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Wednesday, March 26, 2014

SP#7: Unit Q Concept 2: Finding All Trig Functions When Given One Trig Function and Quadrant

This SP7 was made in collaboration with Sarahi Leal. Please visit the other fantabulous, supercalifragilisticexpialidocious posts on her blog by going here.
Solving with Identities:

Solving with SOH CAH TOA: 
The idea we want you to formulate is that there are two ways to find these trig functions: using identities and using SOH CAH TOA. When dealing with identities, it is always best to remember to know WHERE the unknown triangle falls in. As you can see in our very first step, we determined what quadrant the triangle is in, just by knowing what quadrants certain trig functions are positive and such. Another good thing to remember, especially as we deal with identities more and more, is that when you square root a trig function (as seen in #4 of "Solving with Identities"), you will end up with two possible answers because it is both negative and positive. However, we were able to conclude that it was positive because sine and cosecant are positive within Quadrant II. Always remember these little nuances because they do help in making these types of problems easier. 


Wednesday, March 19, 2014

I/D3: Unit Q - Pythagorean Identities

Inquiry Activity Summary: 
1. sin2x+cos2x=1 
This is a Pythagorean Identity (a proven fact/formula that is always true), so, we will be using the Pythagorean Theorem in order to begin deriving it. Originally, our Pythagorean Theorem is a^2 + b^2 = c^2. However, we learned in recent units that we deal with triangles on graphs, specifically when we are dealing with the Unit Circle. The same situation is still present, therefore we want to adjust our formula. The legs of your triangle are measured in how much you go left/right on the graph (x-axis) and how much you go up/down (y-axis). We also remember that our hypotenuse of our triangle is also the radius of the Unit Circle. So we go from a^2 + b^2 = c^2, to x^2 + y^2 = r^2. Now that we have our "new" Pythagorean Theorem, we continue on and think of how on earth we simplify it into our Pythagorean Identity.
Our formula is  x^2 + y^2 = r^2. We want to get to sin2x+cos2x=1. To do that, we divide everything by r^2. If we divide everything by r^2, we get this: x^2/r^2 + y^2/r^2 = 1 (r^2/r^2). Already, we can tell that r^2 divided by r^2 is 1. Now we can look at the other parts; x^2/r^2 is basically (x/r)^2 and y^2/r^2 is (y/r)^2. Immediately, you should notice something about these variables and RATIOS. The ratio for cosine is x/r. The ratio for sine is y/r. *AHA MOMENT*
Now that we are aware of this relationship, let's plug them into our formula. So, (x/r)^2 becomes cos^2x and (y/r)^2 becomes sin^2x. Finally, we end up with cos^2x + sin^2x = 1.

This picture is clarification of what I described in the text ABOVE. 
Don't believe that it actually works? We can use any of the "Magic 3" ordered pairs from the Unit Circle to show that it is true. Look in the picture BELOW to see how I used the 60 degrees ordered pair in order to prove this identity true.

Simple matter of plugging in numbers into our identity and verifying. 
2. Other Pythagorean Identities 
Now that we have our main Pythagorean Identity, we will use it to find our other two. 
  • The first Pythagorean Identity contains secant and tangent. The only way to get that is by dividing by either cosine or sine. For this one, we want to divide everything by cosine because the ratios will simplify into what we want: tangent and secant. We divide sin2x by cos2x and we get tan2x. How do I know? Besides from memorizing this identity, we can use logic. Sine's ratio is y/r and it is being divided by cosine's ratio, which is x/r. Another way we can look at that is y/r TIMES r/x. So as a result we have y/x which is tangent. Cosine divided by cosine is simply 1. And 1 divided by cos2x is sec2x (because secx = 1/cosx; we just powered up because everything is being multiplied in this reciprocal identity). The picture below gives clarification of what you should be left with.
This is what our Pythagorean Identity is. 
  • The other Pythagorean Identity contains cosecant and cotangent. Because we divided the last identity by cosine, we want to divide by sine this time. We divide sin2x by sin2x and get 1. We divide cos2x by sin2x and we get cot2x. (x/r TIMES r/y = x/y). We divide 1 by sin2x and we end up with csc2x. The picture below will have the Pythagorean Identity and work along with it. 
This is what our 3rd Pythagorean Identity looks like. 
Inquiry Activity Reflection: 
1. “The connections that I see between Units N, O, P, and Q so far are…” the continual use of trig functions (and the things we are able to use with them), and the references towards Unit Circles and the triangles within it.
2. “If I had to describe trigonometry in THREE words, they would be…” Ratios. Are. Everywhere.

Tuesday, March 18, 2014

WPP #13 and 14: Unit P Concept 6 and 7: Law of Sines and Law of Cosines

This WPP #13 and 14 was made in collaboration with Tina N. and Sarahi L. Please visit the awesome, fantabulous blog post here! And make sure to check out Tina N.'s awesome blog posts here!

Saturday, March 15, 2014

BQ #1: Unit P Concept 3 and 4: Law of Cosines and Area of an Oblique Triangle

Law of Cosines
1. Why do we need it?

  • When dealing with real life examples of things like triangles, we need to realize that not all of these situations are ideal. There will be triangles that are not right triangles but we still need to know what to do with them. This is where the Laws of Cosines comes in handy. With this, we will be able to find missing parts of practically any triangle, may it be a right triangle or a non-right triangle.
    • We can only use Law of Cosines when we are given SSS (all three sides) or SAS (side, angle, side).
2. How is it derived from what we already know? 
http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm
  • At first glance, it is obvious that this triangle is not a right triangle. As a result, we drop a perpendicular from angle B to create two right triangles (ABD and BCD). We want to identify the perpendicular as "h" (the height) and the distance between angle A to the right angle (D) as "r". 
  • From here, we can begin our derivation. For this, we want to find the distance between angle B and C (side a, basically). If we were to plot this triangle on the graph (beginning at the origin), angle C would be plotted at (b,0) because we move b units on the x-axis and we do not move at all on the y-axis. Angle B would have the ordered pair of (cCosA, cSinA). We are able to get these values with our knowledge of trig functions. Looking at cosA, we know that it is equal to r/c (adjacent/hypotenuse). Now, "r" is the amount of units we go on the x-axis, therefore, we want to change cosA = r/c to r = cCosA. That is why our x-value for Angle B is cCosA. The reason why we have cSinA as our y-value is because sinA = h/c (opposite/hypotenuse), and after we manipulate the equation, we discover that h = cSinA. "H" is the amount of units we go up, therefore, cSinA is our y-value. 
  • Now that we have these two points, we can plug them into our distance formula. We use the distance formula because we are trying to find the distance/length between angle B and C. 


  • In order to have an easier problem we can work with, we square both sides of the equation to get rid of the square root. Now, we can begin substituting certain values for others in order to shape the distance formula into our Law of Cosines. 

http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm
*Keep in mind that these values were plugged in d^2 = (y-y)^2 + (x-x)^2 (just a different arrangement)*
  • Because we are looking for the distance between angles B and C, also known as side a, we change the "d" to an "a", allowing us to be more specific. Now we plug in the values. First, we plug in the y-values: cSinA and 0. Because subtracting 0 does nothing, we leave cSinA by itself in the parentheses. In the next set of parentheses, we plug in the x-values: b and cCosA. That is how we get (b - cCosA). 
  • We want to expand these values next, so we foil them as we would with polynomials. From expanding (cSinA)^2, we get c^2sin^2A. When we expand (b - cCosA)^2, we get -2bcCosA + c^2cos^2A. 
  • It looks overwhelming now, but we can simplify the equation by taking out c^2 and leaving the rest of the problem. At this point, we are left with (sin^2A + cos^2A) within the parentheses. Because of this, we are able to simplify it to 1. 
    • The reason why we can simplify sin^2A + cos^2A is because it is like our Pythagorean Theorem. x^2 + y^2 = r^2. If we divide everything by r^2, we get (x^2/r^2) + (x^2/r^2) = 1 (r^2/r^2). We know from our knowledge of trig functions that cosine is x/r and sine is y/r. Therefore, when we square cosine and sine we get (x^2/r^2) and (x^2/r^2). That is why we are able to have (sin^2A + cos^2A) equal 1. 
  • Now, because we are only multiplying c^2 by 1, we are left with the final equation: a^2 = c^2 + b^2 - 2bc cosA. Keep in mind that we are able to derive this equation for different sides of the triangle using similar processes. If we were looking for side b or c, we would end up with similar equations. 
http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm
*These are the Law of Cosines variations for different sides of the triangle other than side a*
Area of an Oblique Triangle:
1. How is the “area of an oblique” triangle derived?
http://hotmath.com/hotmath_help/topics/law-of-sines/law-of-sines-image015.gif
  • We usually rely on our perpendicular bases and heights in order to find areas of triangles. However, oblique triangles do not have perpendicular bases and heights, so we must create them ourselves, either by dropping a perpendicular, or drawing one ourselves.
  • Our regular formula is A = 1/2(bh). What we really need to know is our height since 1/2 is a given and "b" is usually already known. We are going to be using our trig functions once more. We will use angle B as an example. Sin B = h/c (h being the perpendicular drawn). Remember that we are looking for "h". As a result, we multiply both sides by c to discover that h = cSinB. Now that we know "h", we can substitute it in the area formula. Now we have A = 1/2(bcSinB). This is our formula for the area of an oblique triangle!
  • http://wps.prenhall.com/esm_blitzer_algtrig_2/13/3560/911520.cw/index.html
    • There are different variations of the formula since we can use different angles (depending on what we are given in the problem).
    • We must be given two sides and an angle in between them (SAS) in order to use this.
2. How does it relate to the area formula that you are familiar with?
  • We use the original formula (A = 1/2(bh)) as a starting point to find the area formula for an oblique triangle. We still use the same base, however we have to find our own "h" by using trig functions (sine).
Works Cited: 
1. http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm
2. http://hotmath.com/hotmath_help/topics/law-of-sines/law-of-sines-image015.gif
3. http://wps.prenhall.com/esm_blitzer_algtrig_2/13/3560/911520.cw/index.html

Wednesday, March 5, 2014

WPP #12: Unit O Concept 10: Angles of Elevation and Depression

(http://cdn.c.photoshelter.com/img-get/I0000NJw_WS3ZrAc/s/900/900/cribbar-newquay-20111030-11.jpg)
Kim Nana decided to go surfing since she closed the coffee shop for the day. She gets on her surfboard and begins riding a wave when she notices a larger wave in front of her. She measures the angle of elevation to the wave in front of her to be 38* and the angle of depression (the very bottom of the wave) to 42*8'. If she is 80 feet away from this wave, how tall is the taller wave in front of her? Round to the nearest foot

Please remember to try to do the problem on your own before looking at the pictures. 






Tuesday, March 4, 2014

I/D2: Unit O - Derive the SRTs

Inquiry Activity Series:

1. 30-60-90 triangle
In order to derive this triangle, we first need to begin with an equilateral triangle and go from there. The equilateral triangle's characteristic is that it has the same side measures for each of the three sides. In this case, our equilateral triangle will have side measures of 1. Another thing we should take note of is that an equilateral triangle is also an equiangular triangle which means all of the angles are of the same measure as well. So, because a triangle's angles must always add up to a sum of 180, we can concur that each of the three angles will be 60 degrees (180/3=60).
Now that we remember the basic knowledge of our equilateral angle, we can begin to shape it into a 30-60-90 triangle. The first thing we want to do is split the triangle in half. By doing this, we create a 90 degree angle inside. We also, by dividing the triangle in half, we divide an angle in half. By splitting this angle we create a 30 degree angle on one side of the equilateral triangle. Thus, we technically create two 30-60-90 degree triangles--but we only want to look at one.
Above, we can see the transformation from an equilateral triangle to a 30-60-90 triangle. We have the angles, so now we want to find the actual sides--the important part. As you can see in the photo, 1/2 is written across the 30 degrees. I was able to get the value 1/2 with simple logic. Each side length was 1 and because we split the triangle in half, I also split that bottom side in half. So we know that the side across the 30 degrees is 1/2 and the side across the right angle is 1. Now the question is, what is the side across the 60 degrees, the side we created by splitting the triangle in half. We are going to determine this side with the help of our friend Pythagoras, and his Pythagorean Theorem. 
Pythagorean theorem: let legs be "a" and "b" and hypotenuse be "c"

In the above picture, I used the basic knowledge of the Pythagorean Theorem to find our 2nd leg. After working it out (look at the picture), I discovered that our unknown side has the length of 3/2. Finally, we know all sides of the 30-60-90 triangle!

I placed the letters that the values corresponded with in the Pythagorean Theorem.
We are not completely done quite yet. Dealing with the fractions can be quite annoying at times (correction: all the time) so we want to multiply each side by 2 in order to create nice numbers. Side "a" would be 1, side "b" would be 3, and side "c" would be 2. If you notice, even if we did not multiply by 2, side "c" is twice the amount of "a" and side "b" is basically side "a" multiplied by 3. This is where we begin to derive the 30-60-90 triangle. The relationships I pointed out are found in multiples of these sides. If we add those "n" into the picture, we see this relationship. Our side "a" (across the 30 degrees) would be "n" because it is only 1. Side "b" (across 60 degrees) would be n3 because it is side "a" multiplied by 3 (as we discussed). Side "c" (across the 90 degrees) would be 2n because our hypotenuse is twice the length of side "a". The reason why we place "n" is to emphasize the fact that this relationship is always there with these angles even with different numbers from the ones we found in the above picture. It is like ratios where they do not all have the same numbers but have the same relationship (ex: 1:2 is the same as 2:4). 

The final 30-60-90 triangle derived from the equilateral triangle. 
2. 45-45-90 Triangle
We want to derive this triangle from a square. Refreshing our knowledge of squares, we know that all sides are equal, all angles add up to 360, and each angle is equal to each other. Using quick mental math we know that each angle will have to be 90 degrees (360/4=90). Our square we will be using has side lengths of 1. From here, we want to create a 45-45-90 triangle and the only way to do that is draw a diagonal through the square. Like we saw in the 30-60-90 triangle, we are going to see our angles split. Two angle will be split in half (because of the diagonal) so those angles will be 45 degrees. We now have a 45-45-90 triangle. 
Now, we have both sides across the 45 degrees: they are equal to 1. But now, we want to know what the measure of the hypotenuse (the dotted line) is in order to find the relationship between these numbers. Let us return to the Pythagorean Theorem!
Knowing the two legs allows us to find our hypotenuse, which is radical 2.
We now know each of our sides. Both legs are equal to 1 while the hypotenuse, we just found, is equal to 2. Comparing the sides with one another, we see that the legs are 1 while the hypotenuse is radical 2 multiplied by the legs (1). These are the values when we use a square with the sides of 1. However, the idea we want to understand is that this relationship extends beyond one specific triangle with specific lengths. This is where we substitute "n" into the triangle. We want "n" to take the place of 1 (because we have ONE "n"). So our legs (a&b) are going to be "n". Our hypotenuse (across the 90 degrees), because it is 1 times radical 2 (radical 2 times the length of side "a" or "b"), will be n times radical 2. "N" allows the possibility of different lengths and measures of these sides but still contains the relationship found in the 45-45-90 triangle. 
We found the relationship between the sides of a 45-45-90 triangle!
Inquiry Activity Reflection:

  1. “Something I never noticed before about special right triangles is…” is that they are derived from these sort of shapes (the equilateral triangle and square) in order to justify why we have these numbers (they are not just randomly given).
  2. “Being able to derive these patterns myself aids in my learning because…” if I forget the values of the sides, I can easily use my knowledge of pythagorean theorem to find them.