Explaining How I Learned 8 Trigonometric Formulas: Sine , Cosine, & Tangent: I was introduced to sine and cosine by the worksheet titled: "The Definitions of Sine and Cosine". Sine and Cosine are apart of a mnemonic tool called "SOH-CAH-TOA" which was introduced a bit after the introduction. Using "SOH-CAH-TOA" I was able to remember that SOH means Sineϴ = Opposite/Hypotenuse. CAH means Cosineϴ = Adjacent/Hypotenuse. TOA means Tangentϴ = Opposite/Adjacent. The next set of notes we were given were called "Simple Trig: Obtaining Side Lengths." From these worksheets I learned that Sine is used when you're given two sides and an angle opposite of those two sides on a right triangle. Cosine is used when you're given three side lengths or two sides and the included angle. A Tangent is a line that touches a unit circle at just one point.
ArcSine, ArcCosine, & ArcTangent: ArcSine, ArcCosine, and ArcTangent are called inverse trigonometric formulas which also means the opposite of sine, cosine, and tangent. They're used to find an angle from any of the angle's trigonometric ratios.
Law of Sines & Law of Cosines: The law of sines was explained in the set of notes called "Trigonometry: Law of Sines." The Law of Sines is the formula: A/Sine A = B/Sine B = C/Sine C which means that side lengths are in proportion with the sine of the opposite angle measures. The Law of Cosines was explained by the set of notes "Trigonometry: The Law of Cosines." The Law of Cosines is c² = a² + b² - 2abcosϴ which also means if we know side lengths a, b, and c and the angle in between, ϴ, then we can find side length c. Both of these laws are used when trying to solve a triangle (finding all angles and side lengths).
Proving the Pythagorean Theorem: I've learned that The Pythagorean Theorem can be proved in a few easy steps.
1. To prove that Pythagorean's Theorem is always a²+b²=c², you'd have to start by adding 3 more of those same triangles and arranging them until they formed a square like this:
2. Labeling is always important. In the photo above, c = the hypotenuse of all 3 triangles.
3. Next, find the area of the square that was formed in the center which is: c²
4. Find the area of the triangles then add the area of the square. Which will look like this: 1/2(ab) + 1/2(ab) + 1/2(ab)+1/2 (ab) + c²
Why Proving the Pythagorean Theorem relates to solving measurement problems: This relates to solving measurement problem because it ties into the idea of how to prove your thinking with logical reasoning. In images, it's easy to say two objects look similar or something looks a certain measurement but proving it is always more challenging yet it's always more reliable than believing what you see.
Using the Pythagorean Theorem to derive Distance Formula:
1. First, all of the vertices should be labeled as X₁ and Y₁, X₂ and Y₂, and X₃ and Y₃. and the hypotenuse should be labeled as d.
2. The Pythagorean Theorem says that the sum of the squares length of the legs is equal to the square of the length of the hypotenuse. To find the lengths of the legs (between their coordinates) it's necessary to use absolute value of the differences between their coordinates so adding absolute value lines (|x²−x¹|) shows the variables distance from 0 despite being positive or negative.
3. Solve: (|x²−x¹|)²+(|y²−y¹|)²=d². By solving this, you're going to prove that the Pythagorean Theroem
Why Using the Pythagorean Theorem to Derive Distance Formula Relates to Solving Measurement Problems: This relates to solving measurement problems because the Pythagorean Theorem is the base idea for all right triangles in the way that you can prove that all right triangles are a²+b²=c². Distance formula is an important formula for finding distances between two coordinate points no matter where they are away from each other.
Using the Distance Formula to Derive the Equation of a Circle Centered at the Origin of a Cartesian Coordinate Plane: There's a relationship between the equation of a circle and distance formula.
The equation of a circle is mainly focused on the radial line: x² + y² = r²
While distance Formula is more concerned about the points marked on a coordinate plane that you want to know the distance between: √(x₁)² + (y₁)² = √d²
To derive the equation of a circle that's centered at the origin of a Cartesian Coordinate plane using Distance Formula would be a combination of the two equations which looks like this: (x-0)² + (y-0)² = r²
Why Using the Distant Formula to Derive the Equation of a Circle Centered at the Origin of a Cartesian Coordinate Plane Relates to Solving Measurement Problems: This relates to solvinhg measurement problems because using distance formula to derive the equation of a circle is measuring the distance from a point on the circle to the point of origin. Learning different ways to measure or think about a certain problem can open up the possibility to learning how to solve something a way that makes the most sense to you.
Defining the Unit Circle: A unit circle is a circle with a radius that equals 1. Usually, the unit circle is centered at the origin otherwise known as point (0,0) in the Cartesian coordinate plane.
Why Defining the Unit Circle Relates to Solving Measurement Problems: Knowing the definition of the unit circle can help you better understand it's purpose and anything you might need a unit circle for. Knowing that the unit circle lays on a grid, you can tell that the origin is usually (0,0) from the definition. The unit circle is the base for finding points on a circle and distance formula which is important to know.
Finding points on the unit circle (at 30 degrees, 45 degrees and 60 degrees.
30 degrees: Because a 30 degree triangle with a side length of 1, we know the height of the original triangle is 1/2: y₂ =1/2
x & y coordinates of a 30 degree triangle: (√3/2, 1/2)
45 degrees: Since this is an isocoles right triangle, the two sides (legs) must have the same length. That means: X₁=Y₁. Meaning, we can substitute value Y₁ for X₁ because they're equal. X₁²+Y₁²=1 X₁²+X₁²=1 2X₁²+=1 X₁²=1/2 X₁=√1/2 The only problem is that the radical sign in the denominator. X₁= √1/2 = √1/√2 =1/√2(√2/√2)= √2/2 X₁=y₁=√2/2. So, the X and Y coordinates = (√2/2,√2/2)
60 degrees: A 30,60,90 triangle that was created inside the circle is congruent to triangle that was created when investigating ϴ = 30 degrees. By examining this triangle, we can see that X₃=1/2 and Y₃ =√3/2
This can be labeled on a unit circle as x and y coordinate points: (1/2, √3/2).
Why Finding Points on A Unit Circle Relates to Solving Measurement Problems: Finding points on a unit circle relates to solving measurement problems because with these equations, you're able to tell where a point is on a circle without imagery which is important because it can be very easy to rely heavily on imagery and believe what you see rather than what makes sense, logically.
Using the symmetry of a circle to find the remaining points on the unit circle: Using reflection, you're able to find the remaining points on a unit circle. If your right triangle starts in the positive top right quadrant, and the coordinates are (√2/2,√2/2) you can reflect the triangle downwards across the radial line and the coordinates will change to (√2/2,-√2/2)
Why Using the Symmetry of a Circle to Find the Remaining Points on the Unit Circle Relates to Solving Measurement Problems: This relates to solving measurement problems because this is also a way of proving a circles symmetry and this is also a way to get used to the different quadrants and the x and y value that changes as coordinate points move across the coordinate plane.
Using the unit circle to define sine and cosine (of the angle theta): You can use the unit circle to define sine and cosine through labeling and a little bit of problem solving. It can be proven by the using the mnemonic "SOH-CAH-TOA".
A unit circle is a unit circle because the radial line equals 1 so whatever line that is drawn from the coordinate point (0,0) to the outside circle, is going to have a length of 1. To define sine and cosine using a unit circle, you'd need a right angle triangle with an angle of theta, a hypotenuse of 1, and two labeled side lengths (a = adjacent & b =opposite). After that's done, the proof will look like this:
1. cosϴ=a/1 cosϴ=a The equation above is saying that cosine of theta is equal to the length of point (0,0) to the point where the line intersects with the perpendicular.
2. sineϴ=b/1 sineϴ=b The second equation is saying that the height of the side opposite to the given angle (ϴ) is the same sineϴ
Why Using the Unit Circle to Define Sine and Cosine Relates to Solving Problems with Measurement: This relates to solving problems with measurement because cosine represented how to find the distance between one point to another on unit circle. Sine represented how to show distance, in this case it was the height between the radial line and the tallest point of the right triangle which was the opposite side length.
Defining the tangent function: In a right triangle, the tangent of an angle is the length of the opposite side divided by the length of the adjacent side. (Think of TOA. Tangentϴ=opposite/adjacent).
Why Defining the Tangent Function Relates to Solving Problems with Measurement: Knowing what Tangent means you won't forget how to use it and what it's purpose is. Tangent is used when finding the tangent of an angle. Remembering TOA is what helped me learn and get used to using Tangentϴ = opposite/adjacent.
Using similarity and proportions to derive the general trigonometric functions (sine, cosine and tangent) If you have two similar triangles, with a 90 degree angle, a 45 degree angle, and an angle of theta, you know that the two triangles are similar because of their angles.In other words:
Because of similar triangles, any angle of theta the ratios above are always going to be the same.
Why Using Similarity and Proportions to Derive the General Trigonometric Functions (sine, cosine, and tangent) Relate to Solving Measurement Problems: This relates to solving measurement problems because when you're using Sine, Cosine, and Tangent, you're trying to measure the difference between similar triangles, a missing side length, or a missing angle.
Using the unit circle to define the arcSine, arcCosine and arcTangent functions: ArcSine: If you asked for the sinπ/4 of a right triangle with 2 45 degree angles and 1 90 degree angle that is on a unit circle, the answer would be: x²+ x² = 1² 2x² = 1 x²=1/2 x= 1/√2(√2/√2) = √2/2 sinπ/4 = √2/2
ArcSine is very similar in the way that it's asking what angle it would have to take to get the answer √2/2.
So we know that arcsine √2/2 = π/4. A helpful thing that Arcsine is often used for is restricting the range of outcomes to only the top and bottom right quadrants of the unit circle which now changes it's range to: -π/2</= ϴ </= π/2.
ArcTangent: Just like arcsine and arccosine, arcTangent can be written in two different ways to represent the tangent of any angle is equal to x: tan^-1 x=? tan?=x
If you were to ask for the arctangent of -1. The solution to this problem looks like this: tan-45°= -√2/2/√2/2 =-1 Arctan-1=-45°(πradians/180°) Arctan-1=-45(πradians/180) =-π/4 radians =arctan-1=π/4 radians
ArcCosine: Arccosine, arcsine, and arc tangent follow a pretty clear trend of just being the opposite of cosine, sine, and tangent. Learning how to do this was a little confusing to me because it was never exactly clear what the inverses meant or what they are so it took a little bit of searching for help before I got a picture of what the purpose of inverses are.
If you wanted to find the arccosine of -1/2 using a unit circle, you'd need to draw it out. The right triangle on the unit circle would have a hypotenuse of 1, an adjacent length of 1/2 and an opposite length of √3/2 and those side lengths can be solved using the Pythagorean theorem. (a²+1/2²= 1²)
Now that we know the length opposite to angleϴ is √3/2, we know that angleϴ is 60°. Knowing that the second angle of the right triangle is 60° we know that the last angle of the right triangle is 30°.
Now we need to find the degrees that 60° is supplementary to which is 120 considering the right triangle is on the top left quadrant of the triangle.
The final answer is: ArcCos(-1/2) = ϴ = 2π/3 radians
Why Using the Unit Circle to Define the ArcSine, ArcCosine, and ArcTangent Functions Relates to Solving Measurement Problems: This is relevant because ArcSine, ArcCosine and ArcTangent are often used for finding angles, and side lengths within in a unit circle. Arc means curve, which makes ArcSine, ArcCosine, and ArcTangent so useful for finding angles within a unit circle.
Using the Mount Everest problem to discover the Law of Sines ("taking apart"): Law of Sines: Sin A/A = Sin B/B = Sin/C The Mount Everest problem helped the Law of Sines. The Law of Sines is usually used when given a triangle with 2 angles and 1 side length. The Law of Sines taught me that the ratio between the side angle and the opposite side length are going to be constant throughout the entire triangle.
What the Mount Everest problem taught me to practice was how to take shapes apart which became useful when learning about volume. By taking apart a triangle and looking at it's sides and angles individually to solve for Sines A, B, and C I learned how to narrow down bigger shapes/objects and then rebuild them to make a complicated problem more manageable.
Why Using the Mount Everest Problem to Discover the Law of Sines relates to Solving Measurement Problems: This relates to measurement problems because Mount Everest was asking me to split up a bigger problem into a smaller one and narrow it down to a point that it made more sense. This is a skill that I'm currently building upon when finding the area of bigger shapes that I have to take apart, find the area of and put back together.
Deriving the Law of Sines: If there's a triangle marked A, B and C. You can always drop down a "perpendicular" to create two right angle triangles to solve and then add them together afterwards. We know that Sineϴ is going to be the ratio of its opposite leg to the length of the hypotenuse. You should have a triangle like this:
(Looking for the angles and side lengths of the first triangle with the perpendicular labeled as "h") We know that sine A= h/b We know that Sine B in the second triangle = h/a which h = a sine B =b sine A = Sine B Which means: B/sineB=a/sineA So if, B/sineB=a/sineA that means: C/sineC=B/sineB=A/SineA
Why Deriving the Law of Sines Relates to Solving Measurement Problems: Deriving the Law of Sines relates to measurement because you're measuring two triangles and looking at their ratios. Deriving laws teaches you how to generalize. I didn't understand what sines or cosines were until I learned how to derive them out of a problem because that's what helped me get a better understanding of both and showed me where I should be looking for these Laws.
Deriving the Law of Cosines: Deriving the Law of Cosines is very similar to deriving the Law of Sines. If there was an equilateral triangle, you'd have to drop a perpendicular down the middle to create two right triangles like this (it's important to label your triangles):
Looking at Angle BCD, I know I can use Pythagorean's Theorem: a²=h² + (c-x)² a²=h² + c²-2cx+x² (We'll need to save this equation for later!)
Now looking at triangle ACD (still using Pythagorean's Theorem): b²=x²+h²' h²=b²-x²
Combining the two equations to eliminate the x's and h's: a²=(b²-x²)+c²-2cx+x² a²=b²+c²-2cx
Deriving Cosine: cos(A)= x/b x=bcos(a)
Inserting bcos(a) to a²=b²+c²-2cx giving me the Law of Cosines: a²=b²+c²-2bccos(a)
Why Deriving the Law of Cosines Relates to Solving Measurement Problems: This relates to solving measurement problems because cosines are used when you need to find the length of an adjacent or hypotenuse leg on a triangle. Learning this was difficult for me and I only remembered the Law of Cosines after I remebered where they were derived from.
Measuring Your World Pt.2! Adeline, Simon and I chose to measure the height of the flag pole in front of the middle school because we though it would be a good challenge for us considering we barely knew where to start.
Introduction & Overview: We were given the idea to use a protractor with a paperclip and string attached to it then line up the top of the flag pole along the straight edge of the protractor. Once we had the angle from Simon's head to the top of the flag pole, we were able to make a triangle, from Simon's line of sight, to the top of the flag pole, back to Simon's line of sight to/from the top of it which created the form of a triangle.
With the given 90 degree angle and the other angle we got from our measurements we were able to determine the last angle from the top of the flag pole to Simon. From there, we were able to use Tangent and Pythagorean's theorem to find lengths and the height from Simon's head to the top of the flag pole.
Once we realized we never took into account Simon's height, we decided to add his height to the final height that we obtained from his head to the top of the flag pole.
Finding the volume and area was easier because all we had to do was plug in the measurements we had already to the given equations.
We didn't have to simplify anything considering that the flag pole was just a cylinder rather than a repeating shape like a pentagon on a soccer ball or something of that nature. However, we did dissect the flag pole into below Simon's height to the ground which created another triangle for us to solve.
Mathematical Process: To find the height of the triangle from Simon's head to the top of the flag pole we needed to have all of the angles to use Pythagorean's Theorem for all of the distances that we needed. Pythagorean's Theorem is: a²+b²=c². To find the volume of the pole we needed the radius and the height. Before we found the volume, we needed to find the radius and to find the radius we needed the circumference. To find the radius using circumference divided the circumference by 3.14. After we had the radius and the height we could plug in those values to the equation to find volume which is: V = π r²h. Lastly, we wanted to find the area which was the easiest part because we already had the measurements that were required for the equation: A= πr².
Reflection: Personally, I struggled with the calculations because I had to back track and get a better understanding of Pythagorean's Theorem and SOH-CAH-TOA before I could consider myself helpful during this project. I also spent a lot of time looking at the complications during the mathematical portion of this project like the fact that I realized how Simon was looking at the top of the flag pole from above the pole's starting point in the ground. I did have to do most of the calculations and present the math portion of our presentation so this was definitely challenging but it was good practice. Some habits of a mathematician that I found helpful was generalizing the problem so instead of seeing this problem as a real-life problem, it helped me to think of it as set-in-stone numbers and triangles. We also used the habit "taking apart" when we were finding the part of the flag pole that was taller than Simon and turned it into a triangle and did the same with the part of the pole that was just as tall as Simon. The last habit that we used was "collaborate and listen", every time I had a question, concern, or generally thought I was/we were going in the wrong direction they always heard me out and were willing to help me with whatever I needed. Somethings we would do differently would be: start the project sooner, collaborate more outside of school, and stay on task inside of school