BQ #3 unit T concepts 1-3

 How do the graphs of sin and cos relate to each of the others?

tangent?

      The trig ratio for tangent is sin/cos. Because of this the asymptote for tan can be found whenever cos is equal to zero. When cos is equal to zero the ratio of sin over cos which is equal to tan would be undefined creating and asymptote.

cotangent?

       The trig ratio for cot is cos/sin. Because of this the asymptote for cot can be found whenever sin is equal to zero. When sin is equal to zero cot would be equal to cos/0 and that would be undefined since you cant divide by a zero. An undefined section of the graph will create an asymptote.
secant?

        The trig ratio for sec is 1/cos. Whenever cos is equal to zero sec will be undefined which will create an asymptote. So whenever cos is equal to zero a sec graph will have as asymptote. Sec is also the inverse of cos so the sec graph will stem off from cos graph.
cosecant?

       The trig ratio for csc is 1/sin. This means that whenever sin is equal to zero the related csc graph will have an asymptote. Since csc is the inverse of sin, csc graphs will stem off from the corresponding sin graphs 

BQ #2 Unit T concept: intro

How do trig graphs relate to the unit circle?

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

The period for sin and cos is 2pi because of the patterns they follow on the unit cirlce. Sin follows a ++-- pattern and cos follows a -++- pattern. It takes one full unit circle in order for the pattern to repeat for sin and cos and one full unit circle has a length of 2pi. Since a period is how long the graph takes to reach the point it started it will take a length of 2pi because that's how long it takes the patterns for sin and cos to repeat themselves. The period for tan and cot are both just pi units long. This is because the unit circle pattern for tan and cot is +-+-. Because of that pattern it takes only half the distance of the unit cirlce to repeat itself. Half the distance of the unit circle would be pi, making the period for tan and cot just 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?

sin

 

For trig functions only only sin and cos have amplitudes because of they behave on the unit circle. The limits of a cos on the unit circle are found at 1,0 and -1,0. The +1 and -1 look very familiar. That's because the amplitude of a cos graph is alos equal to one. Since the limits of cos are at those coordinates that has to also be the limits of a cos graph. Since 1,0 is the start and end of a the unit circle that's why cos starts and ends at 1. The same goes for sin only sins limits are found at 0,1 and 0,-1. Since the limits are again at 1 and -1 on the unit circle the same must be the same for the trig graphs.

BQ #4 Unit T

Why is tangent uphill and cotangent down hill?

As you can see tan and cot have their asymptotes in different places which affects how they are drawn. However they both follow the same positive negative pattern according to the unit circle. Since tan has an asymptote at pi where tan is positive the graph must follow close to the asymptote without reaching the x axis which it can only active by going towards positive infinite but then the same graph has to be negative in the next quadrant over with the asymptote being at 2pi so the graph would have to reach towards negative infinity while going along the asymptote. Because of this the tan graph must go follow that shape. The same goes for cot but since the asymptotes are in different places in relation to the positive negative pattern the graph must go towards negative infinity along the first asymptote and towards positive infinity along the second asymptote which will give it the shape it has.

BQ 3 unit T

a. In our trig identities tangent relates to sin and cos in the ratio of tan=sin/cos. With the ratio the asymptotes and where the tangent graph is negative or positive can be explained. Asymptotes happen when the value of the graph is undefined and the only time the value for tan is undefined is when cos is equal to zero. So in turn the asymptotes of a tan graph are located where the relating cos graph touches the x axis. Tan= sin/cos also explains when tan is below or above the x axis. In the first quadrant both sin and cos are positive making the tan graph positive. In the second quadrant sin is positive but cos in negative so tan is negative. In the 3rd quadrant both sin and cos are negative making tan positive and in the 4th quadrant sin is negative but cos is positive so again tan is negative.

b. Cotangent is very similar to tangent the only difference being that cot=cos/sin instead of sin/cos. Because of this the asymptotes of a cot graph will differ from those of a tan graph. Instead of the asymptotes being where cos is zero the asymptotes will be where sin is zero shifting the asymptotes and the overall look of the graph. However whether cot is above or below the x axis stays consistent with tan graphs.

c. Secant graphs are closely related to cosine graphs. In regards to trig ratios sec=1/cos. So with that knowledge the asymptotes of a sec graph will be wherever cos=0. Sec graphs will follow the same positive and negative pattern as a cos graph and will be graphed accordingly, the only difference being that sec graphs are essentially a series of asymptotes.

d. Cosecant graphs are similar to secant graphs except the trig ratio for csc is 1/sin instead of 1/cos. Because of this the asymptotes are found wherever the related sin graph touches the x axis. Csc graphs also follow the same positive negative pattern of a sin graph and once again csc graphs are a series of parabolas.

Reflection #1 Unit Q- verifying trig identities.

1. Verifying a trig identity means using the various ratio and Pythagorean identities in order to simplify long and complicated problems with carious trig functions into something as simple as sin(x). It also means verifying that two different trig ratios are equal to each other which again means simplifying. So all in all verifying trig identities mainly means to simplify big long complicated problems into simple trig functions that are easy to work with.
2. There aren't many tips and tricks to solving these identities. One of the main tips I have is to get everything into related trig functions in order to be able to use a identity. Also you should probably never square both sides because itll make the problem more complicated than it already is.
3. The first thing I do is see if the trig functions currently relate to each other. If they do I look at the multiple possible path and choose the easiest one. After that i keep going using multiple trig identities or factoring along the way if i can. When its to verify I do it until i get both sides equal to each other. When its to simplify I go until there is hopefully just one trig function left

SP7- Unit Q concept 7- finding all trig functions

The biggest thing that needs to be understood is what identities to use. You need an identity that will get you to the information you need but there are usually multiple paths that can be taken. However it doesn't really matter which path is taken because in the end the answers will be the same although the difficulty may be greater for some paths when compared to others.

WPP 13/14 Unit P concepts 6/7

This WPP was made in collaboration with Tommy and William please visit the other awesome posts on their blogs by going here and here'
1. The Ups delivery man needs to make a delivery to Bobs pokemon store as well as Marty's guitar store. The stores are 11 miles apart from each other. The guitar store is S50W from the ups guy and the Pokemon store is S65 E form the ups guy. Find which store is closer to the ups guy.

The guitar store seems to be close at just 5.1 miles away.

2.Marty is sitting in his treehouse, He needs to go to both target and walmart. Target is 6 miles away at a bearing of 310. Walmart is 3 miles away ay 035. How far is Walmart from Target.

3. Bob is taking a break from his pokemon store. He goes to a river and sees a tree with delicious fruit on the the other side. He starts at point A and walks due south for 10 miles to point B. At point A there was a  bearing of S49W and at point B it was N62W. How far is the tree from point B.

BQ 1 Unit P

1. Law of sines: The main reason we need law of sines is to find unknown parts of a non right triangle. How ever in order to use law of sines we must be given either AAS, ASA or SSA.


                                                 http://www.youtube.com/watch?v=gAX_IleqeJQ
 As shown on the video we start with a triangle labeled ACB. We must drop a line from angle C in order to make 2 right triangles. From there we can figure out that the sin of a=h/b. Also that the sin of C=h/a. Then if you divide by ab you get sinA/a= sinC/c with is the law of sines.

4. This video will go over how the area of an oblique triangle is derived.

                  http://www.youtube.com/watchv=H8NEjzbvB6M#aid=P9gOqbIj4Vw
Our regular formula is area=1/2*bh. The new area formula is a=1/2*b(asinC) They relate because essentially they are the same formula the only difference is that we have solved and substituted for h. When we made the oblique triangle into a right triangle we used our knowledge of trig functions to find the h=asinC. This then goes into our normal formula giving us our new formula that now works for all triangles.

WPP #12 Unit O concept 10

http://commons.wikimedia.org/wiki/File:Looking_Up_at_Empire_State_Building.JPG

Bob has recently added a couple stories to his pokemon card shop but he has no idea how tall is new store really is and would really like to know. If Bob is standing 15 feet away and looking up at a 76 degree angle how tall is his store.

Now that Bob has found out how tall his newly improved store is he wants to know know how far the rival card shop is that is down the same perfectly level street that his card shop is on. He's standing on his 60.16 ft tall building looking down at a 33 degree angle.

I/D 2 unit O-how can we derive our patterns for the special right triangles

Inquiry Activity Summary

1.                                                               

We first start out with an equilateral triangle with each side length being 1. Since it is equilateral it is also equilateral so each angle measure is 60 degrees.

In order to create a 30 degree angle to give us our 30-60-90 triangle we must cut straight down splitting the 60 degree angle into two 30 degree angles while also creating a right angle with the base of the triangle.

When you cut an angle in half you also cut the side across from the angle in half. Since all the sides of the triangle started out as 1 when you cut it in half the side across form the 30 degree angle is now 1/2. So far we have the hypotenuse being unchanged and staying at 1 and the base or the side across from the 30 degree angle being cut in half into 1/2

Now that we have 2 of the three side we can now use the pythagorean theorem to solve for the 3rd side. The side across form the 30 will be a the side across from the 60 will be b and the hypotenuse will of course be c. We then plug those values into the formula and then square them. We end up with 1/4+b^2=1. From there we solve for b and end up with rad3/2 for our third side across from the 60 degree angle.

Now that we have each side of the triangle we can get rid of the fractions by multiplying each side by 2. 

By multiplying each side by 2 we get 1 rad3 and 2. This is very similar to our pattern of n n-rad3 and 2n. The only reason we have n instead of one is for it to be a generalization of the pattern for any number instead of just 1

2.

We first start with a square where all the side are equal to 1

From there we cut it diagonally in order to split the right angles and create 4 45 degree angles or two 45-45-90 triangles. Unlike the 30-60-90 triangle the sides stay unchanged at 1 since it was a diagonal cut.

We have the side across from the 2 45 degree angles and we are missing the hypotenuse. Since we have 2 sides and are missing one we can again use pythagorean theorem to find the last side. The work is shown in the picture and we end up finding out that the hypotenuse is equal to rad2

The sides of the triangle are 1 and 1 being the two legs and rad2 being the hypotenuse. This already looks very similar to our pattern and needs nothing done to it except for changing the 1's to n's to create a generalization for any number to fit in its place.

Inquiry Activity Reflection

Something i never noticed about the special right triangles is that they are derived from a square and a equilateral triangle.

Being able to derive these patterns myself aids in my learning because I now have a better understanding of the triangles now that I know where the pattern comes from and just in case i completely forget the pattern i could also just derive the two triangles from the square and the triangle.

RWA Unit m concept 5: ellipses

http://jwilson.coe.uga.edu/EMAT6680/Brown/6690/InstrUnit/DayFive.htm

Mathematical definition: The set of all points such that the sum of the distance from two points also known as the foci is a constant.

DESCRIPTION OF ELLIPSE

equation: (x-h)^2/a^2+(y-k)^2/b^2 for a "fat" ellipse (x-h)^2/ b^2+(y-k)^2/a^2

graphically: refer to above picture

description of key points: center-the center of the ellipse 2 vertices: the endpoints of the major axis Major Axis- the axis where the ellipse is wider crossing through the 2 vertices  Co vertices- points on an ellipse that lay on the minor axis   Minor Axis-axis on graph where ellipse is thinner co vertice to co vertice a is the bigger denominator and the major axis would be 2a units a B: B is the smaller denominator and the minor axis would be 2b units  away.
Foci: The father the foci are from each other the less ellipse will look like a circle and therefor the eccentricity of the ellipse will be closer to 1. ( the eccentricity of an ellipse should be between 1 and 0
(http://www.mathopenref.com/ellipsefoci.html)

REAL WORLD APPLICATION 

One real world application of an ellipse is through an lithotripsy. Extracorporeal Shockwave Lithotripsy is a practice used by doctors in order to get rid of kidney and gall stones without the use of open surgery. This way the patient has less recovery time and less risk of infection. 

 The tool used for this is called a lithotripter. An ellipse is the base of this machine. Shockwaves are shot at one focus and are supposed to reflect to the other focus. A patients kideny or gallbladder should be placed at the other foci in order to recieve the shoackwaves The stone must be at precisly the focus in order for the machine to work. (http://mathcentral.uregina.ca/beyond/articles/Lithotripsy/lithotripsy1.html)

REFERENCES

http://mathcentral.uregina.ca/beyond/articles/Lithotripsy/lithotripsy1.html

http://jwilson.coe.uga.edu/EMAT6680/Brown/6690/InstrUnit/DayFive.htm

(http://www.mathopenref.com/ellipsefoci.html)

http://www.schooltube.com/video/11b0cf3f668f40a1a210/The%20Equation%20for%20the%20Ellipse