## Wednesday, March 26, 2014

### This blog post was made through the infallible alliance that is between the juggernaut forces of Omar "Ice" T. and Michael "Vitamin" C. Any illegal publishing of this copyrighted post without the consent of the above mentioned is strictly prohibited and is punishable by permanent imprisonment or death (whichever comes first). To see various other posts by Omar "Ice" T., feel free to click here.

This is our problem. We are given that cotangent is 7/8 and sine is negative. Also we know that the problem lies in the third quadrant since tangent is positive and cosine is negative according to ASTC. We use what we know about identities to solve for the rest of the trig functions. We will then check our answers by solving with the unit circle, which we learned in Unit N or by using SOH CAH TOA from Unit O.
One of the easiest things to do is find tangent which is the inverse of cotangent. Knowing this, when we use the reciprocal identities we find that tangent is equal to 8/7.
Now that we know tangent we can use a Pythagorean identity to find secant. Secant ends up being equal to radical113 divided by 7.
Cosine is the inverse of secant, which we just found, so we are able to use a reciprocal identity using secant to find cosine. Cosine will be 7 rad 113/ 113.
First here we find sine using a Pythagorean identity and we can do this since we have found the value of cosine. We then find cosecant by doing the inverse of sine, also known as using a reciprocal identity.
We have now found all the trig functions using identities. Note that all the answers except tangent and cotangent are negative because the problem lies in quadrant III.
Here is the same problem done using SOH CAH TOA. Note that the answers are identical with the answers found using identities.

## Inquiry Activity Summary:

1). An identity is a proven fact or formula that is always true. Pythagorean Identities are proven to be true and are one type of identities that we are studying in this concept. They can be easily derived through our knowledge of the Pythagorean theorem and the Unit Circle. The following will show precisely how I have done just that.
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• So basically what I did was I wrote out the Pythagorean theorem and then made it equal to 1 by dividing everything evenly by r squared. I then used what I knew of the trig functions, specifically the ratios of sine and cosine to rewrite the equation. The end result was the first Pythagorean identity.
• The next thing I did was I tested the truth of the equation. I used what I knew of the Unit Circle to get a point (in this case the point for 45 degrees) and plugged in the cosine and sine values of that point into the equation. The equation met expectations, meaning that the equation is indeed an identity.

2). The other two main identities of the Pythagorean Identities can be derived by
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simply by dividing by either cosine squared or sine squared from the first identity.
• After dividing by cosine, the cosines will cancel into a 1 and using the ratio and reciprocal identities we find that we are left with tangent and secant. The result is the second Pythagorean identity.
• After dividing by sine we find that like the other one, something cancels into a one and the trig functions change (this time into cotangent and cosecant). The result this time is the third main Pythagorean identity.

## Inquiry Activity Reflection:

1). The connections I see between Units N, O, P, and Q so far are that the identities are related to the Unit Circle (Unit N) in that we use an r value of 1 for the Pythagorean theorem and we use the sine and cosine of a point on the unit circle to test the equation and that we use the relationships of the various trig functions to do the derivations. We also use the Pythagorean in the other derivations we have done for the various other units.

2). If I had to describe trigonometry in three words they would be: integrated, complex, and understandable.

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## Tuesday, March 18, 2014

### WPP #13-14: Unit P Concepts 6-7: Law of Sines and Law of Cosines

This blockbuster of a blog post was a co-op with the ingenuous swag of Omar T. To view the actual blog post along with some more swaggy content, please do not hesitate to click here.

## Saturday, March 15, 2014

### 1). Deriving the Law Of Sines:

The Law of Sines is used to solve non-right triangles, specifically ones in which we already know or are given two angles and a side either after the two angles or between the two angles (non-right triangles with AAS or ASA relationships). This is useful because most triangles are not right triangles.  The Law of Sines is as depicted below:
﻿﻿
 http://www.mathwarehouse.com/trigonometry/law-of-sines/formula-and-practice-problems.php
We can derive the Law of Sines with a normal oblique triangle. The following are the steps by which this is done:

 http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm

• We first draw a non-right triangle and label the sides and angles respectively.  We then draw a height (h) perpendicular down from angle B, making two right triangles.

• Now, using basic trig functions, we could see that sinA = h/c and sinC = h/a. If we solve for "h" in both equations we get csinA = h and asinC = h. Since both csinA and asinC are now equal to h, they are also both equal to each other (Transitive Property).
• Set csinA and asinC equal to each other. Divide by ac on both sides and we get the Law of Sines (sinA/a = sinC/c). Say we used  angle B instead of angle C, we should get sinA/a = sinB/b , and since they are all equal to each other, the order can be switched around to make the whole Law of Sines: sinA/a = sinB/b = sinC/c.

### 4). Area Formulas: Deriving the Area of an Oblique Triangle formula

The formula for the area of a triangle is A = 1/2bh and this works fine with right triangles since h is just one of the sides. But with an oblique triangle, h, which is the height of the triangle, is a little more complicated. Thus there is a distinct way to find the area of an oblique triangle, in which the value of h is unknown. The following steps walks through how we come up with this distinction in the area formula:
•  First draw your oblique triangle and label the sides and angles. Draw the height of the triangle (h). We know that the area of the triangle is 1/2 times the product of the base and the height, but what's h?
• Similar to what we did for the law of sines, if we take the sine of angle C we see that sinC = h/a. We can then solve for h and we get h = asinC. We now have a value for h, which we can plug back into the area formula.
• We have taken our value for h and plugged it back into the area formula. All you need to fund the area using this formula is two sides and the sine of the angle between them (SAS), in this case our two sides would be side b and side a, and angle C would be between them. We can see that this is essentially the same formula as the original area formula for a triangle; the only difference is we have to find and use a certain substitute value for h because the height is not always given in an oblique triangle.

### References:

First Picture: The Law of Sines: http://www.mathwarehouse.com/trigonometry/law-of-sines/formula-and-practice-problems.php

Second Picture: Deriving Law of Sines: http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm

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## Wednesday, March 5, 2014

### WPP #12: Unit O Concept 10: Elevation and Depression Word Problems

 http://art-sci.blogspot.com/2012/02/statue-of-liberty-gone-wild.html

a). Sir Liftsalot just moved to Newark and is staring intently at the beautiful Lady Liberty from a distance of 19 feet. He looks up at a 71 degree angle in order to see the very top of that attractive statue. How tall is the statue? (not to scale)

b). Sir Liftsalot gets an awesome idea to take selfies from the top of the statue's head so that's exactly what he does. As he sends the selfies, he looks down from the statue at a friend who is 54 feet away from the statue on the ground. What's the angle of depression from Sir Liftalot's eye to his friend? (Assume that Sir Liftsalot's eye is 5 feet above the top of the statue and also be sure to get the height of the statue from the answer for part a).

Part a).
1). First draw a picture of the problem. He is 19 feet away from the statue and he is looking up at 71 degrees. X represents the height of the statue, which we will be finding.
2). We have an angle, its adjacent, and its opposite so we will use tangent of 71 degrees = opposite/adjacent = x/19.
3). We want to solve for x so first thing we do is multiply by 19 on both sides.
4). We then get x is equal to 19 times the tangent of 71. Plug this into your calculator (make sure it's set to degrees!) and we get x = 55.18 (rounded) which means the statue is 55.18 feet tall.

Part b).

1). Draw the picture like so. Since we are using the height from Sir Liftsalot's eyes we will add on another 5ft to the height of the statue for an overall vertical length of 60.18. The horizontal value is 54ft since the friend is 54ft away. X represents the angle of depression, which we will be finding.
2). We know our opposite and our adjacent sides so we are going to use tangent to find the missing degree, x. We set this up as tangent of x = 60.18/54 as shown above.
3). Here is the remainder of the work. We will do the inverse of tangent on both sides to isolate x. We find that x is equal to 48.1 degrees (rounded). This means that the angle of depression was 48.1 degrees.

## Inquiry Activity Summary:

In class we were given a square with side lengths of 1 and an equilateral triangle, also with side lengths of 1. We were given directions to derive the patterns for the 45-45-90 and 30-60-90 triangles from the square and the equilateral triangle using only what we knew of the two shapes (ignoring any pre-knowledge of the 45-45-90 and 30-60-90 patterns). The following is how I accomplished this heroic task:

### 1). Deriving the 30-60-90 pattern with an equilateral triangle:

First things first, you want to draw an equilateral triangle. We know the sides are all equal to 1 and we know that the three angles are all equal to 60 degrees in this equilateral.﻿﻿﻿﻿﻿﻿﻿﻿﻿
Now in order to begin our tedious construction of a 30-60-90 triangle, we will slice the above triangle equally in half right down its altitude (top middle). Several things will change; the first and most obvious being that we will have 2 triangles. Second, the top 60 degree angle will be cut into two 30 degree angles (one for each triangle). And lastly, the bottom side of the original triangle will also split in half, resulting in two horizontal side lengths of 1/2 (again, one for each triangle). Also important is that the intersection cuts the triangle perpendicularly, meaning that you will have two 90 degree angles on each of the two new triangles. [Note: If you haven't noticed by now, we will be deriving not one 30-60-90 triangle, but two.] Take notice that we already have the 30-60-90 degree relationship established by this step.

Next on the agenda would be to find our missing side: the vertical value which is known as "y". We will use only one of the two triangles (doesn't matter which) and the Pythagorean theorem to find that missing side. We do the math and we find that the missing side (y) is equal to (radical 3)/2. This is demonstrated below. We now have derived every part of the 30-60-90 triangle but we are still not done just about yet.

The side values we have on the triangle right now, although they are correct, are not what we are used to seeing in the 30-60-90 pattern. We must do away with the fractions and we do this by multiplying ALL the sides by 2. The results are shown below:

Now all that's left is to place in "n" by multiplying it to ALL sides. We do this because since we are deriving a pattern, we don't want just a set of numbers, we want a set of constants that make up the rule of the formula. The below picture is the 30-60-90 triangle that we have fully derived from the equilateral:

### 2). Deriving the 45-45-90 pattern from a square

Okay. First I drew the square. Its sides will be one and like all squares this one will have 4 right angles.
Next we cut the square equally down its diagonal. The results will be two right triangles, and as shown, two of the right angles will be cut in half into two 45 degree angles. We can already see the 45-45-90 degrees in out triangle, and we have two of the sides; we just need that third missing side again. But also take note that the two sides, like the two angles, are equal; this is because the sides correspond with the angle opposite to it and since they are equal angles, there will be equal sides.

Now we use the Pythagorean theorem to find that missing side. Again this is demonstrated below. We shall find that the missing side (this time "r") is equal to radical 2.
Now we multiply by "n" to get our constant, which represents the overall relationship and pattern that this special triangle follows.
And now we can separate the triangle and see that we have fully derived the 45-45-90 triangle pattern from the square.

## Inquiry Activity Reflection:

1). Something I never noticed before about special right triangles is... that they come from a square and an equilateral triangle, and that the patterns they follow are not just numbers that happen to work every time, but they actually came from what we did in this activity.

2). Being able to derive these patterns myself aids in my learning because... learning something isn't just memorizing something with no meaning; this helps me understand what special right triangles really are.