BQ #3: Unit T Concept 1-3: Graphing sine, cosine, cosecant, secant, tangent, cotangent

 


The above pictures show the relation between sine and cosine with all the other trig function graphs. Notice that sine and cosine graphs are similar in the way they swingle continuously. In contrast to the other graphs, which obtain asymptotes and do not continue in the way that sine and cosine do. Those rest of the trig functions that have asymptotes go on forever in a vertical way. In Mrs. Kirch's words, those trig function graphs require us to raise our pencil to continue to outline the rest of the graph.




You are recommended to go to desmos.com in order to have a full experience on how different values affect the graph.

BQ #5: Unit T Concept 1-3: Graphing sine, cosine, cosecant, secant, tangent and cotangent

Sine and cosine do not have asymptotes, meanwhile all the other trig functions have asymptotes. But, why? Remember that an asymptote forms when there is an undefined ratio/fraction. This means that there is a number divided by 0.
This is based on the trig function's ratios. Sine and cosine have ratios of y/r and x/r. "r" always equals to one in the unit circle so we can be sure that the ratio will never be undefined.
Meanwhile, the ratios for all the other trig functions have a denominator, not of r=1, but of either y or x values, which could happen to be values of 0.

BQ #2: Unit T Concept 1-3: graphing sine, cosine, secant, cosecant, tangent, and cotangent

Trig Graphs relate to the Unit Circle. We must consider the fact that in trig graphs, the Unit circle is expanded into a line form, instead of a circle form, in order to make sense in a graph.

*A period means the graph goes through one cycle while covering certain radian units on the graph. An amplitude are half the distance between the highest and lowest point of the graph.

*Periods.

For instance, sine, cosine, cosecant, and secant have a period of 2 radians. This is because it contains a 4 part repeating unit all the the way around the unit circle (360 degrees=2 radians). This is shown in the unit circle when we look at the pattern of where the sine and cosine are positive and negative.
Sine is positive on Quadrant I and II and negative on Quadrant III and IV. The pattern is positive positive negative negative. And it continues, but we must realize we had to go all around the unit circle in order to continue the pattern. The same with cosine, whose pattern is pattern is positive negative negative positive. The pattern continues but only have we went around the unit circle. Again, around the circle means it is 2 radians.

In contrast with tangent and cotangent, which have a period of only 1 radian. This is because they only have a 2 part repeating unit which just reaches half way through the unit circle at 180 degrees=1 radian until it repeats again. The pattern is positive negative positive negative all around the unit circle. There was one repetition in pattern so the pattern continues every 1 radian.

*Amplitudes

Sine and cosine have amplitudes because they are the ones with the restriction: -1<sine or cosine<1. This means that their highest point is 1 and its lowest point is -1. Relating to the unit circle, we know that the lowest it goes it to -1 and the highest it goes is to one (remember a unit circle has a unit of 1 all around) The sine an cosine ratios are y/r and x/r. The highest the y and x values could be is 1 and the lowest they can be is -1. ("r" is the ration equals to one).
The other trig functions instead go up forever or down forever because they do not have restrictions.

SP #7: Unit Q Concept 2: Find all trig functions given one trig function and a quadrant

The Picture below will show you a problem as well as how to find all trig functions using IDENTITIES (Reciprocal and/or Ration and/or Pythagorean). 







This problem shows that you can find all trig functions by using the identities that we learned about in Unit Q Concept 1. It shows that it is helpful to know these identities in order to have expanded knowledge that would help us solve problems that could become more difficult. It follows that it is a good practice to get to know the identities of trig functions out of the top of your head.
Please notice that we figure out the quadrant that will be used by checking the answers of the trig functions given, whether they are negative answers or positive answers. Also notice on the right how we figured out the answers of  each trig function, whether they were negative or positive, based on the quadrant that we found out will be used.





And please do not forget to to check out the SECOND WAY to find all the trig functions of this problem. Just click on:


                         Kathy's Awsume Second Way

WPP # 13-14: Unit P Concepts 6-7 Word problems using law of sines and law of cosines

TO SEE THE PICTURES SHOWING THE WORK MADE IN COLLABORATION WITH KATHY C. PLEASE CLICK ON: 
                                                                KATHY'S AH-MAZING X10 BLOGPOST

Hershey has stopped at a stoplight and noticed that his best friend, Marlene, is due west of him at the next stoplight, 30 feet away. Both are going to Hershey's Bakery. Hershey walks N 30* W to get there while Marlene goes N 72* E to get there. What is both of their distances to walk over to Hershey's Bakery?




 After having a nice, long conversation with Hershey, they decided to see each other again the next day. They both leave the bakery at the same time. Marlene, in hurry to get to her cousin's Quinceanera, is headed at a bearing of 315* and is traveling 50 MPH. Hershey on the other hand goes home at 30 MPH at a bearing of 078*. How far apart are they after two hours?

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

1. Where does sin^2 x+cos^2 x=1 come from to begin with?

To begin with. you should know that an "identity" is "a proven fact that is always true". Therefore, we assume that a Pythagorean Identity, which is the equation seen above, is called such because that equation can definitely be proven as a fact that is always true. Remember that the Pythagorean Theorem using variables x, y, and r is x^2+y^2=r^2. Now doesn't this look familiar? Yes, we did get it from the a^2+b^2=c^2 equation we've learned before, but notice that just by changing the variables to x, y, and r gives us the exact equation we learned that is of a unit circle.

Now let us play with this equation a little. Who knows what amazing discovery we will come across. If we wanted to set the Pythagorean Theorem (which we just found out is similar to the unit circle equation) equal to 1 as shown in the Pythagorean Identity above, what would we do to it? Oh, of course! Divide everything by r^2 because r^2/r^2 is equal to 1. We are left with x^2/r^2+y^2/r^2=1. You may think, what kind of mess is this? Well, do not fear, Math For Cheese Buckets, like you and me, is here to make it clear! First of all, you should know that the equation can be re-written as (x/r)^2+(y/r)^2=1. 

Moreover, the ratio of cosine is (x/r) and the ration of sine is (y/r). Wait, we've seen these before. Yes, a ton of times in the last 2 units, but wait a sec, we just saw them right now, in the re-re-written equation of the Pythagorean Theorem. We can look at the equation in the previous paragraph and switch the (y/r)^2 with sine^2 x and the (x/r)^2 with cosine^2 x. Guess that we did come across a discovery. We just derived the Pythagorean Identity from the Pythagorean Theorem. 

To show that this identity is true, we will choose one of the "Magic 3" ordered pairs from the unit circle.

• 30 degrees= (radical 3 over 2, 1/2)
• 45 degrees= (radical 2 over 2, radical 2 over 2)
• 60 degrees= (1/2, radical 3 over 2

I will show 30 degrees and 45 degrees (know that 30 degrees is similar to 60 degrees, just switched).

• 30 degrees: (radical 3 over 2)^2 + (1/2)^2 = 1. You cancel out the radical with the squared and end with just the 3 on the top of the first fraction. You square the 2 and end up with a 4 at the bottom of the first fraction, so you have 3/4. For the second fraction you square the 1 and still end with a 1 in the top of that second fraction, and you square the 2 to end up with a 4, so you have 1/4. (3/4)+(1/4)=1.
• 45 degrees: (radical 2 over 2)^2 + (radical 2 over 2)^2 = 1. You can cancel out the radical with the squared for the top of both fractions to end up with a 2 on top of both fractions. Then, square the 2 at the bottom of both fractions and end up with a 4 at the bottom of both fractions: (2/4)+(2/4)=1.

2. Show and explain how to derive the two remaining Pythagorean Identities.

sine^2 x + cosine^2 x = 1. You divide the equation by cosine^2 x first. We know that sine x divided by cosine x equals tangent x by the Ratio Identities. So, we can "power up" through our understanding that it will be the same thing if we squared the sine and cosine we can square the tangent. The first part of the equation becomes tangent^2 x. Obviously, cosine^2 divided by itself is equal to 1. The second part of the equation is 1. On the right side of the equation 1 divided by cosine^2 x will become secant^2 x, because we know by the Ratio Identities that 1 divided by cosine x is secant x, therefore we can again "power up" to show that secant x will become secant^2 x. Our final equation: Tangent^2 x + 1 = secant^2 x.

The picture below can give you a visual. Please ignore the bottom part. Just pay attention up until the equation that is squared.





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Now we divide the equation,  the one at the beginning of the top paragraph, by sine^2 x. Obviously, sine^2 x divided by itself is 1. Through the Ratio Identities we know that cosine x divided by sine x equals cotangent x. Again, we can "power up" to show that cosine^2 x divided by sine^2 x equals cotangent^2 x. On the right side of the equation we show that 1 divided by sine x equals co-secant x through the Ratio Identities, so once again we "power up" to know that 1 divided by sine^2 x will now equal co-secant^2. Our final equation: 1 + Tangent^2 x = Co-secant^2 x.

The picture below will give you the equations I am deriving the above work from.







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BQ #1: Unit P Concept 1-5: Law of Sines, Law of Cosines, Area formulas

#3: Law of Sines
The Law of Sines in a triangle that is SSA, or side-side-angle, meaning we know 2 sides and one angle of the triangle, is an ambiguous case because we can have either one, none, or two answers to the triangle. This is because we only know one definite angle. When we knew at least two definite angles, like in the cases of ASA and AAS, we were able to figure out the third definite angle without wondering if there could be another possible way to draw the triangle. With LIMITED information, it is possible that we can create two different triangles by forming a "bridge" with the side that is given to us for the angle that is not given to us. 

After figuring out the value of the angle using the Law of Sines, we use its reference angle to figure out its other possible value. The reference angle is relevant because you can have the angle in the sine quadrant (quadrant II) besides the all quadrant (quadrant I), since it is less than 180 and could be in the boundaries of the Triangle Sum Theorem.  

You know there is either no triangle when you hit a "wall" in the beginning or only one triangle when you hit a "wall" after figuring out the first triangle. The "wall" could be something that is not possible like sine x being greater that 1 or less than negative 1 and the angles adding up to something greater than 180 degrees.

The picture below will show you a problem that will give you two answers.










#5 AREA FORMULAS


The pictures below will help you see how the area formulas of a triangle work and how by using either of them you will still get the same answers. This will show you that the law of sines works with working out a triangle that is not a special right triangle.