Inquiry Activity Summary:
1.Sin2x+cos2x=1 comes from the identity, Pythagorean theorem, which is a factor or formula that is proven.The Pythagorean theorem, a^2+b^2=c^2 is replaced x^2+y^2=r^2 because of the fact that they hold the same values.
2. To derive the two remaining Pythagorean Identities from sin2x+cos2x=1 you must first, divide everything by r^2 and after doing that we get x^2/r^2+y^2/r^2=1. You can then notice that they x/ r is cosine and y/ r is sine meaning that you can put sin2x+cos2x=1 in place.
Inquiry Activity Reflection:
1. The connections that I see between Unit N,O,P, and Q so far are all related to the unit circle. In order to solve and understand most of them, you must fully understand and keep in mind the unit circle. Having memorized the unit circle is key.
2. If i had to describe trigonometry in THREE words, they would be difficult, challenging, and memorization. Trigonometry really requires a lot of practice and memorization of the unit circle. You can not solve the problems if you do not know the unit circle, formulas, sin, cos, tan, and so much more.
Showing posts with label I/D. Show all posts
Showing posts with label I/D. Show all posts
Wednesday, March 26, 2014
Tuesday, March 4, 2014
I/D#2: Unit O- How Can We Derive the Patterns for a Special Right Triangle?
INQUIRY ACTIVITY SUMMARY
1. 30-60-90 Triangle
This equilateral triangle has all sides equalling one (because given) and the angles equalling 60 degrees. We know that because all angles in a triangle sum up to 180, and since there are 3 angles you divide 180 by 3 giving you 60 for each angle. However, to get our 30-60-90 triangle, we must cut the triangle vertically in half, having the top 60 degrees become two 30 degree angles and the middle bottom of the triangle two 90 degree angles as seen below. As for the sides equalling one, since we cut the triangle in half, each triangle is going to have the bottom side equalling 1/2. To find the other sides, we use the Pythagorean theorem which is a^2+b^2=c^2. In this case were solving 1/2^2+b^2=c^2 which results in b= radical 3 over 2. To make it easier and not deal with fractions we multiply everything by 2 and add a variable, n, to represent any number, meaning the relationship is consistent. We then get a=2n , b=radical 3 over 3, and c= n.
2. 45-45-90 Triangle
Since we want a 45-45-90 triangle, we cut the square diagnolly in half. How do we know we got two 45 degree angles? Well, a square has all 90 degree angles and when we cut it diagonally in half, the two 90 degree angles split into two, meaning the 90 degree angle is divided by 2 as seen below. We are given that all sides equal one but since we don't not know one of the sides because we cut the square we use the Pythagorean theorem which is a^2+b^2=c^2. In this case were solving 1^2+1^2=c^2, giving us c=radical 2. We then add n to each side in which represents any number meaning the relationship is consistent.
INQUIRY ACTIVITY REFLECTION
1. Something i never noticed about a special right triangle is that there will always be two sides that are the same for example a 45-45-90 triangle a and b are the same and in a 30-60-90 triangle.
2. Being able to derive these patterns myself aids in my learning because it gives me a better understanding of how everything connects. If i every forget the rules i can simply derive them.
1. 30-60-90 Triangle
This equilateral triangle has all sides equalling one (because given) and the angles equalling 60 degrees. We know that because all angles in a triangle sum up to 180, and since there are 3 angles you divide 180 by 3 giving you 60 for each angle. However, to get our 30-60-90 triangle, we must cut the triangle vertically in half, having the top 60 degrees become two 30 degree angles and the middle bottom of the triangle two 90 degree angles as seen below. As for the sides equalling one, since we cut the triangle in half, each triangle is going to have the bottom side equalling 1/2. To find the other sides, we use the Pythagorean theorem which is a^2+b^2=c^2. In this case were solving 1/2^2+b^2=c^2 which results in b= radical 3 over 2. To make it easier and not deal with fractions we multiply everything by 2 and add a variable, n, to represent any number, meaning the relationship is consistent. We then get a=2n , b=radical 3 over 3, and c= n.
2. 45-45-90 Triangle
Since we want a 45-45-90 triangle, we cut the square diagnolly in half. How do we know we got two 45 degree angles? Well, a square has all 90 degree angles and when we cut it diagonally in half, the two 90 degree angles split into two, meaning the 90 degree angle is divided by 2 as seen below. We are given that all sides equal one but since we don't not know one of the sides because we cut the square we use the Pythagorean theorem which is a^2+b^2=c^2. In this case were solving 1^2+1^2=c^2, giving us c=radical 2. We then add n to each side in which represents any number meaning the relationship is consistent.
INQUIRY ACTIVITY REFLECTION
1. Something i never noticed about a special right triangle is that there will always be two sides that are the same for example a 45-45-90 triangle a and b are the same and in a 30-60-90 triangle.
2. Being able to derive these patterns myself aids in my learning because it gives me a better understanding of how everything connects. If i every forget the rules i can simply derive them.
Friday, February 28, 2014
I/D# 1: Unit N Concepts 7-9 : How Do Special Right Triangles and the Unit Circle Relate?
INQUIRY ACTIVITY SUMMARY
1. Describe a 30 degree Angle
I first labeled the triangle according to the Special Right Triangle Rules which is the hypotenuse 2x, the horizontal value x radical 3, and the vertical value x. I then had to simplify the three sides of the triangle fairly in a way such that the hypotenuse=1 so i divided each side by 2x (which is the hypotenuse) giving me: r=1, x=radical 3 over 2, and y= 1/2. I labeled the hypotenuse r, the horizontal value x, and the vertical value y. In order to draw a coordinate plane for each angle i first put the ordered pair (0,0) at the 30 degree corner, then since our x= radical 3 over 2, i knew my ordered pair for the 90 degree angle was (radical 3 over 2,0) because it didn't move up the y axis. For the 60 degree angle, i simply already knew that x=radical 3 over 2 and our y=1/2 so i put them as the ordered pair (radical 3 over 2, 1/2) since it moved x units to the right and y units up.
2. Describe a 45 Degree Angle
I first labeled the triangle according to Special Right Triangle Rules which is the hypotenuse x radical 2, the horizontal value x, and the vertical value x as well. In order to have the hypotenuse equal to one as i divided each side by the value of the hypotenuse which is x radical 2 resulting in as r=1, x=radical 2 over 2, y=radical 2 over 2. I labeled the hypotenuse r, the horizontal value x, and the vertical value y. To find the coordinate planes, i applied the same concept as i did for the 30 degree angle above. I put the ordered pair (0,0) at the 45 degree left corner, then since our x= radical 2 over 2, i knew my ordered pair for the 90 degree angle was (radical 2 over 2 over 2,0) because we didn't move up the y axis. For the other 45 degree angle, i simply already knew that x=radical 2 over 2 and our y=radical 2 over 2, so i put them as the ordered pair (radical 2 over 2, radical 2 over 2) because we moved x units to the right and y units up.
3. Describe a 60 degree Angle
I first labeled the triangle according to Special Right Triangle Rules which is the hypotenuse 2x, the horizontal value x, and the vertical value x radical 3. To have the hypotenuse equaling to one, i divided it by itself and the other sides as well having r=1, x=1/2, y=radical 3 over 2. I labeled the hypotenuse r, the horizontal value x, and the vertical value y.To find the coordinate planes, i applied the same concept as i did for the above angles. I put the ordered pair (0,0) at the 60 degree left corner, then since our x= 1/2 , i knew my ordered pair for the 90 degree angle was (radical 2 over 2 over 2,0) because we didn't move up the y axis and moved x units to the right. For the 30 degree angle, i simply already knew that x=1/2 and our y= radical 3 over 2, so i put them as the ordered pair (radical 2 over 2, radical 2 over 2) because we moved x units to the right and y units up.
4. This activity helped me derive the unit circle because it gave me the coordinates, points, and angles that are within the unit circle as seen below. I understood the foundation of it, allowing me to not only have to memorize numbers but know why those numbers exist. I thought it was important to do this activity because it gave me the basic comprehension of what a unit circle is.
5. The 30, 45, and 60 degree angles are within the first quadrant of the unit circle. The second quadrant is a reflection of the first, the third is a reflection of the second, and the fourth is a reflection of the third. In the first quadrant, all coordinates are positive, in the second quadrant, only the x value is negative, in the third quadrant both x and y are negative and in the fourth quadrant only the y is negative having that the first and third quadrant are positive and the second and fourth are negative. For a better understanding look below.
Inquiry Reflection Activity
1. The coolest thing I learned from this activity was how each quadrant is reflected with triangles.
2. This activity will help me in this unit because understanding and memozing the unit circle is a basic need to solve problems in all the concepts.
3. Something I never realized before about special right triangles and the unit circle is that the whole unit circle consists of many right right triangles that give it its coordinates.
1. Describe a 30 degree Angle
I first labeled the triangle according to the Special Right Triangle Rules which is the hypotenuse 2x, the horizontal value x radical 3, and the vertical value x. I then had to simplify the three sides of the triangle fairly in a way such that the hypotenuse=1 so i divided each side by 2x (which is the hypotenuse) giving me: r=1, x=radical 3 over 2, and y= 1/2. I labeled the hypotenuse r, the horizontal value x, and the vertical value y. In order to draw a coordinate plane for each angle i first put the ordered pair (0,0) at the 30 degree corner, then since our x= radical 3 over 2, i knew my ordered pair for the 90 degree angle was (radical 3 over 2,0) because it didn't move up the y axis. For the 60 degree angle, i simply already knew that x=radical 3 over 2 and our y=1/2 so i put them as the ordered pair (radical 3 over 2, 1/2) since it moved x units to the right and y units up.
2. Describe a 45 Degree Angle
I first labeled the triangle according to Special Right Triangle Rules which is the hypotenuse x radical 2, the horizontal value x, and the vertical value x as well. In order to have the hypotenuse equal to one as i divided each side by the value of the hypotenuse which is x radical 2 resulting in as r=1, x=radical 2 over 2, y=radical 2 over 2. I labeled the hypotenuse r, the horizontal value x, and the vertical value y. To find the coordinate planes, i applied the same concept as i did for the 30 degree angle above. I put the ordered pair (0,0) at the 45 degree left corner, then since our x= radical 2 over 2, i knew my ordered pair for the 90 degree angle was (radical 2 over 2 over 2,0) because we didn't move up the y axis. For the other 45 degree angle, i simply already knew that x=radical 2 over 2 and our y=radical 2 over 2, so i put them as the ordered pair (radical 2 over 2, radical 2 over 2) because we moved x units to the right and y units up.
3. Describe a 60 degree Angle
I first labeled the triangle according to Special Right Triangle Rules which is the hypotenuse 2x, the horizontal value x, and the vertical value x radical 3. To have the hypotenuse equaling to one, i divided it by itself and the other sides as well having r=1, x=1/2, y=radical 3 over 2. I labeled the hypotenuse r, the horizontal value x, and the vertical value y.To find the coordinate planes, i applied the same concept as i did for the above angles. I put the ordered pair (0,0) at the 60 degree left corner, then since our x= 1/2 , i knew my ordered pair for the 90 degree angle was (radical 2 over 2 over 2,0) because we didn't move up the y axis and moved x units to the right. For the 30 degree angle, i simply already knew that x=1/2 and our y= radical 3 over 2, so i put them as the ordered pair (radical 2 over 2, radical 2 over 2) because we moved x units to the right and y units up.
4. This activity helped me derive the unit circle because it gave me the coordinates, points, and angles that are within the unit circle as seen below. I understood the foundation of it, allowing me to not only have to memorize numbers but know why those numbers exist. I thought it was important to do this activity because it gave me the basic comprehension of what a unit circle is.
5. The 30, 45, and 60 degree angles are within the first quadrant of the unit circle. The second quadrant is a reflection of the first, the third is a reflection of the second, and the fourth is a reflection of the third. In the first quadrant, all coordinates are positive, in the second quadrant, only the x value is negative, in the third quadrant both x and y are negative and in the fourth quadrant only the y is negative having that the first and third quadrant are positive and the second and fourth are negative. For a better understanding look below.
Inquiry Reflection Activity
1. The coolest thing I learned from this activity was how each quadrant is reflected with triangles.
2. This activity will help me in this unit because understanding and memozing the unit circle is a basic need to solve problems in all the concepts.
3. Something I never realized before about special right triangles and the unit circle is that the whole unit circle consists of many right right triangles that give it its coordinates.
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