BQ # 4 - Unit T Concept 3

Why is a "normal" tangent graph uphill, but a "normal" cotangent graph downhill?

Alright, so a big reason for this would be our asymptotes. As we know, the trigonometric functions are tan(x) = sin(x) / cos(x), and cot(x) = cos(x) / sin(x). As we know, tangent has asymptotes whenever cos(x) = 0, and cotangent has asymptotes when sin(x) = 0. To make this simpler, we know that cos(x) = x / r, and sin(x) = y / r. This means that both x and y have to be 0. For tangent, we know the asymptotes are located at 0 pi, 1 pi, and 2 pi, while cotangent has asymptotes at pi / 2 and 3 pi /2. The reason that tangent goes uphill, however, is because in quadrant two, where the graph begins, is because cosine is negative. This places our first 1/2 of the graph in the negative territory. Tangent goes into the positive section in quadrant three, because both the x and y values are negative in the third quadrant, meaning that you divide two negatives, which gives you a positive. The opposite can be said for cotangent. Since our first asymptote for cotangent is on 0 pi, our graph starts in quadrant one. Since both cosine and sine are positive in quadrant one, we start out in the positive section. As we go into the second quadrant, only sine is positive while cosine is negative, which means that cotangent is negative. Then, we run into our asymptote. As we restart our graph in quadrant three, both cosine and sine are negative, giving us a positive graph. In quadrant four, only cosine is positive while sine is negative, making cotangent negative.

BQ #3 Unit T Concepts 1-3

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

In the graphs shown below, the highlighted sections are Quadrant I (red), Quadrant II (green), Quadrant III (orange), and Quadrant IV (blue). We know this because in the Unit Circle, Quadrant I is 0 pi to pi/2, Quadrant II is pi/2 to pi, Quadrant III is pi to 3pi/2, and Quadrant IV is 3pi/2 to 2pi. This is because, as we know, the Trigonometric graph is ultimately an outstretched version of the Unit Circle. Sine is red, cosine is green, tangent is orange, and cotangent blue.

DESMOS GRAPH CREATED BY MRS. KIRCH

Tangent:
Tangent is similar to sine and cosine because (first of all) tangent is tan(x)= sin(x)/cos(x). Since both sine and cosine are positive in the first quadrant, tangent is also positive because you end up dividing a positive by a positive. Looking at the second quadrant, we see that sine (the red line) is also positive, while cosine is negative. This means that tangent becomes negative. Looking at quadrant three, we see that both sine and cosine are negative, so tangent is positive. In quadrant 4, sine is negative and cosine is positive, so tangent is negative as well. Also, we know that we have asymptotes when when we have an undefined answer, which is basically when cos=0. Cosine=0 at 0 degrees, 90 degrees, and 270 degrees. In terms or radians, those values are, (respectively), pi/2 and 3pi/2. This means that we have a tangent graph whenever sine touches the x-axis, but not when cosine touches the x-axis.

DESMOS GRAPH CREATED BY MRS. KIRCH

Cotangent:
Being the inverse of tangent, the trigonometric identity of cotangent is cot(x)= cos(x)/sin(x). In quadrant one, both cosine and sine are positive, making the blue cotangent line positive (please ignore the orange cotangent line which I forgot to delete from my graph). In quadrant number 2, since cosine is negative while sine is positive, then cotangent is negative in the second quadrant. Cotangent is positive in quadrant three since both sine and cosine are negative, and cotangent is negative in quadrant four since sine is negative while cosine is positive. As for the asymptotes, we know that we have asymptotes when our answer is undefined. This is only when sin=0, which happens at 0 degrees, 180 degrees, and 360 degrees, which are, in terms of radians, o pi, pi, and 2pi.

DESMOS GRAPH CREATED BY MRS. KIRCH

Secant:
We know that the trig function of secant is sec(x)= 1/cos(x). This means that our graph for secant will from, and be the inverse of, cosine. Knowing this, we can predict that the secant graph will sprout from the cosine graph, and this is verified by the graph above, where both our secant and our cosine graphs touch. Wherever cosine has mountain, a secant valley touches it, and wherever cosine has a valley, a secant hill touches it. Looking at quadrant one, we see that since cosine is positive, secant is positive. In both quadrant two and quadrant three, cosine is negative, which means that secant is also negative. Lastly, when we look at quadrant four, cosine is positive, making secant positive as well. Now, let's talk asymptotes. We know that cosine has to be 0 in order to have the undefined answer that would give us our asymptote. Since we know that cosine is adjacent/hypotenuse, we know that this asymptote happens whenever x is 0, and this happens at pi/2 and 3pi/2. We have asymptotes in these two areas, meaning that the secant graph can't cross those x-values.

DESMOS GRAPH CREATED BY MRS. KIRCH

Cosecant:
As we all know, the trigonometric function of cosecant is csc(x)= 1/sine(x). Just like in the secant graph, our cosecant graph sprouts directly from the sine graph. This means that wherever sine has a mountain, cosecant has a valley. Wherever sine has valley, cosecant has a mountain. In quadrant one and quadrant two, cosecant and sine are positive. In quadrant three and quadrant four, both cosecant and sine are negative. Let's go over those asymptotes. We know that sine has to be equal to 0. This happens at 0 pi, pi, and 2pi. These are the locations which the secant graph will NEVER cross

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.

The reason that sine and cosine do not have asymptotes is because their ratios are (y/r) for sine and (x/r) for cosine. As we know from the Unit Circle, the letter "r" is the hypotenuse, which is equal to . This means that we will never divide by zero (or any other number, actually), so we will never have an undefined answer, which, as we know, is the only time we have an asymptote. On the other hand, we have Cosecant, Secant, Tangent, and Cotangent. The ratios for these functions are (respectively): (r/y), (r/x), (y/x), and (x/y) since we can have values of x=0 as well as y=0, we would divide by zero in these four trig functions. This would give un an undefined answer, meaning we have asymptotes.

BQ#2 - Unit T Concept Intro

How do the trig graphs relate to the Unit Circle?

Being near the end of the second semester in this mighty Math Analysis class, we are all familiar with the seemingly endless circle of power that is the Unit Circle. This same Unit Circle can be transformed into the trigonometric graphs we have all come to love greatly by now. We achieve this by simply unwrapping the almighty Unit Circle and placing it in the shape of a simple graph.

-Period? - Why is the period for the sine and cosine 2 pi, whereas the period for tangent and cotangent is pi?
The reason for this is simple. In the sine and cosine graphs, you have your "mountain" that takes up basically quadrants I and II, while the "valley" takes up quadrants III and IV. Since the pattern is technically +|+|-|-|+|+|-|- and repeat for sine, it takes you two negatives to get the back to the double + pattern, thus the 2 pi. In the case of cosine, the pattern is
+|-|-|+|+|-|-|+|. As you can see, it still takes 2 negatives before you can get to 2 positives, thus the 2 pi. In the case of tangent and cotangent, the pattern is +|-|+|-. This means that you only have to get through 1 negative before you get to the positive.

-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 he Unit Circle? Basically, both Sine and Cosine are (respectively) y/r and x/r, which gives is basically y/1 and x/1, thus setting the the restriction of 1 and -1. On the other hand, Tangent and Cotangent are x/y and y/x. This means that we do not have any restrictions because we divide by numbers other than 1 and -1.

Reflection #1 - Unit Q, Concepts 1 & 5

#1: What does it actually mean to verify a trigonometric function?
-Well before explaining how to solve a trig identity, I'll explain WHAT a trig identity is. A trigonometric identity is a "rule" that is always true no matter what. It's kind of like saying that pi= 3.14 or that the constant of gravity on Earth is 9.80 meters per second squared. So basically, in order to solve the identities, we need to use these rules in order to either simply the problem as much as possible, or just prove that the problem given to you is true (depending on what the problem asks for). That being said, there is always a way to solve the trigonometric functions, and you will not have a no solution. You can have one solution, more than one, but you will always have at least one.
#2: What tips and tricks have you found helpful?
-When solving the functions, I have found that it is always helpful to evaluate the problem before you start on it. Looking at what you have and think about what you can plug into the problem is a good way to start. Once you have a specific identity that you can plug in, you just plug them in and then you start to either combine, cancel, or multiply by the inverse, etc.
#3: Explain your thought process and steps you take in verifying a trig identity.  Do not use a specific example, but speak in general terms of what you would do no matter what they give you?

-As mentioned in question #2, the first thing I do, is evaluate the problem. After I do this, I basically just tend to substitute in as much as possible, and then factor out. By doing this, I make my life easier, because I can just eliminate like terms. Then, you can usually re-substitute some of the identities and eliminate further. After this, you are usually done with the problem.