In this unit we focused a lot on triangles and we started with the Pythagorean Theorem. The Pythagorean Theorem is a formula for right triangles to find a missing side length given two other side lengths. The formula is a squared plus b squared equals c squared where c is the hypotenuse and a and b are the opposite and adjacent. A simple proof is the first image to the right. Imagine the side lengths of the large square are c and the side lengths of the small square are b-a. The area of the large square would then be c squared. The area of the small square is (b-a) squared. The area of a triangle is (1/2 )ab and since there are 4 the area of all of them is 4(1/2)ab. Then to find the area of the big square you add the area of the small square and the triangles which is 4(1/2)ab+(b-a)^2. That gets you 2ab+b^2-2ab+a^2 which simplifies to b^2+a^2. That means that c squared equals b squared plus a squared, probing the Pythagorean theorem. You can actually use the Pythagorean Theorem to derive the Distance Formula. In the picture to the right it shows how this is possible. To find the distance of a line, you are basically drawing two lines from either point vertically and horizontally until they meet to create a right triangle. Instead of labeling a side length with a variable you are using points on a plane. A unit circle is a circle centered around the origin of the coordinate plane with a radius of 1 as shown in the third image to the right. The point on the circle is (x,y) since it has a base of x and a height of y. Using the Pythagorean theorem or distance formula we know the following must be true: 1=x^2+y^2. So using these equations we know a point on this circle must satisfy the equation r^2=x^2+y^2. Then we had to find points on the unit circle given the angle. The angles were 45, 30, and 60 degrees. I started out with the 45 degree angle. The fourth image to the right shows the process I took. At first I tried adding a congruent triangle to the other side of the radius to form a square but I realized that wouldn't work so I scratched that idea. Then I just tried looking at what I already had. I saw that the triangle was an isosceles since it had a right angle and a 45 degree angle so the last angle must also be 45 degrees. That meant this triangle was a right isosceles. So x must be equal to y. Then I used the equation we already had to find x and y. The equation was x^2+y^2=1. Since x=y we can substitute y for x so now our equation is x^2+x^2=1. This simplifies to x equals the square root of 1/2. That means that y also equals that value. But there is a fraction inside of a radical so we must simply to remove the radical from the denominator. This simplifies to the square root of 2 over 2, as seen in the image. To check our answer we can plug it into our equation and it is still true. Next we had to find the point for a 30 degree angle. I have it shown on the fifth image to the right. For this one I tried translating off the x axis. This created a larger triangle which was equilateral. I knew this because the original triangle had given angles of 90 and 30 degrees which meant the last one had to be 60 degrees. When we translated it, the 30 degree angle became 60 degrees, the 60 degree angle stayed the same, so that meant the last angle had to also be 60 degrees, making it an equilateral triangle. Using this information we came to the conclusion that every side length was 1 since the radius of the unit circle is always 1, and that was one of the side lengths. Looking at the diagram we can see that y is actually half of one side length, meaning that y=1/2. We can now solve for x using the equation we previously have. The equation now is x^2+1/2^2=1^2. We solve for x and find that x equals the square root of 3/4. But we have a fraction inside a radical so we must simplify it and then we get that x equals the square root of 3 over 2. To check our work we can again plug in our values to the equation and we see that it is true. Now we need to find points for the last angle, 60 degrees. This one is the easiest since we can use our pst knowledge to solve for x and y. I have shown my process on the sixth image. I reflected the triangle off the y axis this time, but I created the exact same triangle as last time. It was equilateral with 60 degree angles. We knew that one side length was 1 so that meant every side was 1. y was half of one side length so that meant y was 1/2 again. And since y was 1/2 again x had to be the square root of 3 over 2 again. We can once again plug these values into our equation and it is true. With these values we can now use symmetry to find the rest of the points on the unit circle as shown in the seventh image to the right. All we do is reflect the points off the x axis then the y axis. We then get a complete unit circle. We can use the unit circle to define sine and cosine. Cosine is what mathematicians use instead of having to say "the corresponding x-coordinate for the point on the unit circle when a radial line makes an angle of theta degrees with the x-axis." It is basically the the x point of an angle. Sine is "the corresponding y-coordinate for the point on the unit circle when a radial line makes an angle of theta degrees with the x-axis." It is essentially the y point of an angle. Cosine is shortened and written as cos while sine is shortened and written as sin. The third one is tangent. The tangent of an angle is the ratio of y/x and we learned that y is equal to the sine of theta and x is equal to the cosine of theta. So that means tangent equals sine of an angle over cosine of an angle. We need to apply it to a real world situation where there may not be a hypotenuse of 1. If you look to at eighth image to the right I show my work. I drew a right triangle with angle theta and label the sides Hypotenuse, Opposite, and Adjacent. Then I drew a similar triangle with a hypotenuse of 1, an opposite of sin theta, and an adjacent of cos theta. After that I wrote ratios with the corresponding sides from both triangles. I simplified it and found a formula for all three trig functions. Sin=O/H, Cos=A/H, and Tan=O/A. Now we can use the three trig functions for triangles of any size. A simple way to remember it is SOH CAH TOA. Sometimes you are given two side lengths and an angle. That is when you use the inverted functions: arcSine, arcCosine, and arcTangent. For example say you are given a triangle with a hypotenuse of 10 and an adjacent of 8. You would use the cos function but it would be cos^-1. Then you would do cos^-1(0.8) which would give you 36.8699. In class we did the Mount Everest problem which connected what we were learning with a real world situation. My work is shown on the ninth image to the right. We were given triangle ABC and we knew the side length of AB which was 170. We were given angle A which was 72 degrees and angle B was 49 degrees. With that information we could find angle C which was 59 degrees. To start we dropped a perpendicular halfway between B and C and called that point D. We then used the sine function to figure out the height of the triangle which gave us 128.30. Using those two side lengths and the Pythagorean theorem we found the length of DB which was 111.53. We did the same thing on the other half of the triangle and found that AC was 149.68 and CD was 77.09. From that we could derive the law of sines which is SinB/b=SinC/c=SinA/a. It is basically saying that the side lengths are in proportion with the sine of their opposite angle as shown in the tenth image to the right. Then we learned the law of cosines which is c^2=a^2+b^2-2ab(cos theta). This law basically states that if we know side length A and B, and the angle in between, we can find side length C. It is shown in the eleventh image to the right.
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