I/D #3:Unit Q Concept 1: Using Fundamental Identities to Simplify or Verify Expressions

INQUIRY SUMMARY ACTIVITY

1. Where does sin^2x + cos^2x=1 come from to begin with?

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An identity is a proven fact and/or formula that is ALWAYS true. The Pythagorean Theorem being itself an identity, as it is a proven statement, that is portrayed in a right triangle within the first quadrant of a unit circle. With using the Pythagorean Theorem, in this case, we use the variables x, y, and r not a^2+b^2= c^2. Instead of it being a, b, and c they would be replaced by x,y,r. 

As worked out in the image above all we had to do to get the Pythagorean Theorem equal to 1 is divide by r^2. By dividing by r^2, we got a ratio used/seen in the unit circle of x/r. Which turns out to be the ratio for cosine.  Another important thing seen is that for the term being added we got y/r, which represents sine when speaking in terms of the unit circle. If we were to substitute those for cos and sin we get: cos^2x + sin^2x = 1. With the substitution we have achieved our Pythagorean identity, through the use of Pythagorean theorem.

I chose one of the "Magic 3" ordered pairs from the unit circle to show that this statement is true. The ordered pair used was the 60* angle, which has coordinates of (1/2, rad3/2).

2. How to derive the two remaining Pythagorean Identities from sin^2x + cos^2x=1

The image above shows that how to derive the identity involving Secant and Tangent. First we have to divide everything by cos^2ø, which would cancel out the cos^2ø making it equal 1. Then we would substitute this ---------> sin^2ø/cos^2ø by tan^2ø. To finish it off we substitute this ---->  1/cos^2ø by this sec^2ø. The answer ending up looking like this: tan^2ø + 1 = sec^2ø. 

In this image the work is shown of how to derive the identity involving Cosecant and Cotangent. The first step taken is dividing everything by sin^2ø. Next we would cancel out sin^2ø and it would be left equal to one. Afterwards we are left with cos^2ø/sin^2ø on one side, what we do with that is substitute it with cot^2ø. Next we would substitute ---> 1/sin^2ø by csc^2ø. Leaving us with our final answer of: 1 + cot^2ø = csc^2ø. 

So now we have been able to tackle the task of deriving each of our three Pythagorean identities. ( tan^2ø + 1 = sec^2ø, 1 + cot^2ø = csc^2ø, and cos^2x + sin^2x = 1) 

I/D #2: Unit O - How can we derive the patterns for our special rights triangle?

Inquiry Activity Summary: 

1.  30-60-90 Triangle 

This is an equilateral triangle from which we will derive a 30-60-90 triangle. In an equilateral triangle all the sides are the same measure and so are the angles. Another thing to keep in mind is that triangles always add up to 180, so we can infer that each of the three angles will be 60*. The first thing to do is split the triangle in half, creating a 90* angle. Another change is that with the split we created a 30* angle. As seen in the image below we have made a 30-60-90 triangle. 

In the image above it's seen how opposite to the 30* angle there is now 1/2, this is because each side length was one and by spliting the triangle in half the bottom side got split in half too. So we know that the side across the 30 degrees is 1/2 and the side across the right angle is 1. Now we will find out what the value of the side across from the 60* angle is by using Pythagorean Theorem. 

The image above demonstrates how to use Pythagorean theorem. Giving us a value for the missing side of radical three divided by two. By now we have successfully derives all the values of the sides of a 30-60-90 triangle. (1/2, 1, and radical 3/2). 

But lastly, so we don't deal with fractions we can multiply each side by 2 in order to get whole numbers. By doing so, side (a) would be 1, side (b) would be radical 3, and side (c) would be 2. Next we go ahead and add the variable of n to all the sides. Now across from 30* angle would be n, across from 60* angle would be n radical 3, and across the 90* angle would be 2n. The reason behind placing the variable n is to show the relationship between the sides, "n" merely just representing any number.

2. 45-45-90 Triangle

Here we have a square with all sides being equal (length of 1), and all angles adding up to 360*,each angle having the same value of 90*. In order to get a 45-45-90 triangle is to draw a diagonal line through the square. As seen in the image below with the split we have acquired 45* angles with the split of the 90* angle. 

Now we know two of our sides and need to find the hypotenuse by using Pythagorean theorem. As seen below after having used Pythagorean theorem the renaming side has a value of radical 2. Now we have derived each of our sides' values ( 1,1,radical 2). 

Finally we need to place "n" into the sides of the triangle. By doing so n takes the place of one, for both sides that have the same value. The remaining side that is across from the 90* angle will be n radical 2. The variable n plays the function of allowing the relationship/ pattern among a 45-45-90 triangle to remain constant, no matter what the value of "n" happens to be.

INQUIRY ACTIVITY REFLECTION

1. SOMETHING I NEVER NOTICED BEFORE ABOUT SPECIAL RIGHT TRIANGLES IS 

that I never really thought that the "n" had a distinct role in special right triangles. I alsways believed it to be just a random variable, but now I see and understand how it helps maintain consistency in relationship among the sides. 

2. BEING ABLE TO DERIVE THESE PATTERNS MYSELF AIDS IN MY LEARNING BECAUSE I can personally figure out the values of the sides if I ever have a brain fart during a test. Not only that, but I now have an a reasonable explanation as to how to get every value of each side.

I/D 1: Unit: N- How do SRT and UC relate?

A)

Here we are dealing with a 30* angle in which the horizontal side is x radical 3, the vertical side x, hypotenuse 2x. We start of by dividing everything by 2x because we need hypotenuse to be one. After canceling out the x we are left with a vertical side of 1/2, horizontal side of radical 3/2, and hypotenuse of one. Afterwards we plot it on the first quadrant deriving the ordered pairs of (radical 3/2,1/2), (radical 3/2,0), and (0,0) as the center. 

B) 

This triangle is a 45* angle and the labeling of the sides differs than those of the 30* angle. Here both the vertical and horizontal side are radical 2/2 and the hypotenuse is 1. We the just plot it on the first quadrant and get the ordered pairs of (0,0), (radical 2/2,0), (radical 2/2, radical 2/2). 

C) 

This is a 60* angle and it is similar to the 30* angle it's just that the sides are switched around. The hypotenuse still being 2x, the vertical side x radical 3, and the horizontal side as x. We first divide everything by 2x and we get a vertical side of radical 3/2, a horizontal side of 1/2, and hypotenuse of one. Next we place the triangle on the first quadrant giving us ordered pairs of (1/2, radical 3\2), (1/2,0), and (0,0). 

This activity helps me derive the unit circle by allowing me to know the ordered pair of a 30*, 45*, and a 60* angle. Which are the basic three triangle used in the entire unit circle. By drawing triangles on the first quadrant we get the positive ordered pairs. When triangles plotted on the 2nd quadrant ordered pairs would have a negative x value. When in the 3rd quadrant both x and y would have a negative values. When in the last quadrant (4) the y value will be negative. Other bottom image shows how the triangles are just flipped around depending on the quadrant it's in. 

 

1. THE COOLEST THING I LEARNED FROM THIS ACTIVITY WAS: that the unit circle primarily consists of these three triangles a 30*, 45*, and a 60*. That they only get flipped around depending on the quadrant they are in. 
2. THIS ACTIVITY WILL HELP ME IN THIS UNIT BECAUSE: it's another way to find the ordered pairs in the unit circle. So in case I forget one of them I can use this tool  to help me find the ordered pair. 
3. SOMETHING I NEVER REALIZED BEFORE ABOUT SPECIAL RIGHT TRIANGLES AND THE UNIT CIRCLE IS:  something that I had not realized about right triangles is that they played an important part I. The unit circle. And something that the unit cicle has is that it involves same points just switched around depending on the quadrant.