y value is negative. 

1. The coolest thing I have learned from this activity was...that the unit circle actually does include the special right triangles and how a lot can come from a simple circle. 

2. This activity will help me in this unit because ... I will not have to literally memorize the entire unit circle.

3. Something I never realized before about special right triangles and the unit circle is... how they can come together and actually make sense. I would've never thought so much could be learned.