BQ: #6 Unit U

1. Continuity is when a graph is predictable, has no breaks jumps or holes, can be drawn without lifting your pencil and has a value that is the same as the limit. Discontinuity is when a graph has any of these things or where the graph's value and limit aren't equal. There are different families of disc. they are removable and non-removable disc. In the removable family we have the point disc. which is similar to a hole, the limit still exists but the value does not unless these is another point on the same x value. In our next family non-removable we have jump which looks like what it sounds like a point jumping to a diff point and here the limit does not exist due to diff. left right. Also in this family we have oscillating which we tend to call wiggly because it is wiggly and the limit does not exist here because it is oscillating. The last type of disc in this family is infinite disc and the limit does not exist here because of unbounded behavior, in this kind of disc. we also have a vertical asymptote.
2. A limit is the intended height of a function, as so wonderfully explained above the only kind of disc in which a limit exists is at a point disc this is why the kinds of disc are divided into diff families. The limit is the intended height while the value is the actual height reached, this is why they can differ and exist while the other does not.
3. We evaluate limits numerically with a number table and place the limit in the center, from the left we subtract one tenth and keep getting it closer to the limit. And on the right we add one tenth and make it get closer to the limit. We evaluate a limit graphically by following the function from the left and the right and meet at the limit, if they don't meet then the limit does not exist. To solve a limit algebraically we use substitution where we plug it in and see what we get, if we get 0/0 then we move on to method two which is factoring where we try to factor something out and try to get things to cancel, and the last method is the rationalizing wherever the radical is and multiply by the conjugate.

BQ 3 Unit T Concepts 1-3

How do the graphs of sine and cosine relate to each of the others?

a. Tangent. 

These graphs are different because we know that sin and cos are y/r and x/r so this is why neither of these graphs have asymptotes, r=1. Tangent on the other hand is y/x therefore wherever x or Cos is equal to zero, there is an asymptote.

b. Cotangent 

Similar to tangent, cot is just x/y so wherever y or sin is equal to zero there is an asymptote, this is because the graph is undefined. 

c. Secant 

Because sec is r/x it will have asymptotes at the same places where tan had them but the two graphs look different because tan is only pi for a period and sec is relates to cos graph which is 2pi for a period. 

d. Cosecant 

Because csc is r/y it will have asymptotes at the same place where tan had them, and once again they are different because of the difference in the period length. 

BQ 5 Unit T Concepts 1-3

An asymptote is when the value of the graph is undefined, we know that something is undefined when it is divided by zero. Now let's take a look at our ratio identities tan is equal to y/x, cot equal to x/y, secant is r/x and cosecant is r/y, if we look at al of these we notice that these mean the numerator are divided by a number ;x, y these could be any number including 0. Meaning that any one of these could have an asymptote. If we look at cos and sin their ratio identities are y/r and x/r r will always equal one and therefore these will never be divided by 0, they won't have an asymptote.

Tan= y/x                Sin= y/r
Cot= x/y                 Cos= x/r
Sec= r/x
Csc= r/y

BQ #2: Unit T Concept Intro

a) Why are the periods for sin and cos 2pi and the period for tan only pi?

If we look at all the signs that we have when looking at a sin graph, we can see that the order of this is; +,+,-,- this pattern does not repeat, therefore we need to have another cycle and then it will repeat, each of these cycles is equal to pi, therefore this graph's period will be 2pi

As for the cosine graph, the cycle is +,-,-,+ so we can see once again that the signs don't repeat themselves yet, therefore we had to once again add another cycle and make this period be 2pi.

When we look at the tan/cot graph however, we see that the cycle is +,-,+,- in this case the signs do repeat themselves so we have our pattern in only one cycle this makes a period of only pi.

b) If we think back to our ratio identities we know that sin is really y/r and cos is x/r we also know that r is always equal to 1 and then we can think about all our other ratios, tan y/x csc r/y ect r is no longer on the bottom, this means that the function isn't restricted to just one. Sin and cos are restricted because they will always be over r and r will always be equal to one, this is also why they have amplitudes of one.

Reflection #1 Unit Q: Verifying Trig Identities

1. To verify an identity means to try and get the left side of a problem equal to the right, to do this we usually have to simplify of use our ratios in order to make things cancel and reduce. Although the problem look like two different trig function when you begin, we are testing to verify that they can be reduced to equal the same thing.

2. Some tips that I have found helpful are to always think of things in terms of sin and cos, things may not seem similar until we convert them to these. I  have also learned to keep in mind your pythagorean identities while working a problem out, so that we are sure not to change the ones we need, or so that we have a trig function we are trying to get to.

3. My first thought when looking at a trig problem is to see if I have any parts for my identities (are things squared, to i have one or two of the ratios needed?) next I check to see if the problem would be easier if I changed it into cos and sin. If these don't work I look to check if anything can be factored out or foiled ect. My final thought is to see if I square the problem would I have a solution or would the problem become easier when squared.

I/D #3 Unit Q Pythagorean Identities

Inquiry Activity Summary

1. So where do we get sin^2x + cos^2x =1? Well if we think back to our triangle, we can see in the picture below that x = cos,  y= sin and r = 1, and if we square this, we and up with our identity above.

   

   http://mathworld.wolfram.com/images/eps-gif/Trigonometry_1001.gif

   2. The two remaining identities can be derived from the identity above by dividing by cos^2x, so it would be sin^2x/cos^2x + 1 = 1/cos^2x we know from our reciprocal identity that 1/ cos^2x is equal to sec^2x and we know from our ratio identity that sin^2x/ cos^2x is equal to tan^2x so we are left with 1 + tan^2x = sec^2x To find the next one we dot he same steps except this time we divide by sin^2x which leaves us with 1 + cos^2x/ sin^2x = 1/ sin^2x after doing out identities, we are left with 1 + cot^2x = csc^2x 

Inquiry Activity Reflection

1. The connections that I have noticed between these previous units, is that we have had to use previous knowledge from our unit circle, the ordered pairs ect. Another thing I've noticed is that we have taken our roots of our ratios, from previous ones such as the pythagorean theorem and we've had to alter the because these are no longer 90 degree triangles. 
2. If I had to describe trigonometry in three words, I would say that it's puzzling, challenging and intriguing. 

SP #7 Unit Q Concept 2

This post was made in collaboration with Carlos Rendon, view other awesome math posts on his blog here

In our problem, we are given the information that sin<1 and also that cot is equal to -1/3
-Our first step was to find tan becasue we already have cot and we only have to do the reciprocal in order to get this. Our answer is -3 as shown in the picture below
-Now that we have tan, we can use the pythagorean identity 1 +tan ^2 = sec^2 so sec^2 is 1+ (9), radical 10
-We have sec so we can do the reciprocal in order to find cos, it is radical 10/ 10
-To find sin we can do 1- cos^2 and get 9/10, after we take the reciprocal and rationalize we should get 3 radical 10/ 10
-And to find csc we do the reciprocal of sin and get radical 10/ 3


                                                                     

In order to solve this using SohCahToa,
-We know sin is y/r which is 3/rad 10 and then it becomes 3rad 10/ 10 after we rationalize
-for csc it's r/y and it is rad10/ 3
- cos is x/r which is rad 10/10
-To get sec it's r/x and we get radical 10
-Next is tan, y/x which is -3
-And to get x/y (cot) it will be -1/3