SP #7: Unit Q Concept 2 (Finding All Trig Functions When Given One Trig Function and Quadrant)

Solving with SOH CAH TOA:

1.

2.Solving with Identities:

When finding actual trig functions, we can use two methods: SOH-CAH-TOA or Identities! As you can see from the first photo, we refer back to our Unit circle and can use the Pythagorean Theorem to get our missing side with the two we already have. In our case, we go to the sin where we know the fraction is y/r. If we place it into the formula, our "x" will be one! The second method is through identities! When using our identities, we are always given two trigs which cuts down a portion of our work! With these two, we first figure out in which quadrant the answers will be either negative or positive in (please refer to the second picture). After that is when we either substitute or solve for the proper trig that we are looking for.

BQ #1: Unit P Concepts 6-7 (Law of Sines and Law of Cosines)

Law of Sines

1. Why do we need it?

• First and foremost, we have to understand that not all triangles are right triangles. This means that not all triangles have a ninety degree angle, meaning we cannot use the Pythagorean theorem to solve for more. Then how will we solve  for irregular triangles you ask? Fear not, my beloved friend because this is where the Law of Sines come in to play.

-The Law of Sines, however, may only be used when given SSA (side-side-angle)

2. How do we derive from what we already know?

• From what we already know, the triangle does not have a right triangle. this is where we draw a Line straight down the middle to make two right triangles. This length is labeled as "h."

 

http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm

 

• If we look at this from a graph perspective, we see that the sin<A will equal h/c because it's opposite over adjacent. If we multiply both sides by "c" then we would get cSin<A=h. Sin<C will equal h/a and multiply by "a" then aSin<C will also equal "h" which will make all angles of the same value.
• This can also be said for angle B therefore we plug Sin<A/value = Sin<B/x and cross multiply! This is how we derive the Law of Sine.

 

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

Inquiry Summary Activity:

A. 30-60-90 triangle
 We can derive a special right triangle of 30-60-90 angles from an equilateral triangle. An equilateral triangle is a triangle that has all three angles of the same degree to add up to 360 degrees which would make them 60 degrees. Along with having all angles of same degree, all sides are of equal length as well. In this case one. We derive this special right triangle by cutting the equilateral. By doing this we create two triangles of 30-60-90 degrees, however, we just need to use one.

Because one side was unaffected by the split, we know that it has a side length of one while we know that the other original side was cut in half giving us 1/2. With these two side lengths, we can use the Pythagorean Theorem to get the missing length of b- theorem being a^2 + b^2 = c^2. By solving, we get the value of radical3/2.

 

After every value has been found, we place them in their sides such as the photo below. In looking at the picture, we can see that fractions can be bothersome. A method to remove fractions would be to multiply all sides by two to become:

 

 

The original values can be labeled as "n" or really any variable because they represent the basis of the Special Right Triangle (SRT). Here is when we begin to see the pattern completely because no matter what the length values will be, n will always be connected to the 30-60-90.

 

 

B. 45-45-90 Triangle

 



The 45-45-90 triangle is much like the one in prior concept of 30-60-90. The difference is, we derive these from a square. Our lengths are already given to us to have an value of one all the way around. Again we are going use a line to cut the shape differently. Please refer to the photo below.

 

 

This would have cut two of the 90 degree angles to make four forty-five degree angles, giving us once again two new angles. This time, our degrees would be a special right triangle of 45-45-90 degrees. Because the outer lengths, also known as the legs (or a and b) were unaffected by the cut, we can once again use the Pythagorean theorem to get the missing leg of the hypotenuse. Your work should be plugged in and solved similarly to the photo below.

 

 

 

Because our leg lengths are once again the value of one, we want to replace it with variable "n." This "n" now represents the basic pattern value of a 45-45-90 triangle because it applies to any length of this type of triangle.  This will give us two legs with the value of n and our hypotenuse with n times radical two. Keep in mind that if the value of the triangle is expanded, it must be multiplied by the original values of n and so on.

 

 

 

Inquiry Activity Reflection:

1. “Something I never noticed before about special right triangles is…” That there is always a pattern in finding values of sides.

2. “Being able to derive these patterns myself aids in my learning because…” they ensure that I will remember the values or the proper method of getting any of my missing values.



 

 

I/D #1: Unit N Concept 7 (identifying degrees, radians, and ordered pairs of a unit circle)

All special right triangles (SRT) may be found not only in the first quadrant of a Unit Circle but all four. Special right triangles either have degrees of 30, 60, 90, or 45, 45, 90. The hypotenuse of either triangle is and always will be 1.

Inquiry Activity Summary:

1. The 30 degree triangle:

The thirty degree triangle is one of the Special right triangles. Assuming we don't know the values of the adjacent and opposite sides (we know the hypotenuse us always one), we can use the one to find value of x. The formula for the Hypotenuse is 2x, for the side where 30 degrees is opening up is x, for the side that 60 degrees opens toward is x-rad3. R is the length of the hypotenuse and as we know, the length of a Unit Circle's Hypotenuse is always one. Since we know 2x = 1, we can find the value of x (and the length of one of our sides) and get 1/2. For the last length, we just evaluate the x to get rad3/2. In the picture below shows the ordered pairs.

2. The 45 degree triangle:

The forty five degree angle has different standards. Although the hypotenuse value of x-rad2 is still equal to 1, the other two legs have a value equal to x. Because both sides have the same value, we only need to find the value once. first, we set x-rad2 equal to 1 to get the value of x. At first we would get 1/rad2 but because the denominator must never have a radical we have to multiply both the bottom and the top by rad2. This will give us an x value of rad2/2 and therefore give us both opposite and adjacent values. In the bottom picture below shows the ordered pairs.



*Correction: In the blue box of the horizontal value it says says rad2/rad/2, I meant to write rad2/2.

3.  The 60 degree triangle:

The 60 degree triangle is almost identical to the 30 degree angle accept they lie differently on the Unit circle. In the 30 degree triangle, the 30 degree angle lies on the x-axis, whereas in the 60 degree triangle, the 60 degree angle lies on the x-axis.  This just means that the x and the x-rad3 values are switched, the hypotenuse stays the same regardless. This also means that when writing the ordered pairs, the x and y values are just switched as well. In the picture below shows the ordered pairs.



4. This activity helped me derive the Unit Circle because not only does it apply to one quadrant but also to the entire circle as well. Through this I discovered the points and ordered pairs to come more automatically to mind. Knowing the special right triangles allow me to understand the bases of a Unit Circle in its entirety.  

5. The Unit Circle is basically made of five degrees known as the "magic five." Three of which we've went through in this activity, the other two being 0 degrees and 90 degrees.  This five angles are repetitive, separating each quadrant in the same angles. This is where conterminals and reference angles come into play because with them we can find many other degrees. As seen in the photo below, there is a 60 degree angle, a 30 degree angle, and a 40 degree angle that can be found also in quadrant II, III, and IV.

Inquiry Activity Reflection:

The coolest thing I learned from this activity is that Unit Circles is really just made up of one portion repeated throughout the circle.

This activity will definitely help me with this unit because there are many things that seem more complicated then they actually seem.

Something I never realized about the special right triangles is how they are all connected in their degrees, plots, and positions.

 

Refer to this website for the picture above:

http://the-crafty-crayon.blogspot.com/2012/09/blog-post_22.html



RWA #1: Unit M Concept 5 (graphing and identifying all parts of an Ellipse)

1. The mathematical definition of an ellipse is "the set of all points such that the sum of the distance from two points is a constant."
-Crystal Kirch
2. Algebraically, the equation of an ellipse may look like this:

http://www.google.com/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&docid=DPKcpdDOYy4itM&tbnid=w0dDhRwnk-AOBM:&ved=0CAUQjRw&url=http%3A%2F%2Fwww.mathwarehouse.com%2Fellipse%2Fequation-of-ellipse.php&ei=vdX6UvLvKs2xqwGvhoAI&bvm=bv.61190604,d.b2I&psig=AFQjCNGfO6poWHy-sP11pQ6Htlr3jjQt6w&ust=1392256786266063
With this equation alone, the center can always be found. The "h and k" are always with the  X and Y terms. If the X and Y squared terms are alone without being in parenthesis it simply means that they are both zero, for example "x^2/4 + (y-4)^2/16," the X term is being simplified or otherwise may be re-written as "(x + 0)^2/4." To find the vertices we would need to look at the bottom portions of the term fractions, generally these would be referred to as "a^2" and "b^2." We can easily tell which is the a because with an ellipse, the A is always bigger than B. If the A is under the X then the Y of the vertices would not change. This would also be associated as the major axis line. If the A was under the Y then it would be the X of the vertices that wouldn't change and then the X be associated (in linear form) as the major axis. The vertices, major axis, and foci will always have this one digit in common, depending on the placement of the A. Along with finding A, we've also found B- together they can be plugged into the a^2 -b^2 = c^2 to get C. With the C we can finish the other half of the foci.

Another way to describe an ellipse is graphically. By having the equation in standard form, which has both terms squared and equal to one, we can infer whether the graph will be "skinny" by having the a under the Y or "fat." by having the a under the X. In other words, if the longer distance is along the x-axis then it would be a horizontal ellipse therefore can be called "fat," now if the longer square root distance goes along the Y-axis then it would be vertical, therefore can be called "skinny."

3. A real life example of an ellipse I found was a race/running track with "designs that help designers take into account top speeds and such depending on shape." According to this website: http://www.barrington220.org/cms/lib2/IL01001296/Centricity/Domain/112/conics%20review%20word%20problemsP2.pdf, the extended circular form of the track. Inside the ellipse are the two  points that are like the foci on the track.

Another thing I found out is that you can use a runner running along the track to find some of its portions. The runner acts as a point on the ellipse because they touch the actual track. If the runner's circling is further away, we can infer that they are currently long the major axis, which as we already know, the longer constant distance that decides the shape of the ellipse.

Please refer to the video below for any further questions:
http://youtu.be/6pDh42E2bbA

4. - http://www.google.com/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&docid=DPKcpdDOYy4itM&tbnid=w0dDhRwnk-AOBM:&ved=0CAUQjRw&url=http%3A%2F%2Fwww.mathwarehouse.com%2Fellipse%2Fequation-of-ellipse.php&ei=vdX6UvLvKs2xqwGvhoAI&bvm=bv.61190604,d.b2I&psig=AFQjCNGfO6poWHy-sP11pQ6Htlr3jjQt6w&ust=1392256786266063
-http://www.barrington220.org/cms/lib2/IL01001296/Centricity/Domain/112/conics%20review%20word%20problemsP2.pdf
-http://youtu.be/6pDh42E2bbA

WPP #10: Unit L Concepts 9-14 (basic probability)

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WPP #9: Unit L Concepts 4-8 (FCP, Combinations, Multiple Events, Permutations etc)

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