Wednesday, March 26, 2014
This SP7 was made in collaboration with Mason Nguyen. Please visit the other awesome posts on their blog by going here
The viewer must pay special attention to identifying which quadrant the triangle is in. This is crucial because it determines which trig functions will be positive and which ones will be negative. Additionally, it is important to pay attention to the denominator, there cannot be a radical in the denominator, it must be rationalized. Lastly, the viewer should be able to recognize that the trig functions were solved in two ways, through the use of identities and SohCahToa, these two methods resulted in the same correct answers! :)
Wednesday, March 19, 2014
INQUIRY ACTIVITY SUMMARY:1. To understand where sin^2x+cos^2x=1 comes from, we must understand that what an identity is. An identity is a proven fact and formula that are always true. The Pythagorean Theorem is an identity because it meets these requirements. The Pythagorean Theorem is a^2+b^2=c^2. However, it can also be re-written as x^2+y^2=r^2. This is because when we look at a triangle in the unit circle, the horizontal leg is x, the vertical leg is y and the length of the hypotenuse is referred to as r. To further simplify the theorem, we want it to equal to one. To accomplish this, we will have to divide both sides of the equation by r^2, because r^2/r^2 is equal to 1. After completing this step, we are left with (x/r)^2+(y/r)^2=1. Isn't this beginning to look very similar? Because we know the unit circle, we know that cosine on it is x/r. We also know that the ratio for sine is y/r. Knowing this we can substitute Cos^2x and Sin^2x for (x/r)^2 and (y/r)^2 in the equation. We are now left with Cos^2x + Sin^2x=1. Now, we can understand that Cos^2x + Sin^2x=1 is referred to as a Pythagorean Identity because it is both derived from the Pythagorean theorem and is also a formula that always proves true.
Let's choose 30* for on of the magic 3 ordered pairs from the unit circle to prove this identity true. We know the ordered pair is (rad3/2, 1/2). We can plug rad3/2 in for cos^2x because cosine is x/r. We can also plug in 1/2 for sin^2x because sine is y/r. When we square both of these, we end up with 3/4+1/4 and this is equal to one, which simplifies to 1=1, proving this identity true.
2. From sin^2x+cos^2x=1, we can divide the entire equation by cos^2x. This leaves us with (sin^2x/cos^2x)+(cos^2x/cos^2x)=(1/cos^2x). Using our ratio identities, we know that (sin^2x/cos^2x) is equal to tan^2x, (cos^2x/cos^2x) is equal to 1, and (1/cos^2x) is equal to sec^2x. We can now rewrite our new equation as tan^2+1=sec^2x.
INQUIRY ACTIVITY REFLECTION:
1. “The connections that I see between Units N, O, P, and Q so far are…” that all triangles are interconnected with the unit circle's right triangles and that the ratios derived can be used to solve for non-right triangles as well. 2. “If I had to describe trigonometry in THREE words, they would be…” Relationships involving triangles!
Sunday, March 16, 2014
1. Law of Sines:Why do we need it?Because most triangles are not right triangles, we must find the sides or angles of a triangle using a different method (Sorry Pythagorean Theorem). In the case that we are given Angle-Angle-Side (AAS) or Angle-Side-Angle (ASA), we can use the law of Sines.
Here is an image of a non-right triangle, labeled A B C for the angles and a b c for the sides.
Recall that the area of a triangle is Area=1/2bh. In order to find the area, we need to find h, the height. From past concepts with right triangles, we know that Sin A = h/b and that Sin B=h/a. To get h by itself, we will multiply the first equation by b and the second by a. The resulting equations would be bSinA=h and aSinB=h. Because both equations are equal to h, we can use the transitive property to have the equations equal each other: bSinA=aSinB. To simplify, we can divide everything by ab to get SinA/a=SinB/b.
We can prove this again by drawing a perpendicular line from Angle A. We now have Sin B=h/c and
Sin C=h/b. We can rewrite both as cSinB=h and bSinC=h. Using the transitive property we have
cSinB = bSinC. By dividing both by bc, we end up with SinB/b and SinC/c. Since SinB/b is equal to SinA/a as formally proved, we can set all of these rations equal to each other. This gives us the law of Sines which is
4. Area Formulas:How is the “area of an oblique” triangle derived?
Recall that the area of a triangle is A=1/2bh. If we need to find the area of a triangle but we don't know what the height is, we can do it using the area of an oblique triangle.
In the image above, we know that SinC is equal to h/a. If we multiply both sides by a we get aSinC=h. If we substitute this value for h into the area of a triangle formula, we get : A=1/2b(aSinC). We can find the area given different information as well. If given side b and c, we need to know angle A, if given side a and c, we need to know Angle B. Similarly to how we solved for the area of the triangle using angle C earlier, we can do the same for angle A and B. Ultimately, we end up with the formula :
How does this relate to the area formula you are familiar with? This relates to the area formula because it is used in deriving the are of an oblique triangle. We basically used sine to find the value of the height and substituted the formula into the equation. Ultimately, the area formula is still being used, it just looks different .
Saturday, March 15, 2014
Wednesday, March 5, 2014
Tuesday, March 4, 2014
1. 30-60-90 triangle: We are given an equilateral triangle with a side length of 1. I drew a vertical line straight down the center of the equilateral triangle to get 2 30-60-90 triangles.
If I look at one of the triangles, I know that the hypotenuse (the side across from the 90* angle) is 1 and the shortest side is 1/2 because we cut the side in half. To find the height , opposite of the 60* angle, I used the Pythagorean theorem, a^2+b^2=c^2.
I plugged in 1/2 for a and 1 for c. Squaring 1/2 gave me 1/4 and squaring 1 is 1. I wanted to solve for b^2 so i subtracted 1/4 from both sides, giving me 3/4. Next, I took the square root of both sides, leaving me with b=rad3/2. To translate this into a normal pattern( without fractions, I multiplied all the side lengths by 2. This gave me a hypotenuse of 2, the shortest leg was 1 and the height was rad3.
Lastly, I added n to all of the side lengths. This is so that this concept can be applied to all 30-60-90 triangles. By adding n, we emphasize the relationship between the sides, so that we can still solve for sides with different values (not just one) by applying the ratio.
2. 45-45-90 triangle: When given a square with a side length of 1, we can derive the patterns for this special right triangle. First, I drew a diagonal line connecting 2 opposite angles to form 2 45-45-90 degree triangles.
Knowing the length of 2 of the sides, which is 1, I used the Pythagorean Theorem (a^2+b^2=c^2) to solve for the hypotenuse. I plugged 1 into a and 1 into b. First I squared them giving me 1+1=c^2 -> 2=c^2. Next, I took the square root of both sides to give me the hypotenuse of rad2.
Now that I have the length for all 3 sides, I added n to each of the lengths so that it can apply to all 45-45-90 triangles when the side length is not 1. Through using this ratio, the relationship of the 45-45-90 degree angles will remain constant with the measure of the sides.
INQUIRY ACTIVITY REFLECTION
1. Something I never noticed before about special right triangles is how the ratios came to be. I never realized that the ratios came from the Pythagorean theorem!
2. Being able to derive these triangles myself aids in my learning because I now conceptually understand the roots of the rules of special right triangles. Through this activity, I am not simply memorizing information, because I now understand where these values came from. Even if I forget the ratios, I will be able to derive these patterns.