I/D #3 Unit Q: Pythagorean Identities

Inquiry Activity Summary:

1. Where does sinx^2+cosx^2=1 come from?

In earlier units we learned about the three basic trig functions which are sin, cosine, and tangent. Sine which means x/r so the y is the side that rises and r is the hypotenuse. R replaces the c^2 in the Pythagorean theorem. So it would really be a^2+b^2=R^2. Cosine we know is y/r   . We know that when looking at the problem above that it really is x/r^2+y/r^2=1. Using the unit circle to help prove that this statement is true is fairly easy we use the 45-45-90 triangle. Which is square root of 2/2, square root of 2/2. When squaring these you get one half plus one half equals one. Which in this case is an identity which means it is a true proven statement. 

2. Show and explain how to derive the two remaining Pythagorean Identities from sin^2x+cos^2x=1

To derive the two remaining identities all you have to do is divide by either cosine or sine. We know that if we divide by the same function we will get one so that is how we get the one on the right side. We will also notice that if we divide by a trig function like sin/cos we simplify that to x/r /y/r we multiply the reciprocal of the denominator to top and bottom to get the new one which is tangent. You do the  same thing for the other identity. 

Inquiry Reflection Activity 

1. The connection I see between Units N, O, and P are that we still have to use the unit circle in some we and that we still use the trig functions even more. 

2. If I had to describe trigonometry in 3 words, they would be challenging, depressing, frustrating. 

I/D #2: Unit O: How can we derive the patterns for our special right triangles?

Inquiry Activity Summary:

For this activity we had to derive the special right triangles not from the unit circle though. We had to derive the two special kinds ,45-45-90 and 30-60-90 right triangles. Each are completely different we had to find out what n was and why n couldn't just be a number. 

How can we derive the 45-45-90 triangle from an square with a side length of 1? 

We know that if we are given a square with side lengths of 1 we will have two right triangles if we split the square in half from the corners. For a 45-45-90 triangle we know the base and height are the same,1. If we do the Pythagorean theorem we will see that the hypotenuse will equal radical 2. N is there to be any given value, without n the triangle sides cannot be altered to match the initial constants, it is multiplied by the initial constants. For example if n were to equal 2 then the base and height are 2. The hypotenuse will equal 2 radical 2. 

How can we derive the 30-60-90 triangle from an equilateral triangle with a side length of 1?

To derive the pattern for the 30-60-90 triangles we have to cut an equilateral triangle in half straight down the middle. Each side length of the triangle is one so when we split the triangle in half the base turns into 1/2. We also split the triangle in  half to get 30 degrees as one of the angles. Then we will have a 30-60-90 triangle. Then we notice that we do not have the height of the triangle, we have to use the Pythagorean Theorem. Once you get the constants they can be altered to get rid of the ugly fractions by multiplying by two ( look at picture). The constants are the same because they are all proportional and everything was multiplied equally. 

Inquiry Activity Reflection: 

 Something I never noticed before about special right triangles is how we have to tweak equilateral triangles and squares to get the sides for n. 

Being able to derive these patterns myself aids in my learning because now I can refer to this square or triangle if I ever forget what the sides of the triangle equal. 

I/D #1: Unit N: Concept 7: How do special right triangles and the unit circle compare?

1. The 30* Triangle

The 30* triangle has three different sides to it: adjacent which is x, opposite which is y, and hypotenuse which is r. The side opposite the hypotenuse will always be x.The hypotenuse must be one if you want to derive the unit circle from the triangle. If you want this to happen you have to divide each side by 2x. Once you do this you will get x=radical 3 divided by 2, y=1/2 and r=1. We can use these simplified values as coordinates to determine where 30 degrees lies in a quadrant on the unit circle. We know that 30 degrees on the unit circle is located on radical 3/2, 1/2. This can be used for 150 degrees, 210 degrees, and 330 degrees. The only difference is that they are located on different quadrants and there will be negatives and positives.

2. The 45* Triangle

The 45* triangle has two sides that are the same length which are x and y the hypotenuse is r. To derive the unit circle from the triangle we have to get the hypotenuse to equal 1. In order to do this we divide every side by  x radical 2. Once we get one on the hypotenuse we can get r and figure out the points. For the angle of 45 degrees we plot radical two over two, radical two over two. This is where it will lie on the unit circle. This will also stand for 135, 225 and 315 degrees. The only differences are the quadrants the negatives.

3. The 60* Triangle

 

The 60 degree triangle is the same as the 30 degree triangle it has three different sides. In order to derive this triangle from the unit circle is to divide the hypotenuse to get one. This is the same as the 30 degree triangle because you divide by 2x. When you get your final answers you will be able to plot the points. The points for 60 degrees are 1/2, and radical 3/2. These rules also apply for 120,240, 300 degrees.

4.
This activity helps me derive the unit circle because the triangles reflect different points on the unit circle throughout all of the four quadrants. Each of these triangles can be found in all of these quadrants and are all the same the only exception is that there are negatives and positives and they are located in different quadrants too.

5.
The triangle in this activity lies in quadrant one both the x and y values are positive which means that it is in quadrant one.

Inquiry Activity Reflection:
1. The coolest thing I learned from this activity was how you can find the points on the unit circle by using the special triangles.
2. This activity will help me in this unit because it can help me memorize where different points are and where some points lie on different quadrants.
3.Something I have never realized before about the special right triangles and the unit circle are that both these are in relation to each other when the hypotenuse is equal to one