BQ#3 – Unit T Concepts 1-3

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

We know that both sin and cos are seen in the ratio identities of the other 4 trig functions. Before examining their relations to each other, we need

Tangent?

The ratio for tangent is sin/cos aka y/x. So the asymptotes are located when tangent is cosine, this would be where cos is equal to 0. Because cos is the x-value, x=0 at pi/2 and 3pi/2this is where the asymptotes are.  Knowing this, we will be looking at the relations between the graphs of tan, cos, and sin. Let's begin with the first quadrant. We know that on the unit circle, Q1 begins at 0 and ends at pi/2. According to ASTC, we know that all trig functions in the first quadrant are positive. However, let's pretend that we don't know that tangent is positive in the first quadrant, how can we find out where tangent belongs? We can use the trig ratio for tan: sin/cos. Both sin and cos are pos. in the first quadrant, when you divide one by the other, you end up with a pos. value for tangent. For Q2, from pi/2 to pi, we know that sin is pos. and cos is neg. This means tangent is neg. in the second Qaudrant because pos/neg=neg. Q3 is from pi to 3pi/2, in which sin and cos are both neg. A neg/neg gives us a pos. value for tan. In Q4, from 3pi/2 to pi, sin is pos and cos is neg. A pos/neg gives us a neg. value for tangent in the 4th quadrant. Thus, the graph for tangent is pos, neg, pos, neg. When we graph it within its period of pi/2 to 3pi/2, the graph will be uphill. In its 1st period, it will start below the x-axis close to the asymptote at pi/2 and move uphill when it enters into Q3 where it is positive. This pattern will continue for every tangent graph. 

https://www.desmos.com/calculator/hjts26gwst

Cotangent?

The trig ratio for cotangent is cos/sin. So the asymptotes are located where sin (y) is equal to 0.Y is equal to 0 at 0 and pi, so this is the location of the asymptotes. We know that on the unit circle, Q1 begins at 0 and ends at pi/2.To find out how the graph for cot looks, we can use the  trig ratio for cot: cos/sin. Both sin and cos are pos. in the first quadrant, so when you divide one by the other, you end up with a pos. value for cot. For Q2, from pi/2 to pi, we know that sin is pos. and cos is neg. This means cot is neg. in the second Quadrant because neg/pos=neg. Q3 is from pi to 3pi/2, sin and cos are both neg. A neg/neg gives us a pos. value for tan. In Q4, from 3pi/2 to pi, sin is pos and cos is neg. A neg./pos. gives us a neg. value for cotangent in the 4th quadrant. Thus, the graph for cotangent is pos, neg, pos, neg. When we graph it within its period of 0 and pi, the graph will begin close to the asymptote above the x-axis.  When it enters Q2 it will be negative so it will start going downhill. This pattern will continue for every cotangent graph.

https://www.desmos.com/calculator/hjts26gwst

Secant?  

The trig ratio for secant is 1/cos. So the asymptotes are located where cos (x) is equal to 0. X is equal to 0 at pi/2 and 3pi/2, so this is the location of the asymptotes. We know that on the unit circle, Q1 begins at 0 and ends at pi/2.To find out how the graph for sec looks like, we can use our knowledge of  trig ratio of sec: 1/cos. Because cos is pos. in the first quadrant, sec is also positive. For Q2, from pi/2 to pi, we know that cos is neg. making sec neg. Q3 is from pi to 3pi/2, in which cos is neg. 1/neg gives us a pos. value for sec. In Q4, from 3pi/2 to pi, cos is pos so sec is also pos. because a 1/pos. gives us a pos. value for secant in the 4th quadrant. Thus, the graph for sec t is pos, neg, neg, pos. Notice that this is the same pattern for cos! When we graph sec, we know that it takes 2pi to have an entire period. Our graph will look like an upside down parabola below the x-axis in Q2 and 3, with he vertex touching with the cos graph. However, in Q4 and 1, the parabola is above the x-axis and shares it vertex with the top of the cos graph.

https://www.desmos.com/calculator/hjts26gwst

Cosecant?

The trig ratio for cosecant is 1/sin. So the asymptotes are located where sin (y) is equal to 0. Y is equal to 0 at 0 and pi, so this is the location of the asymptotes. We know that on the unit circle, Q1 begins at 0 and ends at pi/2.To find out how the graph for csc looks like, we can use our knowledge of  trig ratio of csc: 1/sin. Because sin is pos. in the first quadrant, csc is also positive. For Q2, from pi/2 to pi, we know that sin is pos. making csc pos. Q3 is from pi to 3pi/2, because sin is neg. 1/neg gives us a pos. value for csc. In Q4, from 3pi/2 to pi, sin is neg so 1/neg. gives us a neg. value for csc in the 4th quadrant. Thus, the graph for csc is pos, pos, neg, neg. Notice that this is the same pattern for sin and csc! When we graph it withing Q1 and 2, the graph will look like a parabola above the x-axis in which it shares its vertex with the sin graph.  When it enters Q3 and 4 it will look like an upside down parabola below the x-axis, sharing its vertex with sin as well.

https://www.desmos.com/calculator/hjts26gwst

BQ#4 – Unit T Concept 3

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

We know that the ratio for tangent is y/x. This means the asymptote is where x is equal to 0 (making the ratio undefined). The x-value is equal to O at (0,1) and (0,-1) which is pi/2 and 3pi/2, which could also be referred to as -pi/2. When we draw these asymptotes located at pi/2 and -pi/2 on a graph, the period consists of Q4 and Q1. According to ASTC, Tan is negative in Q4 and positive in Q1. Because of this, the only possible way to draw this is with an uphill graph. It will begin below the x-axis in Q4 and then move above the x-axis when it enters into Q1. For cot, which is x/y, the asymptote is where y is equal to 0. This occurs at (1,0) and (-1,0), which is 0 and pi. When we draw the asymptotes, the period consists of Q1 and Q2. Because cot is pos. in Q1 and neg. in Q2, the shift is from pos to neg. The only way to draw this downhill. The cot graph will begin near the asymptote above the x-axis and then go downhill to below the x-axis when it reaches Q2. Through this reasoning, a normal tangent graph is uphill while a normal cotangent graph is downhill. 

 The image below will help clarify the difference between tan and cot in reference to uphill or downhill.

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.

In order for a trig function to have an asymptote, it must be undefined. We know that the ratio for sin is y/r and the ratio for cos is x/r. In a unit circle, the measure of r will always equal 1, thus, the value of sin and cos will never be undefined. As a result, there are no asymptotes. However the other trig functions do have asymptotes. Let's look at why this is. Cosecant has asymptotes where sine is equal to 0.This is because it is the reciprocal of sine, 1/sin. This means that when the value of y is 0, the value of cosecant is 1/0, making it undefined. Similarly, Cotangent has an asymptote when sin = 0 (y=0) because it's trig ratio is cos/sin. Because both cosecant and cotangents asymptote depend on when y = 0, they share the same asymptotes at (1,0) and (-1,0) which would be 0* and 180*, also referred to as 0 and pi in radians (the 2 points on the unit circle where the y values are 0). Secant is undefined when cos is equal to 0 (x=0). The value of secant would be 1/0 , making it undefined. Likewise, tangent is y/x , so whenever the x value= 0 , there is an asymptote. The x-value equals 0 at (0,1) and (0,-1) which can also be written as pi/2 and 3pi/2. Because of this, secant and tangent share the same asymptotes, both relying on their denominator of x. 

Please refer to the picture below that identifies the different unit circle ratios.

BQ#2 – Unit T Concept Intro

How do the trig graphs relate to the Unit Circle?

When you straighten out the unit circle, it is easier to decipher how it is related to trig graphs. With our former knowledge of trig functions, we know that each trig function has its positive and negative values, depending on which quadrant it is located in. According to ASTC, we know that sine is positive in the first and second quadrant and negative in the third and fourth(+ + - -). We can take what we know from the unit circle and use it to help graph our trig functions by even using the points on the unit circle ( 0, pi/2, pi, 3pi/2, 2pi). Because sin is positive in the first and second quadrant, this part of the graph will be above the x-axis (the distance from 0-pi). Likewise, the part of the unit circle in the third and fourth quadrant will be in the negatives, below the x-axis(the distance from pi-2pi). Similarly to the unit circle, trig graphs can also be divided into "quadrants". For Cosine, it is positive in the first and fourth quadrant and negative in the second and third quadrant(+ - - +). As a result, when a cosine graph is plotted, it will be above the x-axis for the first quadrant, below the x-axis for quadrant 2 and 3 because it is negative. Then it rises above the x-axis again when it enters quadrant 4. for Tangent/Cotangent, it is positive in the 1st, negative in the 2nd, positive in the 3rd, and negative in the 4th. The pattern is basically + - + -. So when graphing , quadrant 1 and 3 will be above the x-axis while 2 and 4 will be below.

Please Refer to this picture as you read through the text to help clarify on what I mean by the Quadrants and how it ultimately affects the graph. Additionally, a sketch of a sin, cos, and tan graph are provided in relation to the quadrants.

a) Period?

To understand why the period for sin and cos is 2pi, we must define a period. A period is one time through the cycle, one repeat of the pattern as you can say. When looking at a sin graph, we know that it is above the x-axis for the first 2 quadrants (from 0 to pi) and below for the last 2 quadrants (from pi to 2 pi). Thus the pattern is pos-pos-neg-neg, from 0 to 2pi. As a result, one repetition requires 2pi. Similarly, the pattern for cosine can be described as beginning positive from 0, once it reaches pi/2 the graph remains negative until reaching 3pi/2. From there it enters the fourth quadrant in which it is positive to the point 2pi. As a result the pattern can be simplified to pos-neg-neg-pos.This means that cos also requires 2pi for a period. However, tangent shifts from being positive beginning at 0 - pi/2. then it becomes negative from pi/2 - pi. From there, it becomes positive once again till it reaches 3pi/2. Then it becomes negative till 2pi. Thus the pattern is pos-neg-pos-neg. However, we see that the pattern repeats itself twice within one unit circle. Thus, one period will only need to be half of the unit circle which is pi.

b) Amplitude?

Sine and Cosine have amplitudes because they are restricted to values ranging from -1 to1. We know that sin is y/r and cos is x/r. R can only be one because the radius of a unit circle is always 1. Because the radius is 1, the y and x values can not go beyond 1 meaning it has to be equal to or less than 1 or negative 1. When dividing the max value for y , which is one, by r, which has to be one, the greatest value we can have is 1. For these reasons, there are asymptotes, restricting the functions. As for the other trig functions, there are no amplitudes because there are no restrictions based on their ratios. For example, csc is 1/sin. The value of sine can range from -1 to 1. If the value was 0.1, then if we divide 1 by 0.1 we get 10. For these reasons, csc can't be restricted. The following applies to the rest of the trig functions. Tan is sin/cos, and the value of cos can range from 1 to -1. Thus we have an unrestricted value for tangent. This is the reasoning behind why there are limits on sin and cos but no restrictions for the others.

Reflection#1: Unit Q: Verifying Trig Identities

1. What does it actually mean to verify a trig identity?

 Verifying a trig function means to prove that one side is equal to the other side, to prove that it is true. This can be accomplished through manipulating one side of the equation to make it look like what we want it to. This is usually making the left side look like the right side of the equations. To verify, identities are used to simplify the equation as much as possible in order to get to the point we want them to be.

2. What tips and tricks have you found helpful?

 Since there are no set steps to verifying trig functions, it can be a bit confusing, like a puzzle. However, the biggest tip I have, is to practice, practice, practice! This is a concept that takes a lot of practice problems in order to improve. Practice is very beneficial in the fact that it helps you memorize the identities as well as recognize patterns. Depending on how the problem is formatted, different approaches need to be taken. Although there are no set ways to approach a problem, there are indeed patterns that are useful when recognized.

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. 

When I first begin verifying a trig functions, I check to see if I have solved a similar problem before hand. If it is entirely new, there are a couple approaches that I find useful. When I am stuck, I find it helpful to look for the greatest common denominator. If there is none, I check to see if I can substitute for and identify or if changing everything to sin and cos. If this is not possible, I check if multiplying a conjugate will be helpful. If all these are not useful, I use  various other simplification methods that include combining fractions with a binomial denominator, separating fractions with monomial denominators, or factoring.