BQ #7: Unit V: Derivatives and the Area Problem

1.  Explain in detail where the formula for the difference quotient comes from now that you know! Include all appropriate terminology (secant line, tangent line, h/delta x, etc).

The difference quotient is a common expression in calculus. It is used during the concept of derivatives. The formula is 

(http://www.jcu.edu/cspitzna/joma/livemath/diffquo.htm)

But I bet you're wondering how this equation is derived! Well it's easy!

(http://cis.stvincent.edu/carlsond/ma109/diffquot.html)

The difference quotient is used to find the value of the slopes and curves of all tangent lines. Remember that the slope formula is (y-y)/(x-x), this is also known as the rise over run when looking directly at a graph so we can find it much easier. There are also two types of graph which is one with a tangent line, only touches the graph once, and a secant graph, touches the graph twice! In the above picture, it is a secant line graph in which we must get it to a tangent line where only one spot is touched. We get the formula by subtracting the furthest point to the closer one. The y-axis is the f(x) while the x axis is noted as x. "H" is the change of x which can be also noted as "Δx". Our first point is (x, f(x)) while our seond one is (x+h, f(x+h). (We get this point from adding the change of x to x for the total distance on the graph and since y is just f(x), we plug in x which is x+h into it to get f(x+h). So now, plug it into the slope formula of (y-y)/(x-x) and you get [f(x+h)-f(x)]/ [x-x+h]. Once you simplify the denominator (where the x's will cancel out) we get the difference quotient of [f(x+h)-f(x)]/h!

Once we find the derivative by using the difference quotient, it is written as f'(x) or "f prime of x". When you have a derivative, you can find anything, from the value at a certain point to even the tangent line equation, and even the slope value.

To get a full understanding of this concept and example of solving a derivative, please check out IntuitiveMath's video below!

Referenceshttp://www.jcu.edu/cspitzna/joma/livemath/diffquo.htmhttp://cis.stvincent.edu/carlsond/ma109/diffquot.htmlhttps://www.youtube.com/watch?v=iMaJDAV7as0

BQ #6- Unit U

1. What is a continuity? What is a discontinuity?

A continuity is a function that is predictable, has no breaks in the graph, no holes, no jumps and can be drawn without lifting your pencil. 

A discontinuity is distinguished in two different groups: removable discontinuities and non-removeable discontinuities. There is only one discontinuity known under the removable one which is called the point discontinuity. At the point of the open circle on the function, it is known as a hole, while the other point (which is closed) is on the same vertical line near it. 

(http://image.tutorvista.com/content/feed/u364/discontin.GIF)

As for the non-removable discontinuity, there are three known as the jump discontinuity, oscillating behavior, and infinitie. In a jump, it's exactly how's named. The graph jumps from one point of a function to another. However, we need to keep in mind that an open/open and closed/open circles work, however closed/closed do not. 

(http://web.cs.du.edu/~rjudd/calculus/calc1/notes/dis6.png)

In an oscillating behavior, there are wiggly lines.

 

(http://web.cs.du.edu/~rjudd/calculus/calc1/notes/dis3.png)

For infinite discontinuities, it is known as an unbounded behavior that occurs when there is a vertical asymptote.

2. What is a limit? When does a limit exist? When does a limit not exist? What is the difference between a limit and a value?

A limit is the intended height of a function. A limit exists when the same height is reached from both left and right hand limits. It also exists when a grpah does not break at a given x-value. Last, it can also exist if your ultimate decision is a hole in the graph (point discontinuity). However, a limit does not exist when there are different left and right hand limits (jump discontiuity). It also does not exist when it is unbounded because infinity is not a number. Last, it doesn't exist when it doesn't approach any single value. The difference between a limit and a value is that a limit is the intended height of a graph while a value is the actual height. When you look on a graph, it can be represented by open or closed circles. Most of the time, the intended height will be represented with open circles while the actual value will be with closed circles. However, remember that a limit and value can be at the same spot on a function!

(http://curvebank.calstatela.edu/limit/grid3.gif)

For example, from this graph we know at x=2, that the INTENDED height is 3, however the actual value is 2.

3. How do we evaluate limits numerically, graphically, and algebraically?

Numerically- We can evaluate limits numerically with a table. By having our given middle value, which we can take from the "x approaches a number", that number will be our middle. All we have to do is either add or subtract one tenths depending on which way the limit is coming from. The two answers we get will be our outside answer and all we have to do is add 0's in or 9's depending where you are. To find the value of each number, all we have to do is plug in the equation in the "y=screen" and hit graph and trace the x-values to find the actual values. 

(This example is taken off the SSS packet and the template off the given one online)

Graphically- To solve a limit graphically, we will obviously need a graph. First you need to put your finger on a spot to the LEFT and to the RIGHT of where you want the limit to be evaluated. If your fingers meet, then the limit exists, however if it's does not, then the limit does not exist. This method can be the most useful because there's a visual representation of it which will help some people see it, like the points and different kinds of discontinuity.

Algebraically- When solving algebraically, there are three different "shortcuts" to evaluating your limits. The first method is direct substitution which is exactly what is says it is. You directly substitute the given value of x into the equation and solve. The four possible answers you can get are 1) a numerical answer 2) 0/# which is 0 3) #/0 which is underfined and 4) 0/0 which is the indeterminate form (this means we'll need to use another method to solve it).The second method is dividing/factoring out which can be used if AFTER direct substitution results in an inderterminate solution (0/0). All you have to do is factor out the nomials and hopefully cancel out any common terms to remove the zero in the denominator. Last, there is the rationalizing/conjugate method! Wherever there is a radical, we just take the conjugate of it and multiply it on both the numerator and denominator. The only time we FOIL is with the conjugate denominators. We DO NOT FOIL the non-conjugate denominator because it will eventually cancel out (That's what we want to do!)

REFERENCES: Unit U SSS Packet

BQ#4 – Unit T Concept 3

Why is a “normal” tangent graph uphill, but a “normal” Cotangent graph downhill? Use unit circle ratios to explain.

Both tangent and cotangent's graph direction is due to their asymptotes. Asymptotes lie on the x-axis where x=0. Finding the direction of the graph can be determined by the trig ratio and unit circle.

*color stands for each quadrant in order from 1-4*
(template taken from Unit T Intro Desmo Activity & drawn on by self) 

For tangent, the ratio is y/x. To find the asymptote, x would have to be 0 to be undefined. So when x is 0, we find that the asymptotes lie at (0,1) and (0,-1) which in radians would be pi/2 and 3pi/2. Knowing the asymptotes, we can start on the direction of the graph. Back to the Unit Circle, we know that tangent is only positive in the first and third quadrant and negative in the second and last quadrant. In between the asympotes that we found are the second and third quadrant. Since we know that it's negative in the second, then positive in the third, it goes uphill. The graph will eventually start to repeat itself as the period goes on.

However, for a cotangent graph, it goes down hill. The ratio of cotangent is x/y. Again, asymptotes have a denominator of 0 to make it undefined so we have (1,0) and (-1,0). In radians, it's 0pi and pi. So on our x-axis, these points would be where the asymptote lines are. We know that cotangent is just like tangent in it's positive and negative location. Quandrant I and 3 are positive while quadrant 2 and 4 are negative. We see that the asymptote boundaries are in the first and second quadrant. So it will go from a positive position, down to a negative in the second quadrant. This is how we get our downhill shape.

BQ#3 – Unit T Concepts 1-3

How do the graphs of sine and cosine relate to each other the others? Emphasize asymptotes in your response.

Sine and cosine are both part of the trig ratio and reciprical identities. All the functions have sin/cos in them so that's how they ultimately relate to each other. When a value in the denominator is equal to zero, we know that is undefined and that is where our asymptote is since asymptotes are undefined. 

Tangent? We know the ratio identitiy is sin/cos. On a Unit Circle, we know that the graph is positive in the first and third quadrant while it is negative in the second and fourth. This leads the the graph to have asymptotes at pi/2 and 3pi/2 since that's where cos=0 and we know anything over 0 is the asymptote. From there, the second quadrant starts with a negative and then in the third quadrant, a positive period. This gives tangent the uphill graph.

Cotangent? As for cotangent, the ratio is opposite of tangent (cos/sin). When sin is 0, we find our asymptotes at 0 and pi on the graph. The period start from being positive in the first quadrant to ending in the negative by the second quadrant on the graph. This gives cotangent its downhill graph. Since both tangent and cotangent only cover up pi units in the x-axis, we see it repeats itself right after an asymptote is drawn.

Secant? Secant is the reciprocal of cosine and has a ratio of 1/cos. So when cos is 0, that's where the asymptote lies. This means it would lie at pi/2 and 3pi/2. In the Unit Circle, we know that secant is only positive in the first and last quadrant. That means in the second and third quadrant, it would be negative. To get from a negative to a positive in an instant, we know that an asymptote will be there when the denominator (cos) is zero. This graph follows the pattern of cosine, however, the differences are the asymptotes.

Cosecant? Cosecant is the reciprical of sine and has a ratio of 1/sin. When sin is equal to zero, we know there will be asymptotes. These would be located at 0 and pi. In the unit circle, we know cosecant is negative in both the first and second quadrant and is positive in both the third and fourth quadrant. This graph follows the pattern as sine, however, the differences are the asymptotes.

BQ#5 – Unit T Concepts 1-3

Why do sine and cosine NOT have asymptotes, but the other four trig graphs do? Use unit circle ratios to explain.

Sine and cosine do not have asymptotes unlike the other four trig graphs. When we refer back to the Unit Circle and find the ratios, we see that sine has a ratio of y/r and cosine has x/r. We know that an asymptote is undefined so that would only mean the denominator has to be 0. However in this case, our denominator will always be r which will never be 0. That is why sine and cosine do not have asymptotes.

For the other trig functions, tangent, cotangent, cosecant, and secant, these have asymptotes. For tangent and cotangent, we know that it deal with x/y or y/x. It is possible that the denominators of the two can be 0 which will make it undefined and result in an asymptote. As for cosecant (r/y) and secant (r/x), we see that the radius is put as the numerator while x and y are on the denominator. Again, they can value 0 which will make them asymptotes. You can notice that when tangent and cosecant both have y as their numerator they land at (-1, 0) or (1, 0). As for cotangent and secant, they both have x for their denominator and the possible points are at (-1, 0) or (1, 0). This is why the other four trig functions are asymptotes.

BQ#2 – Unit T Concept Intro

How do the trig graphs relate to the unit circle?

Trig graphs relate to the unit circle by the quadants and the positive and negative values. The unit circle is bascially just unraveled onto a line. Since we know that each quadrant as a corresponding value according to it's trig function, the direction of the graph will follow. For example, sine in a Unit Circle is only positive in the first and second quadrant, then in the third and fourth quadrant, it is negative. When it is drawn out on a graph, we know the position of each parts of the line will either be on the positive side or negative. This means that sine will be on the positive the first two parts, then it will be on the negative side. The image below indicates the positive and negative values in each quadrant for each trig function. Following the color coordination and +/- values, the line drawn below shows where each part of the quadrant exists.

(https://docs.google.com/file/d/0B4NSkh2FgPbXQ2p6dmxpdUVLa1k/edit)
*Drawing and labeling done by me*

Period? - Why is the period for sine and cosine 2pi, whereas the period for tangent and cotangent is pi?

The period of sine and cosine is 2pi because it would take a whole circle to complete the line. We see that both either stay on the positive side or negative side for a good two quadrants. Then, it ends up at where the starting point is. Since the cycle goes through all four quadrants, we know that in a Unit Circle, a 360' would be 2pi. That is why the period for sine and cosine is 2pi. 

The peiod for tangent and cotangent differs. It is just pi. In the image above, it shows how tangent would be graphed on the line. It goes from a positive value, then dips all the way down (representing by an asymptote) towards negative and goes back up the the positive, then back down. From this, we only need to go up to the second quadrant because the cycle repeats itself from there on. In the Unit Circle, we know that the second quadrant goes up to 180', which in radians is just pi. This is why tangent and cotangent's period is pi.

Amplitude? – How does the fact that sine and cosine have amplitudes of one (and the other trig functions don’t have amplitudes) relate to what we know about the Unit Circle?

We know that sine has a ratio of y/r and cosine has a ratio of x/r. The "r", or radius, will always equal to 1. Another thing to remember is that sine and cosine can't be greater than 1 or less than -1 or it would be "undefined". So the only way to divide a number from 1 and get either 1 or -1, is 1. This is why sine and cosine have an amplitude of 1.

Unlike sine and cosine, tangent, cotangent, cosecant, and secant don't have an amplitude. These all have ratios of either x/y, y/x, or r's over x and y. There is no limit to them being 1 since the radius is not on the bottom. In tangent, y/x, the value of the varibles can vary and it is not restricted; it doesn't matter whether it equals to  -1/1. That is why these trig functions do not have amplitudes. 

Reflection #1 - Unit Q: Verifying Trig Identities

1. To verify a trig function means to check if the equation is true. To get this, we must get our left side equal to the right side of the equation. Meaning, we already know what the right side is, we just need to get the left side to the same value to verify that the equation is a true statement.
2. The one important tip I found helpful is using your resources. That means all the videos provided, friends around you, and Mrs. Kirch are there to help you in figuring out trig identities. So even if you think you have the answer, your classmates are a big advantage in helping you check if your solution is correct. Another tip that I found helpful is actually memorizing or getting the jist of the types of identities. The ratio, recipricol, and Pythagorean identites all play a significant role and lays a foundation throughout this unit.
3. To verify a trig function, there are many steps that can be taken that can differ from one way to another yet still end up with the same results. The first thing I make sure is that my denominators are the same. If they're not, I have to get them to be the same by multiplying the denominator to the opposing parts or multiplying by the conjugate. Another thing I look at is whether I can convert one thing to a different identity to help me verify/solve the trig identity. You always need to make sure to know which identities can power up or power down or do neither! Itis highly suggested that the trig functions in the identities are similar so that it may be crossed off or simplified easily.

SP #7: Unit Q Concept 2: Finding all trig functions values when given one trig function and quadrant (using identities)

Please see my SP 7, made in collaboration with Sol S., by visiting their blog here. Also be sure to check out the other awesome posts on their blog!

I/D#3: Unit Q- Pythagorean Identities

INQUIRY ACTIVITY SUMMARY:

• First of all, we must make the connection that the Pythagorean Theorem is an identity. An identity is "a proven fact or formula that is always true". So in the PY, a^2+b^2=c^2, it's a proven statement. On a graph, instead of using "a,b,c" we use "x,y,z" which we can refer back when we were graphing special right triangles within the Unit Circle. So how do we get from x^2+y^2=r^2 to the pythagorean identity of sin2x+cos2x=1? We simply just divide r^2 on both sides to the x^2 and y^2. So now we have (x/r)^2+(y/r)^2=1. This is the moment where things just connect together and we can all turn on our lightbulbs! We see the ratio x/r and y/r and we know the trig functions that go with it. So when we plug that in we get sine from (y/r) and cosine from (x/r)! Since it's squared, we get the PI of sin2x+cos2x=1.
  
• How can we be sure that it actually works? We can use one of the Magic 3 pairs from the Unit Circle! Below is an example using 60 degrees:
  

• Here are the remaining Pythagorean Identities are derived from sin2x+cos2x=1
  
  

• ID w/ secant and tangent
  When we derive secant and tangent from our base Pythagorean Idenity, sin2x+cos2x=1, there are two things we know from memorizing the chart! We know that to get tangent, we would need sin2x/cos2x. Also, to get secant, we know the recipricol identity will be 1/cos2x. So how do we get this? Simply divide cos2x from everything and we are left with our Pythagorean ID of tan2x+1=sec2x!
  
  
  
• ID w/ cosecant and cotangent
  Similar to the previous problem, there are something we already know from the chart! To get cosecant and cotangent, we must divide sin2x from everything. From here, we know that cos2x/sin2x is the ratio identity for tan2x. As for 1/sin2x, we know that is the recipricol identity for csc2x. After we plug in the identities to the corresponding one, we get our Pythagorean ID of 1+cot2x=csc2x.

INQUIRY ACTIVITY REFLECTION:

• The connections that I see between Units N, O, P, and Q so far are that the Unit Circle from Unit O will always be our base/foundation. From it, we can use trig functions that we found from the special right triangles in the circle. The SRT also used the Pythagorean Theorem! So from Unit Q back to N, it all connects in the way that the identities are made up from ratio, recipricol, and pythagorean.
• If I had to describe trigonometry in THREE words, they would be tricky, complicated, "oOoOoOoOh" (when I understand the concept)!

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

*This WPP #13-14 was made in collaboration with Anthony S. Please visit his blog for more awesome posts by clicking here

Introduction: The Bakery Collaboration

(http://rainydaydesserts.files.wordpress.com/2013/03/macarons.jpeg)

(http://sallysbakingaddiction.com/2012/08/21/soft-andes-mint-chocolate-chunk-cookies/)

The Problem:

Law of Sines 

1.) Hannah decides that she wants to pair up with Juana, the owner of Juana's bakery, to sell her macarons! While Hannah walks towards the bakery, she sees that Juana is across from her in parking lot at a distance of 24 feet. Juana is N26*W from the bakery while Hannah is N43*E from the bakery. What is the distance bewtween Juanna and the bakery?

Law of Cosines 

2.) The two meet up at Starbucks to talk then decide to head over to the bakery. Both leave from the same point. Juana heads 027° driving at 32 mph while Hannah heads 146° driving 46 mph. If they drive for 2 hours, what is the distance between them? 

The Solution:

Solution to #1

Solution to #2

BQ #1: Unit P Concepts 2 & 4: Law of Sines and Area of an Oblique Triangle

2. Law of Sines

Why is SSA ambiguous?

• SSA is has three possible solutions: one, none, or two. When solving a problem for this triangle, you need to solve it as if you're going to get two possible solutions. The only way you know there's only one or no solution is when you "hit a wall". A wall can either be when sine is less than -1 or more than 1. Remember that the trig function sine should be between -1 and 1 to be possible. Another wall can be if the angle of your triagle adds up to be greater than 180 degrees. We know that a triangle can only be 180 degrees so if two angles already add up to that amount, we will hit a wall. It is possible that when we find the two possible value of an angle (one acute in the first quadrant; one obtuse in the second quadrant), one can still work for a possible of one solution!

Two possible triangles-

When given the triangle, we match each angle to its corresponding side and we can indicate the "magical pair" and "bridge". The "magical pair" is the one with both given angles and side while the bridge is the one with either a given side or and angle. We use the law of sines using the pairs of B and C to find the value of angle B.

In order to make sure a wall was not hit, the Triangle Sum Theroem can be used. After finding the possible values of B, we add them to the given angle of C and make sure it does not go over 180 degrees. Afterwards, we can find the degree of angle A and A'. Afterwards we use the same maigcal pair and use the Law of Sines but with the unknown variable of side a. 

Our answer results with two possible triangles, one acute and the other obtuse.

One Possible Triangle-

As usual, we find the magical pairs and bridge. Then we use the Law of Sines to find the value of angle C. However, only one of the angles work in this problem because we have hit a wall. The value of C' is 159 and with that added to 84, it exceeds 180 degrees. So now we know that angle won't work for a triangle but we must continue to find the rest of the other triangle.

Using the Triangle Sum Theroem again, we find the value of angle B. Then we use the Law of Sines to find side b.

Our answer results in one possible triangle.

No Possible Triangle-

In this problem, when we find sin C, we result with 1.96 Remember that sine can only be -1< sine Ø < 1. Since sine is over 1, there is no solution and this results in no triangle.

4. Area Formulas

How is the “area of an oblique” triangle derived? 

• The area of an oblique triangle is derived from the area of a triangle formula.

Since we know the area of a triangle is A=1/2bh, we can make a substitution for h that will help us find the area of an oblique triangle. If we are looking for the angle C, we know that the sine formula will be h/a sine it's opposite over hypotenuse. After we multiple a to both sides, we get h alone and this is where "a sin C" can be subbed into h from the regular area of a triangle. From there on, we can find the area of an oblique triangle by plugging in the two sides and the sine of the included angle. 

It can also look like these options depending on what sides and angle is given:

(http://wps.prenhall.com/esm_blitzer_algtrig_2/13/3560/911520.cw/index.html)

How does it relate to the area formula that you are familiar with?

• This relates to the area of a triangle (A=1/2bh) because it is derived from it. There is both the base and height multipled together and divided by two, however, the oblique formula also has the sine of an angle multiplied.

Works Cited-

1. http://wps.prenhall.com/esm_blitzer_algtrig_2/13/3560/911520.cw/index.html

WPP 12: Unit O Concept 10: Solving angle of elevation and depression word problems

Introduction: Hannah's Hike

 
(http://www.destination360.com/north-america/us/vermont/vermont-hiking)

The Problem: Hannah's Hike

After all the bow business and shopping, Hannah takes time to appreciate nature by going on a hiking trip with her friends. She is determined to hike up the cliff and be able to look down the side of the cliff to see the ledge where most people jumped off after such a long hike! (Don't worry it's safe even though it's 300 feet!)

1) Before they start their hike, Hannah wants to know the height of the cliff. The pole where she is standing from is 600 feet away from the cliff itself. She knows from the base of the pole to the base of the flagpole, there is an angle of elevation of 45 degrees. What is the total height of the cliff? (Round to the nearest foot!)

2) After reaching the top of the cliff, Hannah lies down next to the flag pole and looks down towards the tip of the ledge. She estimates the angle of depression from where she is lying to the tip of the ledge to be 21 degrees. If the ledge is 300 feet higher from the ocean floor, what is the distance she is looking at from the base of the flagpole to the tip of the ledge? (Round to the nearest foot!)

The Solution: Hannah's Hike

(This is a visual of what Hannah's hike looks like)

First we need to find the height of the cliff (we'll need this piece of evidence for problem #2). We know we must use trig functions to be able to find the missing side we're looking for. Remember that in an angle of elevation, there will be a horizontal line then the angle will go upwards.

Next, we can use the height we found earlier and other clues to find the side labeled "opp." to help us solve the whole problem. Again, we use trig functions to find the desired side. Remember that in an angle of depression, there is a horizontal line but the angle will go downwards.

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

INQUIRY ACTIVITY SUMMARY:

1.) 30-60-90

(Deriving a 30-60-90 triangle from an equilateral triangle)

In a 30-60-90 triangle, we know that it is derived from an equilateral triangle. In an equilateral triangle, all the angles add up to 180, 60 at each angle. Since we know that the sides are the same in an equilateral, we can label them all the same. We are given 1 as our side length in this activity. Once we split the triangle in half, we get a 30-60-90 triangle. The value of a is 1/2 since it was split from 1 and c (the hypotenuse) stays the same as 1. To solve for the last side, b, we must apply the Pythagorean Theroem.

(finding the missing vairable, b using the Pythagorean Theorem)

We know that the Pythagorean Theorem is a^2+b^2=c^2. If we plug in our known variables, which will be a and c, we can find the missing variable, b. After we plug it in and solve, we are left with b=(rad)3/2 after finding the squareroot of (rad)3/4.

(Triangle after all sides are found)

Once we figure out the side length for each side, we are left with the above picture. Dealing with fractions can be such a hassle so it's easier to just multiply everything by 2 and get a non-fraction answer for each side. We are left with a=1, b=(rad)3, and c=2.

(final 30-60-90 triangle derived from an equilateral triangle)

When we are using SRT's it doesn't necesarily always have to have a hypotenuse of 1. By putting the variable n where 1 would be, it allows us to plug in any given value and keep the ratio at the same time. So in the above picture, we are left with the variables n(rad)3, 2n, and n. When given a value for n, we can easily plug it in and figure out the value of each side.

2.) 45-45-90

(Deriving a 45-45-90 triangle from a square)

When deriving a 45-45-90 triangle from a square with a side length of one, we know that the sides will be 1 all around. We know that in a square, all the angles are 90 because the total degree of a square is Once we split the square in half, our angles will be 45-45-90. Our a and b value will be 1 and to solve for side, c (the hypotenuse), we can apply the Pythagorean Theorem.

(Using Pythagorean Theorem to solve for missing variable, c)

To find the missing sides, c, we plug in our known variables into the Pythagorean Theorem. We end up with c=(rad)2. 

(45-45-90 triangle with all the sides found)

After we find the missing piece, our 45-45-90 triangle will end up like this. Unlike the 30-60-90 triangle where we had to deal with fractions, we can just leave the numbers alone since they're whole numbers.

(final 45-45-90 triangle derived from a square)

After finding all the sides, we can do the same thing as in the previous trianlgle and plug n into where the 1's are. We are left with n(rad)2, n and n. The n is used to plug in any given value while keeping the ratio, which represents the relationshp between the three sides of a special right triangle.

INQUIRY ACTIVITY REFLECTION

1. Something I never noticed before about special right triangles is how it derived from a square (for a 45-45-90 triangle) or an equilateral triangle (for a 30-60-90 triangle).

2. Being able to depreive these patterns myself aids in my learning because now I can derive it on my own and I know where the values came to be. 

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

INQUIRY ACTIVITY SUMMARY

Before moving onto the next concept, we were given this chart to figure out on our own (with the help of Google) how to label the special right triangles. The first thing I did was use Google, of course, and found the following images below. For the activity we had to simplify the sides so that the hypotenuses would equal to 1. Next was labeling the sides with variables. The hypotenuse (the longest side- opposite of the 90°) is labeled by "r", the horizontal value by "x", and the vertical value by "y". After doing this, we had to draw a coordinate plane for each triangle making it so that the triangle lies in Quadrant I. Our final step was labeling all three vertices of each triangle as ordered pairs.

(http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U19_L1_T3_text_final_3_files/image022.gif)

1. Describe the 30° 

In a 30° SPT triangle, we know that all three angles will equal to 180°. Since 30° is given and we know that it's a right angle (90°), then we know that the last angle will be 60°. We focus on the 30° angle and we label accordingly. The opposite side from the 30° is x/2x. The side adjacent is x√3. The hypotenuse is always across from the 90° angle and it is 2x. 
To get the hypotenuse to equal to 1, we have to divide all the sides by 2x. By doing so, the hypotenuse will be the desired value of 1, the x-value we will have x√3/2x which when you cancel out the x's, will be √3/2. For our y-value (vertical value), we divide 2x from x and we will end up with 1/2. 
          To find the ordered pairs, we just find the points plotted on the circle. The UC revolves around it's center point (0,0) which is one of the vertices. On a graph, we use (x,y) for our points and now we can just apply the same concept. Since the x-value is √3,2 and our y-value is 1/2, our first vertices at the top corner will be (√3/2, 1/2). Our last point is where the 90° angle is. Since it lies on the x-axis only, we know there won't be a y-value so our ordered pair is (√3/2, 0). 

          

(http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U19_L1_T3_text_final_3_files/image035.gif)
(http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U19_L1_T3_text_final_3_files/image036.gif)

2. Decribe the 45°

In a 45° SPT trianlgle, we know that the two angles will be 45° because they add up to 90° and our right angle will make up the total of 180°. Since the length of the two sides are equal, they're both labeled "x" on the horizontal and vertical axis. The hypotenuse is x√2. 
          Since we are working with a unit circle, we want our radius (ultimately our hypotenuse) to be 1. To do this, we have to divide x√2 to all sides. So our hypotenuse will end up with 1, just how we wanted. When we divide x√2 from x, the x's will cancel out and we will be left with 1/√2 and since radicals aren't allowed to be in the denominator of a fraction, we have to mulitply it by √2 on the top and bottom so our answer will be √2/√4, but the √4 will be square rooted to 2. Our x-value (horizontal value) will be end up to be √2/2 and our y value (vertical value) will also be 2√2.
         We know that our origin will always be at (0,0) and that is our first vertices. Since we know our x and y-value are both √2/2, our ordred pair will be (√2/2, √2/2). Our third vertices will lie on the x-axis and it will end up to be (√2/2, 0)
      

(http://www.sparknotes.com/testprep/books/sat2/math2c/chapter6section2.rhtml)

(http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U19_L1_T3_text_final_3_files/image108.gif)

3. Describe the 60°

In the 60° angle SRT, it's exactly the same as the 30° angle one but the value of the sides are switched around. The hypotenuse stays the same with 2x, however the horizonatal and vertical value switch. So for the x-value, it is now just x, while the y-value is x√3. To get the hypotenuse to equal one, we do the same thing as before and divide everything by 2x. We end up with the x-value to be 1/2 and the y-value as √3/2. 

          Just like the other triangles, we have a point at our origin, (0,0). Our x-value is 1/2 and our y-value is √3/2 so our point will be (1/2, √3/2). The third point is on the x-axis as well and it is (1/2, 0).

4. By doing this activity, we derive how to find the ordered pairs in a unit circle. We also notice that the triangles are in the circle so it's easy to understand where we got the numbers from instead of just memorizing and not having the background knowlege of where the numbers came from. We can easily find out the value of the sides for the ordered pairs by just referring back to the special right triangles. It helps us build a better understanding of the circle. 

5. In the activity, we focused our trianlges to be in the first quadrant so it could easily be understood. The values start to change when we move it into different quadrants.

(http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U19_L1_T3_text_final_3_files/image100.gif)

In the picture above, it shows a 30° angle in the II, III, and IV quadrant. The only thing that changes is the negative signs. In the second quadrant, the x-value bececomes a negative and that's the only thing that changes. However, in Quadrant III, both of the values are negative. In the fourth quadrant, only the y-value of the ordered pair is quadrant.

The pictures on the right show a 45° angle in the II, III, and IV. The only thing that changes in the second quadrant is the negative sign on the x-value. In the thirf quadrant, both numbers are negative, while in the last one, only the y-value is negative.

(http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U19_L1_T3_text_final_3_files/image036.gif) ^original picture but I editted it into the other three quadrants.

(http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U19_L1_T3_text_final_3_files/image118.gif)

(http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U19_L1_T3_text_final_3_files/image111.gif)

Just as the other pictures, in the above picture is an example of the 60° SRT in the 3rd and 4th quadrant. In an example of the triangle in the 2nd quadrant (not shown), the x value would be negative. In the third quadrant, both x and y-values are negative. In Quadrant 4, only the y-value is negative.

INQUIRY ACTIVITY REFLECTION

1. The coolest thing I learned from this activity was how the SRT and UC connected together and were in a way dependent on each other for our learning.
2. This activity will help me in this unit because it helped me better understand how to find the values of the sides which ultimately gave me the ordered pairs of the circle. Instead of memorizing numbers, by connecting the two, it helped me understand and showed me how I can easily find the value if I forgot the memorization parts.
3. Something I never realized before about special right triangles and the unit circle is how it both went together. I never saw the triangles in the circle until we did the activity before entering Concept 7.

WORKS CITED

1.http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U19_L1_T3_text_final_3_files/image022.gif
2. http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U19_L1_T3_text_final_3_files/image035.gif

3.http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U19_L1_T3_text_final_3_files/image036.gif

4.http://www.sparknotes.com/testprep/books/sat2/math2c/chapter6section2.rhtml)

5.http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U19_L1_T3_text_final_3_files/image108.gif
6. http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U19_L1_T3_text_final_3_files/image100.gif
7. http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U19_L1_T3_text_final_3_files/image036.gif
8. http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U19_L1_T3_text_final_3_files/image118.gif
9.http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U19_L1_T3_text_final_3_files/image111.gif