Reflection #1: Unit Q- Verifying Trig Functions

1. In this Unit, Unit Q, we learned how to verify a trig function. This actually means that we must prove that both sides are equal to each other because we know our answer already. So we must use our identities, which we all memorized to make the left side equal to the right side. WE NEVER TOUCH THE RIGHT SIDE. We are just verifying that is proven to be right. 

2. There are many tips and tricks that I have found helpful in this unit to help me understand it way more. One huge tip that I learned that if you can, change EVERYTHING into sine and cosine. By doing this it will make our lives easier because they will either cancel out or be equal to one. Another tip is to always work on the complicated side first that has a lot going on. By doing this first we are able to make it easier and get the hard side out of the way. A third tip is if you can, look for the greatest common factor. This will allow us to make our problems simplier and closer to our answer. The biggest tip I can give is to ALWAYS identify what you did in each step and make it neat. This will help you see where you went wrong if you did not get the answer right. It will also look nicer as well. 

3. The first step in my thought process is to see which identies I have in my problem. I always look to see if I have sine and cosine together, secant and tangent together, or cosecants and cotangents together because they pair up well with their idenities. Next I look for to see if anything cancel or if I have a greatest common factor so I can simplify it even more. If none of that works, the next step I do is look for a common denominator. Once I have done one or all of those steps, usually the identies work out the way I want it to and give me the same answer as the right side. Concepts 1 and 5 are tricky, but if you practice it will all make sense. Just memorize the identities. 

SP #7: Unit Q Concept 2

This SP#7 was made in collaboration with Kelsea Del Campo please visit their awesome blog by clicking here.

Using Identities:

Using SOCAHTOA: 

 In this student problem me and my partner made our own example from Unit Q Concept 2. In this first picture we showed how to find the values using Ratio Identities, Reciprocal Identities, and Pythagorean Identities. In the second picture we showed how to find the values usng SOCAHTOA. As you can see, we can find the values using both ways. In the first picture, using identities, you can see that we found all our values using different Identities and just substituting them into the problem. We used one identity to find another and so on. For the second picture, using SOCAHTOA, we just used the Unit Circle ratios from the previous unit. By doing these two pictures we have proven that identities can be associated with SOCAHTOA. 

I/D #3: Unit Q Concept 1- Using Fundamental Idenities to Simplify or Verify Expressions

1. An "identitiy" are proven facts and formulas that are ALWAYS true. The Pythagorean Theorem is a type of "identitiy" because we get proven facts and formulas that end up being true for any numbers we plug into it. With using the Pythagorean Theorem we use for this unit we use x, y, and r. We use these instead of a, b, and c because on a graph it when we graph it becomes x, y, and r.

As you see in the picture above all we had to do to get the Pythagorean Theorem equal to 1 is divide by r^2. Once we had r^2 it became equal to 1. From dividing by r^2 I noticed that we got x/r which is the ratio for cosine we used for the unit circle. I also noticed that we got y/r which is the ratio for sine we used for the unit circle. If we subsitiute those two for cos and sin we get cos^2x + sin^2x = 1. We can conclude that cos^2x + sin^2x = 1 beigins with the Pythagorean Theorem. It is referred to the Pythagorean Ideinity because we use the Pythagorean Theorem to get it.

As you can see in the picture above, I chose one of the "Magic 3" ordered pairs from the unit circle to show that it is true. The ordred pair used was the 60* angle one which is (1/2, rad3/2).

2.

The picture above shows that how to derive the identity with Secant and Tangent. First we had to divde everything by cos^2ø. Then that would cancel out the cos^2ø to make that be 1. Then we got sin^2ø/cos^2ø which we subtitute with tan^2ø. Lastly we have 1/cos^2ø and we substitute that with sec^2ø. Our final answer will look like tan^2ø + 1 = sec^2ø. 

The picture above shows that how to derive the idenity Cosecant and Cotangent. First we had to divdie everything by sin^2ø. Then that would cancel and make it be 1. After we got cos^2ø/sin^2ø which we substitute with cot^2ø. Lastly we have 1/sin^2ø which turns into csc^2ø. Our final answer will look like 1 + cot^2ø = csc^2ø.

WPP#13-14: Unit P: Concept 6 and 7

Please see my WPP 13-14, made in callboration with Kelsea Del Campo, by visiting their blog here. Also be sure to check out the other awesome posts on their blog!

BQ #1: Unit P Concept 2 and 4: Law of Sines SSA and Oblique Triangles

2. Law of Sines- Side Side Angle (SSA) is an ambigious case not like AAS or ASA. When dealing with SSA the three angles are not all known as easily as the others because we only know ONE angle out of the three. It is amigious because it can be three different types: one triangle, two triangles, or no triangle at all.

One Triangle 

As seen in the picture above there is only one possible triangle for this problem. We know there is not a second triangle because angle A and Angle C add up to 325.9 which is way past 180 degrees so therefore there is no second triangle because it is greater than 180.

Two Traingles 

As seen in the picture above this example has two possible triangles. We know there are two triangles because the law of sines means there is one angle in the first quadrant and another angle in the second quadrant. We find the second angle (the prime angle) by subtracting the first angle we got by 180.

No Triangle 

As seen in the picture above this example has no possible triangles. We know there are no possible triangles because once we used the law of sines with SSA, we got sinC: 1.75 and we know from the previous unit that sin can not be greater than 1 so therefore that leads to no solution. Another reason we know a triangle has no solution is when there is more than one obtuse angle.

4. Area Formulas- The area of an oblique triangle is derived from the formula for the area of a trianlge which is A=1/2bh. It area of an oblique triangle is one-half of the product of two sides and the sine of the angle the problem gives you. So basically the three types of equations can be A=1/2bcSinA, A=1/2acSinB, and A=1/2abSinC. It relates to the formula we are familar with by substituting in our h in the normal equation with the a side and sine of an angle given. We just have to make sure when we have our equation, all the letters are different.

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

My new puppy Lady 

A. From a dog named Lady on the grass, the angle of elevation to the top of a tree is 22*25'. If the base of the tree is 625 feet from Lady, how high is the tree? (to the nearest foot). 

B. Lady the dog is just about to dig down underground. She estimates the angle of depression from where she is now to the bone in the dirt to be 35*. Lady knows she is 250 feet higher than the bone, how long is the path she will dig? (to the nearest foot).

I/D #2: Unit O Concept 7 & 8- Special Right Triangles

Inquiry Activity Summary 

1.  We get a 30-60-90 triangle by cutting an equilateral triangle in half. We start off by using the Pythagorean theorem, which is a^2 + b^2 = c^2. By using this we know that our a is going to be 1 and our b is still unknown. Our c is going to be 2 as seen in the picture above.  Since we know our a and c we can plug those numbers into the Pythagorean theorem for our b. As seen in the picture above after plugging in those two values we find our b which is rad3.

As seen in this picture above we found a pattern with the 30-60-90 triangles. We added a variable at the end of each value side because we noticed that it can work for any number possible. We just start off with n to get a, then times it by rad3 to get b, and by 2 to get c. We multiplied each variable by 2 to get rid of the fractions so it can be simpler as well. 

2. We get a 45-45-90 triangle by drawing a slanted line through a square. We start off using the Pythagorean theorem, which is a^2 + b^2 = c^2. Since it is a square all sides are the same. For this example our sides are going to be 1. So we must plug in 1 into a and b to find c. As seen in the picture above, the work shows that our c is rad2.

As seen in this other picture above, we found a pattern with the 45-45-90 triangles. We added a variable, which is n to each value because the sides will not always be 1. This can work for any number because it is derived from the square. We put “n” in the pattern because it can be any number if it follows the 45-45-90 pattern. 

Inquiry Activity Reflection 

1. Something I never noticed before about special right triangles is where these patterns for them actually came from. After doing this activity I found out the origins of the patterns, which makes sense to me now.  I never knew the 45-45-90 came from a square and the 30-60-90 came from an equilateral triangle.

2. Being able to derive these patterns myself aids to my learning because it kind of ties everything together now. It even ties in together to unit circle because of the vales of the triangles. It also helps me remember the patterns now thanks to the Pythagorean theorem.