I/D#3: Unit Q Concept 1: Pythagorean Identities

INQUIRY ACTIVITY SUMMARY:
     Pythagorean identities come from both the Unit Circle and the Pythagorean Theorem. First to clarify, an identity is "an equation that is true no matter what values are chosen". The Pythagorean Theorem is seen as an identity because no matter what two values you have and when you solve it it will give you the third value. If you check your work by using the found value and a value given, the answer will be that same given value you did not use. To get the Pythagorean identity, sin^2(theta) + cos^2(theta) = 1, I will work it out using the Unit Circle and the Pythagorean Theorem and I will show and explain it in the following pictures:

First I'll show it by using the Unit Circle using the first quadrant since sine, cosine, tangent, co-secant, secant, and cotangent are all positives. If we plot a triangle in the quadrant we can already see the connection between the two. Th height will be "y" since it follows the y-axis, "x" will be base since it's on the x-axis, and the hypotenuse will be "r" because the hypotenuse is the radius of the circle. When it gets plugged in to the Pythagorean Theorem "a" will be "y", "b" will be "x" and "r" will be "c".

When I use "x,y and r" in the Pythagorean Theorem it will make sense if it's written out like y^2 + x^2 = 6^2 because when we try to make it equal one then it will make sense. If I take the sine of angle A it will be y/r, the cosine of it will be x/r, and "r" will stay as "r". Now I can show how (y/r)^2 + (x/r)^2 = r^2 can equal one but there is a faster and easier way to get one. If I write out "opp" as opposite, "adj" for adjacent, and "hyp" for hypotenuse and then in Pythagorean form, it's opp^2 + adj^2 = hyp^2. If I divide both side by hyp^2 then the hyp^2 will be 1 and it will be opp^2/ hyp^2 + adj^2/ hyp^2 = 1. If I pug it in to the other equation I said, opp^2/ hyp^2 is the cosine of angle A and adj^2/ hyp^2 is the sine of angle A. So cos^2(theta) + sin^2(theta) = 1 is one of three Pythagorean identities.

To find the second one, which is the identity with secant and tangent, I will use cos^2(theta) + sin^2(theta) = 1 and divide both sides by cos^2. The cosines will divide to be 1, sin^2/cos^2 is the Ratio identity of tangent(theta), and 1/cos^2 is the Reciprocal identity of sec(theta). So we end up with 1 + tan^2(theta) = sec^2(theta).

    

To find the third Pythagorean identity that has cotangent and co-secant, it will be just like the second one except that I will divide it by sine^2. If I divide both side by sin^2 to both sides then cos^2/ sin^2 will be the Ratio identity of cotangent, sin^2/ sin^2 will divide to 1, and 1/ sin^2 is the Reciprocal identity of csc(theta).

INQUIRY ACTIVITY REFLECTION: 
1) The connections that I see between Unit N, O, P, and Q so far are the angles that can be found in the quadrants of the Unit Circle and how the triangles that are made in the quadrants can be used with the Law of Sine And the Law of Cosine to find any missing angle or side length.
2) If I had to describe trigonometry in THREE words, they will be hard, understandable, and progressive.

I/D #2: Unit O- Derive the SRTs

INQUIRY ACTIVITY SUMMARY
     To get the patterns for a 45-45-90 triangle, it's better to use a square that has equal side of 1. From there we can split the square diagonally and get two 45-45-90 triangles. From there we can use the Pythagorean theorem to get the hypotenuse of the triangle and from there we can see the pattern starting to form. Now we use "n" to represent the ratio of each side and as a variable to represent any number that can take it's place. These next few pictures will show how we can derive the pattern of the 45-45-90 triangle.

THESE SET OF PICTURES ARE FOR A 45-45-90 TRIANGLE ONLY:

Here we have a perfect square with each side being 1 and each corner being 90 degrees. To get the 45-45-90 triangle we are going to do one step to get the triangle and the Pythagorean theorem to get the missing side.

To get our triangles we just split the square diagonally but we are just going to use the triangle highlighted in green. We already have two sides, the horizontal and the vertical side, and each side is 1. The triangle is still incomplete because we need the hypotenuse and to find it we'll use the Pythagorean theorem to find it.

It doesn't matter which side is a or b because both sides are the same. We plug it in the the Pythagorean theorem and we have 1^2 + 1^2 = c^2. We square the 1's and add them and we get 2 on that side. Now it's 2 = c^2, we square root each side making 2 into radical 2 and c^2 to c. So radical 2 will be our hypotenuse for the 45-45-90 triangle. We nearly done but we need to plug in "n" to each side, so the vertical and the horizontal sides will be "n" and "n-radical-2" for the hypotenuse side. 

 THESE SET OF PICTURES ARE FOR A 30-60-90 TRIANGLE.

Here we have an equilateral triangle with each side length of 1 and each angle being 60 degrees. To get a 30-60-90 triangle we split the triangle in half and get two of them but in this case we are only going to use the one highlighted green. The hypotenuse is already there so it's length is 1, the horizontal is also there but it's not 1 because we split it in half so it's actually 1/2. The third side will be found by using the Pythagorean theorem.

In this case, side "a" will be the horizontal side and side "b" will be the vertical side. When we plug it into the Pythagorean theorem, 1/2 will be squared and it will be 1/4 and 1^2 will just be 1. So we then subtract 1/4 to both sides and end up with b^2 = 3/4 but we need to square root both side to make b^2 to just b. When we square root 3/4 it applies to the top and the bottom, which means that 3 will be radical-3 and 4 will be 2 since the square root of 4 is 2. Our final answer for side"b" will be radical-3/4.  

This is how a 30-60-90 triangle is suppose to look, the hypotenuse side being 2n, side "a" being n, and side "b" being n-radical-3. Well when I left off in the other picture it was suppose to be radical-3 for side "b", 1/2 for side "a", and 1 for the hypotenuse. To get it to be the derived pattern we just multiply 2 to each side, so the hypotenuse will be 2, side "a" 1 because the 2's cancel each other, and radical-3 for side "b" because the 2's also cancel each other. The "n's" are put in to show that any number can take it's place.

INQUIRY ACTIVITY REFLECTION
 1. Something I never noticed before about special right triangles is how they are derived form other shapes like the square and the equilateral triangle.
2. Being able to derive these patterns myself aids in my learning because it shows that I know how to derive them and how they came to be a 45-45-90 and 30-60-90 triangle.

I/D #1: Unit N SRT and UC

INQUIRY ACTIVITY SUMMARY
    SRT really do connect with the UC, because they point out the ordered pairs for that specific angle and it tells us how we get that ordered pair. We have three ordered pair in a quadrant and it does not include the quadrant angle. For 30 degrees, 45 degrees, and 60 degrees they all have different ordered pairs. For the set of pictures that are going to be presented, they will be labeled as a SRT should be  and how to get it's ordered pair.

Hypotenuse or "r" will be in blue.
Horizontal side or "x" will be in pink.
Vertical side or "y" will be in green.

THESE SET OF PICTURE IS FOR A TRIANGLE WITH AN ORIGIN POINT 30 DEGREES ONLY.

Labeled as a SRT, "r" is 2x, "y" is x, and "x" is radical 3; but the hypotenuse but be 1.

 

Since "r" has to be 1, the only way to make it 1 is to divide 2x by itself. If we do that to one side, all the sides must be divided by 2x. As we divide, the x's will cancel for the "y"side, x is really 1x and it's divided by 2x so it will equal 1/2. For the "x" side, the x's cancel and we are just left with radical 3/ 2 and that will be our x value.

We have the triangle in it's new form with it's new values for "r", "y", and "x" and it's on a coordinate plane. The origin, which is 30 degrees, will be (0,0). As we find a point for the triangle on the coordinate plane, the hypotenuse and "y" intersect and that is where our order pair lies. Since radical 3/2 is "x"'s value and 1/2 for "y"'s value, that (x,y) point on the coordinate plane will be the same. So our ordered pair for a triangle with the origin of 30 degrees will be (radical 3/2, 1/2).

THESE SET OF PICTURE IS FOR A TRIANGLE WITH AN ORIGIN POINT 45 DEGREES ONLY. 

This is how a triangle with the origin point of 45 degrees looks like when it's labeled as a SRT. Just like the one before it, the hypotenuse has to be 1.

The hypotenuse of the triangle has to be 1 so what we do is multiply the reciprocal of x. X is not by itself, it's x/1 so when we multiply the reciprocal of it, which is 1/x, we'll get our 1. We'll have to multiply the 1/x to the other two sides. For the "y" side, it's x radical 2/ 2 and we multiply 1/x to it. The xs' cancel and we end up with radical 2/2 and that its value for the "y" side. The same goes for the "x", we do the exact same thing and radical 2/2 will be the value for "x" also.

This is the new version of the 45 degree triangle with its new side values. As we plot it to a coordinate plane, the 45 degree angle will be on the origin which is (0,0). With the ordered pair being (x,y) it will be just like the other angle before this one. The x value be radical 2/2 because it goes radical 2/2 to the right and radical 2/2 will be y because it goes up radical 2/2. So (radical 2/2, radical 2/2) will be our ordered pair.

THESE SET OF PICTURE IS FOR A TRIANGLE WITH AN ORIGIN POINT 60 DEGREES ONLY.

These triangles are just like the 30 degrees triangles that I did in the beginning, except that they have some parts switched around. This is how it looks like when it's labeled like a SRT, the hypotenuse is still 2x. The "y" side is now x radical 3 and the"x" side is just x.

Like the other two, "r" has to be 1 and we will have to divide 2x by itself to get 1. We will also have to divide the other two by 2x. For the "y" side, x radical 3 will be divided by 2x. The x's will cancel leaving us with just radical 3/2 and that is our answer for that side. The "x" side just has x and being divided by 2x the x's will just cancel and it will leave us with 1/2. X is not just x by itself, it's 1x so that's how we get 1/2.

Here it is on on a coordinate with 60 degree angle as the origin point. To get the (x,y) ordered pairs we will have to use the values of the "x" and "y" sides of the triangle. Since we go 1/2 to the right because that is value of the "x" side, 1/2 will be our x. Then we will go up, the value of the "y" side is radical 3/2 and that will be our y. So our ordered pair for this triangle will be (1/2, radical 3/2).

     From this activity and the explanation that I gave for each type of Special Right Triangles, helps derive each ordered pair that is on the Unit Circle. It shows how we get an ordered pair for a 30 degree, 60 degree, and 45 degree angle that is on the Unit Circle.
     For these types of angles and their ordered pairs, they are just in the first quadrant of the unit circle. Now the following pictures will show the changes of these triangles it the other three quadrants:

30 degree angles will be in blue.
45 degree angles will be in green.
60 degree angles will be in pink.

This is how all of the look like in the first quadrant, each degree with it's ordered pair. All of the ordered pair are all positive because they are on the positive side of the Unit Circle. Well in the Unit Circle we are just using a circle with a radius of 1. So x and y in quadrant 1 are both positive which means they are also positive.

 

In quadrant 2 I will just show the 30 degree angle and its ordered pair. In the second quadrant, x is now a negative. Which means that radical 3/2 is now a negative but the y is still a positive so 1/2 stays the same. This applies to the other two angles, what they have as x will be negative while the y stays positive.

This is in quadrant 3 and I'm using the 45 degree angle to show how its ordered pair changes from the first quadrant to the third. In the third quadrant both x and y are negative so the ordered pair of the 45 degree angle are negative. Both x and y on the coordinate plane are negative so every angle with an ordered pair in the third quadrant will be negative.

In the last and final quadrant, quadrant 4, I will use the 60 degree angle and its ordered pair to show the changes from quadrant 1 to quadrant 4. Well in this quadrant the x is positive and the y is negative. So 1/2 is x and it will stay positive while radical 3/2 is y, which means that it will be negative. This applies to the other two ordered pairs, their x's will be positive while their y's will be negative.

INQUIRY ACTIVITY REFLECTION 
1. The coolest thing I learned from this activity was the fact that the SRT tells us how these triangles get their ordered pairs and how change in each quadrant.


2. This activity will help me in this unit because it reveals where the ordered pairs are in each quadrant depending on their angle and which quadrant they lie in to determine if they are positive or negative.


3. Something I never realized before about special right triangles and the unit circle is that they both need each other to work. With out the SRT, there will be no ordered pairs in the UC and with out the UC the SRT will just apply to triangles and nothing else.