SP #7: Unit Q COncept 2

In this SP I will show how to write and solve  a problem using two methods: The Identities, ( Ratio, Rational, Pythagorean) and the famous SOHCAHTOA.

SOHCAHTOA

In the image above you can see the work for finding the 6 trig functions using the method of SOH CAH TOA. To find the 6 trig functions we use a certain method of SOH which the OH in SOH is represented opposite over hypotenuse, and we substitute it in for the values/ numbers. TO find the CAH or TOH we use the same logical reasoning. As a result we were able to find the reciprocal to find the other 3 trig functions, which where CSC, SEC. COT.

IDENTITIES

In the identity method to find the 6 trig functions, we had to use the ratio, rational, and Pythagorean identities to solve. First we started of with tangent, which was SINE over COSINE, and basic substiitution of numbers,from there we were able to plug in COT, which equaled to 1/ tangent. The we substituted and we got our answer. Then we used the Pythagorean identity of 1 + tan ^2 theta = Sec ^2 of theta, then we substituted in for tan ^2 theta and then from there we combined like terms to get our answer. To find CSC, we used another Pythagorean identity. Which was 1 + COT ^2 theta = CSC ^2 theta and simplified the same way as finding for SEC.Finally we used the reciprocal identity of CSC and SEC to find SINE and COSINE, from there we basically simplified.

I/D #3: Unit Q: Pythagorean Identities

Inquiry Activity Summary

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

Back in Unit N, we learned about the Unit Circle. In the Unit Circle we learned that the ratio for cosine is (x/r) and the ratio of sine is(y/r). Now, we all know the Pythagorean Theorem which is a^2 + b^2 = c^2,  but with the Unit Circle it would be x^2 + y^2 = r^2, but in the Unit Circle R always equals 1 and to make this true and correct, we would divide by r^2 on both sides, leaving us with (x/r)^2 + (y/r)^2 = 1.  So we can safely say (x/r) and (y/r)  are cosine and sine.  They're squared and because they are squared, we have Pythagorean Identity. This means it cannot be "powered up"or "powered down", also because it is a Pythagorean Identity it must be squared and no power greater or less. Also because the Pythagorean Theorem is a proven fact and the formula is always true, it is called an identity. We can also prove this by demonstrating one of the "Magic 3" ordered pair from the Unit Circle (30*, 45*, 60*). We'll use 6* now theta of 60* is (1/2, radical 3/2) with 1/2 being x and radical 3/2 being Y. To prove our derivation of x^2 + y^2 = r^2 we're going to use cos^2 +sin^2 = 1, so (1/2)^2 + (radical 3/2) ^2 =1.

To derive the remaining two Pythogorean Identities we have to divide the whole thing by cos^2x to find the tangent derivation which will lead us to tan^2x + 1 = sec^2x

Now all we divide the original by sin^2x to find the cotangent derivation leading us to 1 + cot^2x = csc^2x.

Now all this information was found from page 1 of our Unit Q SSS packet, which is information we should Memorize!

Inquiry Activity Reflection

1. The connections that Iv'e seen between Units N, O, P, and Q are:

1. First the role of sine and cosine in Unit Q from the Unit Circle and its properties which can be seen in the reciprocal identities which are similar to the inverse of these trig. functions from Unit O and P. Also it helps to see how all the units are coming along together. 
2. In Unit P Concept 3 when we are using the distance formula for Law of Cosines with SSS or SAS, one of the formulas is a^2 = c^2(sin^2A + cos^2A) - 2bcCosA + b^2 and we know that cos^2A + sin^2A = 1. Helps everything make sense :D 

2. If I had to describe trigonometry in THREE words, they would be: 

challenging, tiring, and confusing. 

WPP #13 and 14: Unit P Concepts 6 & 7: Applications with Law of Sines & Law of Cosines

On a cool December morning, David decided to go skiing out on the alps. He knew that his buddy Thomas had skii lodge and equipment. So Thomas welcomed David and the two went skiing. After a few hours  they decided to head back, but they had gone so far off course, they were lost. Thomas called the Police station, but the heavy blizzard was interfering with the connection. The police needed the two friends exact location but it was so cold and snowy they couldn't give an exact location. Police station B is due south of Station A and they are 100 miles apart. From what David can tell, they are S 45 degrees W of Station A, and N 65 degrees W of Station B. How far is each station to the lonely skiiers?

Here is how the triangle should look like. This a ASA problem since we give two angles and one side in between them.  Since we give two angles ( 65 and 45), we add them up which equals 110 degrees. Then we subtract that value from 180 degrees, to find our missing angle value, of 70 degrees.

By using the Law of Sine to solve for side "a". Angle C, which is where the skiiers are, will be our bridge to solve for sides "a' and side "b". Sin 70/ 100 = Sin 45/ a, we can cross multiply to give us a(sin 70) = 100(sin 450). Then we divide sin 70 to both sides to cancel sin 70 on one side so that it will be a = 100(sin 45)/ sin 70. Station A is 75.2 miles away.

As said before, sin 70/ 100 is our next bridge and it will be used to fine side "b". So sin 70/100 = sin 65/b, we cross multiply to give us b(sin 70) = 100(sin 65). We then divide sin 70 to both sides so that sin 70 cancels on one side so that we can have b = 100(sin 65)/ sin 70. Station B is 96.4 miles away from the skiiers.
Once they got rescued and taken back to Police station A they are taken back home. David and Thomas both leave the station at the same time, they diverge an angle of 100 degrees. If David is 4 miles away from the station and Thomas is 5.5 miles away, then how far away are they from each other?

This picture hopefully sums up the rest of the work. We will use the Law of Cosine to find the missing side. Make sure all the work is right up to this point! Then you write your equation as a^2 = 4^2 + 5.5^2 - 2(4)(5.5) cos 100. After that, we just plug that in to our calculator and we get our answer of 7.3 miles. David and Thomas are both 7.3 miles apart.

CONGRATULATIONS!!:D You have solved the problem {=~^0^~=}

BQ #1: Unit P Concepts 1 and 4: Law of Sines and Area of Obliques

Concept 1: LAW OF SINES 

 The Law of Sines is crucial when we are working on a triangle that is not a right triangle. When we have that we use the Law of Sines to solve it but only if the the triangle is an AAS or an ASA. How to derive the Law of Sines will be shown in these following pictures:

Here we are given a triangle the A,B, and C as their angles and a, b, and c as their sides we can just split the triangle in half to form two triangles.

Lets focus on angle A, angle C, side a, side c and the height, h, I can show you how they can equal each other. If we take the sine of angle A, it will be h/c but since this is a part of a triangle we don't know what h is. So c is then multiplied to both sides and it will give us h = c(sin of angle A). The same thing will happen if we do the sine of C, except it's h/a and we multiply a to both sides and we'll get h= a(sine of angle C). We can get the height in two ways but in the end the height will be the same for the both of them. Since c(sine of angle A) = a(sine of angle C), and we cross multiply, then ( sine of angle A)/a =  (sine of angle C/c.

By staying with the same angles and side but with a different height then we see how sine of B can be the same as the other two.

In this triangle we'll use the sine of angle B and it is k/c but we don't know what k is so we multiply c to both side and we'll get k= c(sine of angle B ). On the other triangle it's the sine of angle C and that's k/ b. Just like the other one we multiply b to both side and get k= b(sine of angle C). But if  c(sine of angle B ) = k and b(sine of angle C)= k then they are both the same. Which mean that if we cross multiply them then (sine of angle B )/b = (sine of angle C)/c and if that is true then (sine of angle A)/a will be the same and all three will work and we'll get the same answer.
Concept 4: AREA OF OBLIQUES.
     The "area of obliques" is derived from the area of a triangle which is A = 1/2 bh. The base is b and the height is h in a right triangle. But if we don't have a right triangle then it will almost be the same but that depends on the side and angle given to us. If we have angle A, side b, and side c given to us then the area formula will be A = 1/2 b(c(sine of angle A). The formula can be rewritten to work with the two other angle. The following images will show which formula to work with for that problem.

WPP #12: Unit O Concept 10

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

INQUIRY ACTIVITY SUMMARY

     To obtain patterns for the 45-45-90 triangles, it's better to use a square given that its side is equal to 1. From there we split the square diagonally and get two 45-45-90 triangles. Now we apply the Pythagorean theorem to get the hypotenuse of this triangle and from there we can see the pattern forming. Now we use the variable "n" to represent the ratio of each side and because it's a variable it can represent any number. The images below will demonstrate how we can derive the pattern of the 45-45-90 triangle.

Here is an image of a perfect square with all sides being equal to 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 orange. 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 into the Pythagorean theorem and we should  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. Almost done c: but we need to plug in "n" to both sides, 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 ( meaning that all sides are equal) 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 down the middle, and get two of them but in this case we are only going to use the one highlighted in pink. 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 3/2. 

This is how a 30-60-90 triangle should look like, the hypotenuse side being 2n, side "a" being n, and side "b" being n-radical-3. Well the values are supposed 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. PLus ther amount of logcial reasoning it took to solve these types of problems.
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, It also helps me review some of my math skills.

I/D# 1: Unit N: Concept 7: How does the Special Right Triangles and Unit Circle relate?

Before we get into all the crazy fun math; we must remember the Rule of Special Right Triangles, which is: The special value of these triangles in their ability to yield exact answers.

Inquiry Activity Summary

#1: The 30º Triangle

In this class activity, we learned where and how to get the plotted points of the Unit Circle's Magic three triangles located on the 1st Quadrant. The triangles are the (30,45,60 degrees) by using the Rules of Special Right Triangles. To start the activity we first are required to label all three sides.This includes getting the Hypotenuse to equal R, the Horizontal value to be X and the Vertical Value to be Y. In this case, the Hypotenuse is equal to one as you can see in the picture we have 2x/2x (as instructed) which gives us our answer of 1 since we can cancel out the 2's leaving us an answer of 1. Then we have to find the other two sides. Which takes us back to knowing  the special rules that apply to this particular triangle. This that the x side is x-radical 3 and the y side is just x. Since we since we divided by 2x in the hypotenuse, we are obliged do the same through out the triangle and then simplify. So x radical 3 over 2x turns to radical 3/2 which is your x value since we can cancel the x values.. Then we do the same for our y value and x/2x turns to 1/2, again by canceling the value of x. Then we draw a graph on the triangle starting from (0,0). So were the 30º opens up will be (0,0), where the 90º open up will be (radical3/2,0) and where the 60º opens up, the value will be (radical3/2,1/2). The whole point of this activity is supposed to make you realize that the unit circle is just the 3 "magical triangles" that repeat it self in different quadrants. So instead of viewing it as a circle, you view it as three triangles. 

#2: The 45º Triangle

Now, on to the 45º triangle, First we do the same basic concept. Label the sides accordingly to the special rules of the 45º triangle Since it is a 45 degree triangle it varies just a little bit. This means that even though the hypotenuse still equals to 1, we must now divide by radical 2 in order to get it to 1. Since this is a 45º triangle, two side are the same so that means both the horizontal and vertical value is equal x. When you divide by a radical 2, you take out the x and leave it as 1/radical 2. Once you have that you must multiply by radical 2 on top and bottom. THERE CAN BE NO RADICAL IN THE DENOMINATOR.  So you end up with radical 2/2 . Again we do the graph that connects to get 45º to equal (0,0), the 90º to equal (radical2/2,0) and the second 45º to be (radical2/2,radical2/2). This again relates to the unit circle because as you go into different quadrants, you're going to notice that they're the exact same points just the (x,y) have a negative/positive sign.

#3: The 60º triangle 

The 60º triangle is very similar to the 30º one. It uses the same rules its just switched around. The Horizontal value is now x and the vertical value is now x-radical 3. With this, we use the exact proceedures as before making 2x/2x=1 (for our hypotenuse) so we divide each side by 2x. If you notice, its the same exact thing with a 30 degree triangle so you basically did the work already (yayy! {=~^0^~=}) However as i mentioned before, it is flipped so our x value is now 1/2 and our y value is now radical3/2. Once you do the graph you realize it is the same points as a 30 degree triangle (but the x and y are just flipped). so where 60º opens up is (0,0), where 90º opens up is (1/2,0), and where 30º opens up is now (1/2,radical3/2).  

4. How does this help derive the unit circle?

 This helps me because instead of trying to kill yourself learning radical this radical that, (which i have been doing with great success but way to much time), you are able to use just 3 different triangles and simply change whether the x is +/- and if the y is +/- which can be easily determined by the quadrant. Plus it makes it more fun to understand that every unit has all these other angles, and sides and with just one quadrant, you know the whole thing!

 5. How do Quadrants affect the values? 

Simple. It all depends on each quadrant, since each quadrant is like graphing a simple point. If the the triangle lies within that specific quadrant, for example; Quadrant II, than the the values will be (-x,+y), if it lies in Quadrant III, then the values will change to  (-x,-y). And if they lie on Quadrant IV, then the values change to (x,-y).

Inquiry Activity Reflection

A: The hardest thing was actually the most fun and that's that I've learned from this activity where all the points in the Unit circle came from. This was especially important because last year as a sophomore, I had surgery and was absent from school for about a month. So i never learned the UNit Circle, or Logs or anything of second semester, so to learn something like this without prior knowledge is important to me c': . But now, as a junior in Math-analysis, it all  makes so much sense rather than just being scared of weird numbers x) 

B: This activity will definitely help me in this units test because, i can simply refer to the Special Right Triangle rules to help me remember anyhting i forgot and finish the unit circle and as long as i can complete quadrant I, i can complete them all. 

C: Something I never realized before about Special Right Triangle and the Unit Circle is that they were both related to each other. Plus I didn't know they existed to be honest. When I first heard of this activity i was a bit scared to fail because i'm not the best at math so for me to remember  all these radicals, and numbers, and degrees it felt like so much >_< but with practice and constant support from the teacher, i was able to figure them out and feel confident in myself and my performance on my test.