Pythagoreans Theorem to Distance FormulaDistance formula to Beginning of Unit Circle |
Trigonometry Narritive In the beginning if this unit our class started with proving the Pythagorean theorem (a^2+b^2=c^2.) With this, the hypotenuse always represents c (the hypotenuse) and the remaining two sides represent a and b. This formula is used to find a missing side lengths of any right angle triangle by plugging in know side lengths into the equation. We used the theorem in many different scenarios by splitting a regular triangle into two right triangles. Our class was able to prove that this formula works by using the deriving the distance formula (shown on the top left).
We were able to find this formula by putting a right triangle on a grid, creating a unit circle (using the hypotenuse as a radius). This is also known as a Cartesian coordinate plane. All the points given in this circle are equidistant from the radius. This leads us into plot the points on the grid giving us the distance formula and a transition into the unit circle (shown on the bottom left). |
The way we found the point for a 30˚ line on a unit circle was by taking that right angle triangle and reflecting it off the x-axis to create an equilateral triangle. Using our new triangle we know that each side has a length of 1 (making the side y 0.5). Which gives us the information needed to solve for side x and giving us the point (x,y) (√3/2, 1/2.)
The way we found the point for a 45˚ line on a unit circle was first knowing that it was an isosceles triangle. Therefore you only have to solve for one side because x is equal to y, giving us (√2/2.√2/2). Finding the point for the 60˚ line was very simple because it is the same angles as the 30˚ triangle, except it is reflected off the 45˚ line. So all we needed to do is switch the x and y coordinates getting (1/2,√3/2). |
I was able to see the connection between the functions Sine and Cosine by using the unit circle and by making mistakes on previous work. On a previous worksheet I had substituted the wrong angle for sine and cosine (both using the same angle of 20˚) this mistake showed me that both functions were proportional/ similar. Our class was able to use the unit circle, proportions and similarity when looking at Sine, Cosine, and Tangent because Sine is equal to S=O/H (sinϑ=opposite/hypotenuse), Cosine is equal to C=A/H (cosϑ=adjacent/hypotenuse), and Tangent is equal to T=O/A (tanϑ=opposite/adjacent).
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Unit Circle |
The Unit Circle is a tool we used to determine what coordinates are for any triangle placed on a grid. This works because all circles are proportional/similar when the center is used as the origin.
We used the information on the right to discover the remaining point of the circle by reflecting the point that your looking for over each axis. The sine function is used in the unit circle to find the y coordinate, using the angle on the x-axis. Since the cosine function is the opposite of sine, we are able to inverse the function for Cosine. Therefore we are able to use the angle on the y-axis to find the x coordinate. Our Class was able to use these steps (on a unit circle) to find the tangent function. This function was found by reflecting the angled line off the x-axis, this shows that it gives a perfect 90˚ angle. |
The next functions we found were ArcSine, ArcCosine, and ArcTangent. These functions are the opposite of Sine, Cosine, and Tangent. where you are solving for the angle and only the side lengths are given. The Arc-Functions are writen like so: Sine ~ ϑ=sin^-1(O/H), Cosine ~ ϑ=cos^-1(A/H), and Tangent ~ ~ ϑ=tan^-1(O/A).
The way we studied the Law of Sines was with the Mount Everest Problem. This for this problem we were given a scalene triangle (with no right angles), the length of one side, and the angles. The challenge was to find the missing lengths. |
Our class was able to solve this problem by dropping a perpendicular from the A point then B point, this gave us a 90˚ which made it easy to solve. We were able to derive the formula Sin(A)/a = Sin(B)/b = Sin(C)/c which is the law of sines. Along with the Law of Sines we were able to derive the Law of cosines, this function helps us find the missing angles on a non-right triangle, when you are only given two side lengths and one angle. The formula was (c^2=a^2+b^2 - 2abcosϑ).
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Measuring an Expo Marker
Presentation Reflection:
I think that we did a very good job on our presentation me, and my partner did a lot of hard work to get all of the measurements done on time, and also accurate. I think that it went very successfully. I am very proud of how it all turned out.
Project Explanation/Intro:
For this segment of this project we had to choose an object, and measure each part to eventually find the areas then the volumes of each piece. In the end you add it all up to get the "Total Volume." Our object we chose was a Expo Marker. We chose this because not only does it have all of the various measurements we needed but it also was perfect for our time constraint. On top of that it is also something we use every day.
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We used many different habits of a mathematician but narrowing it down to three we chose Generalize, Stay Organized, lastly Collaborate, and listen. If I were to do something different I would have chosen something more complex, but time wasn't on my side for this one. I think that this was a very successful project, and I would love to do it again.