BQ#2: Unit T Intro-How do the trig graphs relate to the Unit Circle?

Terms to help for the reading below:

         PERIOD-graphs are cyclical and once completion of one cycle its called a period.

         AMPLITUDES-are half the distance between the highest and lowest points on the graph.

Photo-credit: Fabulous and Beautiful Victoria Ventura

Photo-credit: Fabulous and Beautiful Victoria Ventura

A. Period?-Why is the period for sine and cosine 2pi, whereas the period for tangent and cotangent is pi? Interesting question. As seen from the lovely drawn pictures from above, we notice that both in sine and cosine we have repeating patterns: +,+,-,- and +, -, -,+. In order to complete a period we must go all around the unit circle. Keep in mind our graphs are just the unit circle unfolded and repeated with the same period. From 0 degrees to 360 degrees is 2pi. So that is why sine and cosine are 2pi. Seen from the picture below, we see that the pattern is +,-,+,-. The reason why tangent and cotangent's period is only pi is because the pattern is already completed by pi, unlike sine and cosine, in which one has to go all around in order to complete the period. Life is much easier doing tangent and cotangent, to be honest. 

Photo-credit: Fabulous and Beautiful Victoria Ventura

B. 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 the Unit Circle? In the Unit Circle all the other trig functions could be greater than one (YAY, for them) but for sine and cosine, they couldn't be greater than one. Which makes complete sense why the amplitude for sine and cosine is restricted to one and the others CAN be greater than one (such rebels.)


ANY QUESTIONS FEEL FREE TO COMMENT BELOW, THANK YOU FOR YOUR TIME.

Reflection #1: Unit Q Verifying Trig Identities

1. What does it actually mean to verify a trig identity?
What verifying a trig function actually means is to prove that the answer given can actually be solved out from the equation given. There are multiple ways to verify a trig function. It might, at first, seem easy but because we are given an answer, we have to make sure to get the exact answer. There are multiple ways to verifying a trig identity but one must be cautious in what other trig identity is being used.

2. What tips and tricks have you found helpful?
Definitely memorizing all trig identities is very helpful. Using reciprocal identities come very handy when solving. If you see a sin, cos or tan squared, figure out if you can use a Pythagorean identity! It will honestly save your life. Sometimes ratio identities can save your life too. Although it is important to know the identities off the top of your head, while doing pq's, or homework it is also helpful to have a list of the identities out in front of you.

3. Explain your thought process and steps you take in verifying a trig identity.
First thing is DO NOT FREAK OUT. I freak out all the time and instead of doing the identity I stare at it and how hard it looks instead of doing it! Then I figure out I can use the reciprocal identities, cross multiply and BOOM, I get the answer. A trig identity might look extremely complex but all it might take is replacing one of the parts with a Pythagorean identity, another with a reciprocal identity and that can lead to your answer. Remember, do not freak out, be calm and solve your life away!

SP#7: Unit Q Concept 2-Finding trig functions using identities

This post was made in collaboration with Jorge Molina. Please visit the other awesome posts on his blog by going here.

Hello there fellow students. Before starting the problem there are a few things you must be aware about. First of all, make sure not to solve this problem straight using trigonometry. We will be using identities for this problem, including: reciprocal identities, ratio identities, and Pythagorean identities. Secondly, solving identities are much easier than we think of them to be. Make sure to look at the clues to help you solve for the problem. Lastly, make sure when rationalizing to be careful what you are multiplying and always look for simplification. Other than that, relax, enjoy and solve your life away!

Photo-credits to the fab-u-lous Victoria Ventura

Photo-credits to the fab-u-lous Victoria Ventura

Photo-credits to the fab-u-lous Victoria Ventura

Photo-credits to the fab-u-lous Jorge Molina

Photo-credits to the fab-u-lous Jorge Molina

Photo-credits to the fab-u-lous Jorge Molina

Photo-credits to the fab-u-lous Victoria Ventura

WPP#12: Unit O Concept 10-Solving angle of elevation and depression word problems

Pooky's great adventure in Disneyland!:

One beautiful Tuesday afternoon, while roaming around looking for her parents, Pooky stumbled upon the Sleeping Beauty Castle at Disneyland. She was admiring its beauty when all of the sudden she was kidnapped by the troublesome Tinkerbell! GASP. (Here's the deets: Tinkerbell was jealous that as Peter Pan flew past the castle, he admiringly stared at Pooky.) Unfortunately, Tinkerbell placed Pooky in the highest tower, making it difficult for Pooky to be rescued.

While Pooky waited for Prince Charming to save her from the tower, she decided to measure Matterhorn's mountain height. Being the smart-alec she was, she figured out something incredible.

She measured the angle of elevation to the mountain across to be 29 degrees and the angle of depression (to the base of the mountain) to be 40 degrees. If the two, the castle and mountain, are 75 feet apart, how tall is the Matterhorn Mountain? (Round to the nearest foot) 

Here's a lovely picture of Pooky admiring Disneyland:

Picture made and taken by the lovely Victoria Ventura

Here's the real business and work for the problem:

Picture made and taken by the lovely Victoria Ventura

ANY QUESTIONS? FEEL FREE TO COMMENT.

I/D#2: Unit O-How can we derive the patterns for our special right triangles?

Inquiry Activity Summary:

30-60-90

Credits: To the wonderful and beautiful Victoria Ventura

So you may be asking yourself, "what is this hoopla that I am looking at?" Its not as hard as you think or at least how it looks. Let me simplify it for you. The triangle is an equilateral triangle all the sides equal 1. But this means all angles are 60 degrees? So what do I do? You split them in half, making it into two 30-60-90 degree triangles. We know for a fact from our unit circle the bottom "x" side would be 1/2 and the "y" side would be radical3/2. And the hypotenuse would be one.

Now how could this be applied to all different triangles that simply don't have the hypotenuse as one? We'd simply add a variable to make it applicable to all side lengths possible ever! This is the pattern if you haven't caught up...

Below is a picture of an example to help you further on your understanding of 30-60-90 triangles:

Credits: To the wonderful and beautiful Victoria Ventura

45-45-90

Credits: To the wonderful and beautiful Victoria Ventura

Much simpler than the 30-60-90 triangle is the 45-45-90 triangles. Just as we did before with the the equilateral triangle, we would have to split the square in half. Splitting directly through two 90 degree angles. Thus, you'll get two 45-45-90 triangles. Fabulous, I know. All sides are equal to one. But what would the hypotenuse equal? This is where our Pythagorean theorem would take play. 1^2 + 1^2=2 which in fact would be radical2. To make it relate-able to other 45-45-90 triangles, just as we did with 30-60-90 triangles, we would multiply all sides with n.

With this added variable, it would be a great pattern to follow and find other missing sides, knowing limited information.

Here's a fabulous example further showing solving 45-45-90 triangles having only limited information!:

Credits: To the wonderful and beautiful Victoria Ventura

Inquiry Activity Summary:

Something I never noticed before about special right triangles is how simple they can actually be. I made my life misery last year I over-thought things and kind of just got tired so I memorized the sides. I never put thought into how the sides were derived. This I/D really helped me to see that in a clearer light and understand how these triangles actually got the sides.

Being able to derive these patterns myself aids in my learning because I'm learning to be able to solve things on my own without help of anyone else. Of course, just like any other human on planet earth, I get a little help from my friends from time to time but, only when most needed. Just like the previous I/D, it really helps students to learn to actually practice thinking, not just doing careless work and understanding why, what and how and a problem can be solved, such like this I/D. It really helps us see our strengths and weaknesses and is a major help to students.

example: Unit O Concept 6

I/D#1: Unit N Concept 7:The Unit Circle and the Magic Five

Inquiry Activity Summary:

Angles on the first quadrant

30 degree angle

Figuring out the 30 degree angle triangle was very simple, though at first it was a bit difficult. The reason being because I didn't know the measurements of each side. But once I logged onto legoogle.com, I was able to find all the sides. Side R = 2x, side Y = x and side X = xradical3 (as seen in the picture.) Our hypotenuse/sideR had to equal one, therefore, we divided out 2x from the 2x and we got side R to equal 1. Simple right, are you done yet? Nope. In order to divide out the 2x from sideR we had to do that to all the sides. SideX when we divided and simplified it gave us radical3 over two. For sideY we did the same and the x's cancelled therefore leaving us with 1/2.

The second part of the 30 degree angle was finding the points as if it were on a graph. Taking that initiative we drew out the x and y axis. Our first point on the graph was the center point and that was simply just (0,0). The second point on the the x-axis, on the right, would turn out to be (radical3/2, 0) because of the information we had solved earlier. According to rise over run, our last point would have to be (radical3/2, 1/2). Why? Because we went over radical3/2 and rose 1/2. Lastly we filled in the blanks for r, x, and y.

45 degree angle

Figuring out the 45 degree angle was much simpler than the first angle we had to figure out because we already had an idea of how to solve it out. According to our sources, google.com, sideR was xradical2, sideX and sideY were both x. Just as we did for angle 30, sideR had to equal 1. Therefore, we divided out xradical2 from xradical2 to get one. We divided out xradical2 from all the sides. When dividing out xradical2 from x we ended up with 1/radical2, of course we all know, that no radical can be in the denominator, so we multiplied radical2 to both the bottom and top. Our final result was radical2/2. From there we drew the x and y axis and figured out the points, resulting with what is above in the picture. Lastly, we filled in the r, x and y blanks.

60 degree angle

While we all solved this angle we slowly realized that everything was the exact same as the 30 degree angle except everything was reversed. SideR was the same. But the X for the 60 degree angle was 1/2 instead of being radical3/2 and sideY was radical3/2 instead of 1/2. As for the points as well! Instead of going through all the work, since we already knew how to do it from the previous 30 degree angle, we just reversed everything and finished our 60 degree angle in seconds.

How did the activity help us derive the Unit Circle?

Well, glad you asked. These three angles are key to finding the rest of the Unit Circle. All we need to know is each and every angle (referring to 30, 45,  60 degree angle) and we can find the rest of the unit circle.

http://fac-web.spsu.edu/math/edwards/1113/unitcircle.htm

As seen from the Unit Circle above, the only points that are repeatedly seen are 1/2, radical3/2 and radical2/2. That's because all angles have the same common thing, they all have reference angles of 30, 45, and 60. 30 and 150 reflect off one another and have the same point except for the negative x, but that makes sense since it is on the negative side of the x-axis. This method goes to the rest of the angles as well they all reflect one another and tie back to either a 30, 45 or 60 degree angle. The only thing that changes throughout the Unit Circle is the degrees and radians but that's simply because as we go around the circle the angle gets bigger and bigger.

Quadrant II (30 degree angle)

This angle above in fact is the 150 degree angle. But it has the reference angle of 30 degrees. The only thing that really changes compared to the original 30 degree angle is that the 150 degree angle has a negative radical3/2 and of course the degree and radian are different as well.

Quadrant III (45 degree angle)

The angle above is actually 225 degrees. This angle has a reference angle of 45 degrees. The only thing that really changes compared to the original 45 degree angle is that the 225 degree angle's the x and y points are negative. To conclude, the degree and radian are one of the other factors that are different.

Quadrant IV (60 degree angle)

The angle above is actually 300 degrees. This angle has a reference angle of 60 degrees. The only thing that really changes compared to the original 60 degree angle is that the 300 degree angle's x and y points are negative. Lastly, the degree and radian are one of the other factors that are different.

Inquiry Activity Reflection:

The coolest thing I learned from this activity was that I finally understood how to solve the Unit Circle! Last year, in Algebra II, when we were learning this, I was so lost. I wanted to love the Unit Circle but the Unit Circle didn't love me. It was a very depressing time for me. But now me and the Unit Circle are on the same terms. I get it and the circle gets me. It's a happy world isn't it? Yes, very.