BQ #7 - Unit V Concepts 1-4: Origin of the difference quotient

How is the difference quotient derived?

With cheery songs and an able memory, one can easily memorize the difference quotient formula. However, knowing how this formula is derived can reveal so much about functions and how they, along with their derivatives, work. The difference quotient formula is derived from the slope of a secant line that touches the function at two points.

Finding the slope of a secant line, a line that touches a function at two points, can be easy and applicable to any function if we use variables to represent points and distances. In the function depicted below, the pink secant line traverses the blue function at two points. The first point has a distance of x, and its y value can be denoted as f(x). The second point touched by the secant line can be written as the distances of x and h combined. This is because the distance is measured at 0 and includes x along with its measurement. Its y value can be denoted as f(x+h). 

Having two points of a line is all that is required to use the slope formula and find the slope of the secant line.

The "any point in a function" refers to the distance between point 1 and point 2. That distance is also referred to as change in h (or deltah). As that distance gets closer and closer, nearing zero, it can be called a tangent line. When the difference quotient is evaluated at zero, it becomes the derivative.

BQ #6 - Unit U Concepts 1-4: Continuity, discontinuity, limits, and their evaluation.

Continuity, Discontinuity, Limits, and How to Evaluate Them

1.) Continuity is being able to draw a function without lifting your pencil off the paper. There are no breaks, jumps, or holes in a continuous function. It is also predictable, the function goes where you think it should go, this means that the limit (intended height) will be the same as the value (the actual height). For example, if the limf(x) x->3= 5 and f(x)=5, then that function is continuous. A continuous function can be thought of as a bridge, you can't fall through (no hole/point discontinuity) and you can't fall off (no jump discontinuity. A discontinuity can be categorized into two sections, removable and non-removal. They are separated because the limit only exists at removable and the limit DNE at non-removable. The only removable discontiinuity is called a point discontinuity and it is represented by a hole. In the non-removable category there are jump dicontinuity, infinite discontinuity, and oscillating behavior. In a jump discontinuity, the limit DNE exist because of different left and right limits. If the limits are different on opposite sides, then it is unclear what the intended height is. Similarly, in an infinite discontinuity an asymptote causes the graph to go up or down in infinite directions, thus resulting in a limit that DNE due to unbounded behavior. Oscillating behavior results in a limit that DNE because the function does not approach any single value.

Types of Discontinuities: https://docs.google.com/viewer?a=v&pid=sites&srcid=ZGVmYXVsdGRvbWFpbnxkb2NzZm9ya2lyY2hzY2xhc3Nlc3xneDoxYWY2ZjAyMWIwMTI5YWJj

Here the limit is the same as the value, thus the function is continuous. - https://docs.google.com/viewer?a=v&pid=sites&srcid=ZGVmYXVsdGRvbWFpbnxkb2NzZm9ya2lyY2hzY2xhc3Nlc3xneDoxYWY2ZjAyMWIwMTI5YWJj 

2.) A limit is the intended height of a function (Mrs. Kirch). It is the y-value that is reached the nearest when approaching a specific x-value of a function. Limits exist at holes, which is why point discontinuity has limits. this is because the limit is not the actual height, but the intended height. A limit exists when you reach the same height from both the left and right hand sides. In class we have been using the analogy of driving to a diner to evalute limits graphically. If your fingers meet at the same spot from left and right directions, then the limit exists. To evaluate this algebraically, we use left and right hand side limits and denote them with + for right and - for left. If these limits are the same, a limit exists. A limit does not exist if the values in the left and right hand limits are different. A limit is the intended height of a function whereas the value is the actual height of a function. When the limit and the value are the same we have a continuous function.

An example of a graphical expression but also different types of limits. The value of a function is undefined at a hole but the limit still exists. The value would be wherever the hole is shaded. This is unless it is a jump discontinuity, as seen in the image. - https://docs.google.com/viewer?a=v&pid=sites&srcid=ZGVmYXVsdGRvbWFpbnxkb2NzZm9ya2lyY2hzY2xhc3Nlc3xneDoxYWY2ZjAyMWIwMTI5YWJj

3.) Numerically, we evaluate limits by using a t-table with approximate x-values and use a graphing calculator to find the corresponding y-values. Whatever value is reached the closest would be our limit. Graphically, we use our fingers and approach a certain value from the left and right hand sides. If our fingers meet, then the limit is at that height. If they do but the circle is not shaded, then it is a hole and the limit still exists. Similar to our diner analogy, this means the diner is burned down but we still reached the destination, a.k.a. the intended height. Algebraically we use three methods to evaluate limits: direct substitution, dividing out, and rationalizing. In direct substitution, we plug in the exact x-value and solve. When that results in 0/0 thenw e must resort to the dividing out method in which we look to factor numerator and denominator and hope something cancels so we can then substitute directly. If nothing factors, then use the rationalizing method; in this method, we use a conjugate to cancel out a part of a function in order to solve for the rest. What results would be the value of our limit.

Evaluating a function numerically, here the limit would be 6, as it is the nearest approached value. - https://docs.google.com/viewer?a=v&pid=sites&srcid=ZGVmYXVsdGRvbWFpbnxkb2NzZm9ya2lyY2hzY2xhc3Nlc3xneDoxYWY2ZjAyMWIwMTI5YWJj 

This algebraic solution shows how using a conjugate can help cancel out an x-value. That followed by direct substitution gives you the limit. - https://docs.google.com/viewer?a=v&pid=sites&srcid=ZGVmYXVsdGRvbWFpbnxkb2NzZm9ya2lyY2hzY2xhc3Nlc3xneDoxYWY2ZjAyMWIwMTI5YWJj

REFERENCES:

Mrs. Kirch's SSS packet - https://docs.google.com/viewer?a=v&pid=sites&srcid=ZGVmYXVsdGRvbWFpbnxkb2NzZm9ya2lyY2hzY2xhc3Nlc3xneDoxYWY2ZjAyMWIwMTI5YWJj

BQ #4 - Unit T Concepts 1-3: Why is tangent uphill while cotangent downhill?

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

This picture represents the different quadrants and the graph of cosine. We graph cosine because we know based off of the unit circle ratios that tangent is undefined when cosine equals zero (tan=sin/cos). That means there are asymptotes where the graph of cosine hits the x-axis (pi/2, 3pi/2). Due to this, the graph of tangent must be broken between quadrants 1 and 2, it cannot continue because of that asymptote. However, it continues between quadrants 2 and 3, as there is no asymptote at pi (cosine does not touch the x-axis/equal zero). This means it can follow a downhill then uphill shape, giving the normal tangent graph an "uphill" look.

Cotangent is undefined when sin equals zero since the unit circle ratio of cotangent is (cos/sin). This means there are asymptotes where sin=0, a.k.a. where it hits the x-axis (0, pi, and 2pi). Since there is no asymptote between quadrants 1 and 2 (sine does not touch x-axis/does not equal zero), the graph of cotangent can be continuous. The only way we can graph this is by following a downhill pattern, giving cotangent a "downhill" look.

As seen by the images, the graph of tangent and cotangent differ because of the location of the asymptotes. In tangent, the pattern is "broken" by an asymptote, causing it to have an uphill look. Cotangent is not broken by asymptotes and thus ensues a downhill look.

REFERENCE: All graphs from Desmos.com and template from Mrs. Kirch. YOu can view more of her blogs here.

BQ#3 – Unit T Concepts 1-3: How do sin/cos graphs relate to other trigonometric graphs?

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

Tangent:

We know from our trig identities that the ratio for tangent is sin/cos. In the first quadrant, we see that sine and cosine are positive; therefore, tangent must also be positive. 

 The second quadrant features cosine as negative. Based off of tangent's ratio (sin/cos), when cosine is negative tangent must also be negative.

 Both sine and cosine are negative in quadrant three. If the ratio is sin/cos and they are both negative, then the ratio becomes positive. That it why the graphs of sin and cosine are negative while the graph of tangent is positive.

We know that asymptotes result when we have an undefined value. That undefined value comes when we divide by zero. If the ratio for tangent is sine/cosine, we would get undefined when cosine is zero. Based off our graph, cosine is zero at pi/2 and 3pi/2. We can tell based off of tangent's graph that it does limit itself to those asymptotes. Our plotted points show the locations where cosine is equal to zero.

Cotangent:

 If we look at quadrant one, both sine and cosine are positive. The ratio for cotangent is cos/sin. If they are both positive, then the graph of cotangent must also be positive. Our graph proves that that is true, cotangent is positive in quadrant one.

Cosine is negative in quadrant two. If the graph of cotangent is a ratio that contains cosine (cos/sin), then the graph for cotangent must be negative also if one of its component is negative.

The graphs for sine and cosine are both negative in quadrant three. If a ratio has negative in both numerator and denominator, then the ratio becomes positive. Because both sine and cosine are negative and the ratio for cotangent is (cos/sin), the graph for cotangent is positive.

In the last quadrant sine is negative. That makes the graph for cotangent negative since it would make the ratio for cotangent (cos/sin) negative.

If the ratio for cotangent is cos/sin, we would get undefined when sine is equal to zero. Whenever we get undefined we have an asymptote, so we should see the graph limit itself at those points. The plotted points mark where sine is equal to zero (0, pi, and 2pi). Those points are where sine is zero (where the graph of sine touches the x axis).

Secant:

The ratio for secant is the reciprocal of cosine. That means it is 1/cos. If cosine is positive, then secant must also be positive. Since it is a reciprocal, the graph goes up really high (we are dealing with small fractions).

Because cosine is negative, the reciprocal also becomes negative. This causes the graph of secant to be negative.

Cosine is also negative in quadrant three, causing the graph of secant to be negative again. Since there is no asymptote between quadrant two and three (the graph of cosine does not touch the x-axis), we are able to continue the graph of secant from the last quadrant.

Cosine is positive in quadrant four, making its reciprocal also positive.

If secant is the reciprocal of cosine, it will have asymptotes wherever cosine equals zero. Cosine is zero when its graph touches the x-axis. We see that the symptotes are at pi/2 and 2pi. The graph limits itself as it nears these asymptotes.

Cosecant:

The first quadrant presents cosecant as positive. The ratio for cosecant is 1/sin (the reciprocal of sine). If sine is positive, csc's graph will also be positive.

Sine is positive in quadrant two also, making its reciprocal positive. Since there is no asymptote in quadrants one and two (sine does not touch the x-axis), the graph of the parabola continues.

Sine is negative in quadrant three, causing its reciprocal to be negative. 

Quadrant four features sine as negative, making the reciprocal also negative. Since there is no asymptote in quadrants three and four (sine does not touch the x-axis in between),. the parabola continues.

Cosecant is the reciprocal of sine. That means it will have undefined values whenever sine is zero. Sine is zero at the places where it touches the x-axis, which are outlined as plotted points in this graph. That means the graph will limit itself around these points.

REFERENCE: All graph images obtained from Desmos.com and template from Mrs. Kirch. See more of her posts here.

BQ#5 – Unit T Concepts 1-3: Graphing trig functions

Why no asymptotes for sine/cosine?

The trig functions that we have come to know off the unit circle can also be expressed through ratio identities. These identities allow us to make sense out of multiple trigonometric functions. We know that when a ratio has a denominator of zero, that value is undefined. This undefined results in asymptotes when graphing trig functions.

The ratios for sine and cosine are opposite/hypotenuse (y/r) and adjacent/hypotenuse (x/r) respectively. If we were trying to find asymptotes for those two graphs, we might try to find an instance where r equals zero. However, r is not zero when observing a unit circle because the radius is always one. 

If r can not be zero, then sine and cosine can never be undefined. Because they can not be undefined, sine and cosine do not have asymptotes.

BQ#2 – Unit T Concept Intro

How do trig graphs relate to the Unit Circle?

The period for sine and cosine is 2pi whereas the period for tangent and cotangent is pi because that's the distance it takes for the patterns of sine and cosine to repeat. It takes them a whole revolution of the unit circle (2pi) to repeat their patterns of ++-- for sine and +--+ for cosine. In tangent and cotangent it only takes half a revolution of the unit circle (pi) to repeat the patterns of +- for tangent and +- for cotangent. The rest of the revolution is not counted as a period because it is the same as the first pair.

Sine and cosine have amplitudes of one because that is as far as the unit circle extends. The unit circle has a radius of one and therefore any value of sine and cosine must be between the parameters of -1 or 1. This means that is the highest/lowest values cosine and sine can have since their ratios are both over r. Other trig functions don't have amplitudes because they don't have restrictions. Meaning theta of any of the other trig functions can be equal to any numerical value due to their ratios that are not over r. 

Reflection# 1: Unit Q: Verifying Trig Identities

Verifying Trigonometric Identities

1. To verify a trigonometric identity means to simplify an expression until it looks like what they want you to verify it to. This is not the same as simplifying because in verification, you already know what you want your expression to look like; you are simply doing everything you see fit in order to make the expression match it. Verification often involves simplifying an expression so that it equals a simpler expression or it could also verify to the number 1 and many other variations.

2. The tips and tricks I have found helpful are manipulating expressions so that I can find pythagorean identities. When I identify a pythagorean identity, my problem usually goes in the right direction. When I get something that looks similar to a pythagorean identity, I take out a one and the result is usually something found on my Unit Q SSS cover sheet. When a problem involves different trigonometric functions, I like to convert everything to sine and cosine because usually something cancels out.

3. My thought process and steps I take in verifying a identity involve splitting the fraction if it is a polynomial numerator with a binomial denominator. I also like to get everything to equal 0 when I am working on concept 2 and 4. I also like to look at my verification's trigonometric function so that I know what I am trying to have in the end. This helps me convert everything to the appropriate function and also which things I need to cancel out.

SP# 7: Unit Q Concept 2: Find all Trig Functions when given one trig function and quadrant (using identities and SOH CAH TOA)

Finding All Trig Functions (with identities and SOH CAH TOA)

This SP7 was made in collaboration with Julian T. Please visit the other awesome posts on their blog by clicking here.

Given secx=-rad185/8 ; tanx= >0 . Find all the trigonometric functions.

Solved with Pythagorean/Ratio/Reciprocal Identities (Unit Q Concept 2): *note: The identity work is written in red letters.

First, we need to figure out in what quadrant our triangle is in given the clues. We must know this so we get the sign of our trigonometric functions correct. Based on our clues, our triangle lies in quadrant 2 meaning only tangent and cotangent will be positive, the rest will be negative.

Given secant, it is easy to find its inverse, cosine. What we do is use the reciprocal identity of secx=1/cosx.

Using the Pythagorean identity of 1+tan^2x=sec^2x we can find tangent.

Another Pythagorean identity which states that sin^2x+cos^2x=1 allows us to find cosine. We were then left with a value that could be positive or negative but from what we did in the beginning, we know sine must be negative.

Using the reciprocal identity of cscx=1/sinx we can find cosecant because we have already found sine.

We can then use the tangent which was already found in green writing to find cotangent using the reciprocal identity of cotx=1/tanx.

Our findings correspond with the prediction that the triangle lied in quadrant 2 and the signs correspond also.

Solved with right triangles [SOH CAH TOA] (Unit O Concept 5) *note: The SOHCAHTOA work is explained along with each picture.

Knowing Secant is -rad185/8 we can find cosine because we have both hypotenuse and adjacent. Our answer is negative, which corresponds with what we found in the beginning (that our triangle is in quadrant 2) since cosine is negative there.

From here we can use Pythagorean theorem to find our missing side, opposite. We find that it could be 11. 

Then we can find sin, now that we found opposite and were given hypotenuse.

Its inverse would be cosecant and is found by setting sine over one and rationalizing.

Tangent and cotangent are easily found because they are opposite over adjacent for tangent and adjacent over opposite for cotangent and we have already found the values for adjacent and opposite using what was given and the Pythagorean theorem.

From our answers and what we concluded given the clues we can conclude that our triangle looks something like this:

I/D# 3: Unit Q Concept 1: Pythagorean Identities

Pythagorean Identities

INQUIRY ACTIVITY SUMMARY:

Fig 1 (http://www.regentsprep.org/Regents/math/algtrig/ATT9/pythagoreanid.htm)

1. Where does sin2x+cos2x=1 come from?

a) An "identity" by definition according to Mrs. Kirch, is a "proven fact or formula that is always true". The Pythagorean theorem is an identity because it will work with any triangle that is legitimately a right triangle. Because right triangles reoccur in the unit circle (in 30, 60, and 45 degree angle smaller triangles), the Pythagorean theorem can be found all throughout.

b) The Pythagorean theorem is side x squared plus side y squared equals the hypotenuse (c) squared. These terms only work when we have a right angle, and luckily, we do in all 3 types of our unit circle triangles (30, 60, and 90).

Fig 2 (http://image.tutorvista.com/cms/images/38/pythagorus-theorem.JPG)

c) We can make the Pythagorean theorem equal to 1 simply by dividing both dies by c^2. This can be then simplified by writing the squared outside parenthesis, signifying everything inside is squared. What's left inside the parenthesis should look very familiar :)

What's left after simplifying is (x/r)^2+(y/r)^2=1

d) The ratio for cosine on the unit circle is (x/r),with x being "adjacent" and r being "radius". In the unit circle, r is always 1 so cosine can also be represented as simply "x" (x/1).

e) The ratio for sine on the unit circle is (y/r), with y being "opposite" and r being "radius". In the unit circle, the radius is always equal to 1 so sine can also be represented as simply "y" (y/1). 

f) The inside of the parenthesis in part c can be represented by the ratios we found.  (x/r)^2+(y/r)^2=1 becomes (cosine)^2+(sine)^2=1. I can conclude that the Pythagorean theorem is always true because it is a THEOREM, therefore the relationship I just came up with after simplifying is also true.

Fig 3(http://www.regentsprep.org/Regents/math/algtrig/ATT9/pythagoreanid.htm)

g) sin2x+cos2x=1 is referred to as a Pythagorean identity because when a circle is formed out of the unit circle with a radius of 1, the legs of that triangle can have ratios that are over 1 (cosine and sine) and thus represented simply as x and y. The picture above depicts this sort of triangle in the unit circle Because it is a right triangle, x^2 (leg 1 and also cosinex) + y^2 (leg 2 and also sinex) = 1 (radius in unit circle). 

h)Our "magic pair" of a 30 degree (rad3/2, 1/2) triangle can prove the validity of the Pythagorean identity.

2. You can derive the two remaining Pythagorean Identities from sin2x+cos2x=1 easily with both secant and tangent (a) and cosecant and cotangent (b).

a)

 Fig 4 (http://www.regentsprep.org/Regents/math/algtrig/ATT9/pythagoreanid.htm)

 Using our fundamental Pythagorean identity of sin2x+cos2x=1 we can derive a second equation that contains tangent and secant. Simply divide the formula by cos^2x. According to our ratio identities, sinx/cosx=tanx. We also know that the inverse of cosine (1/cosx) is secantx. By plugging in these new values we get tan^2x+1=sec^2x.

b)

Fig 5 (http://www.regentsprep.org/Regents/math/algtrig/ATT9/pythagoreanid.htm)

Starting off we our fundamental pythegorean identity of sin2x+cos2x=1, we can divide by sinx^2x. We know cosx/sinx=cotangentx because it is a ratio identity. We also know the inverse of sine (1/sinx) is cosecant x. Therefore, if we plug in those new values we know 1+cot^2x=csc^2x.

INQUIRY ACTIVITY REFLECTION:

1. The connections that I see between Units N, O, P, and Q so far are... the same trigonometric ratios, the same validity tests that can be proven with our magic pair triangle coordinates, and the repeated use of the unit circle (with radius of 1 of course).

2. If I had to describe trigonometry in THREE words, they would be... ratios, identities, and derivations.

REFERENCES:

Fig 2: http://image.tutorvista.com/cms/images/38/pythagorus-theorem.JPG

Fig 1,3,4,5: http://www.regentsprep.org/Regents/math/algtrig/ATT9/pythagoreanid.htm