SP7: Unit Q Concept 2: Finding all six trig functions using trigonometric identities.

This SP7 was made in collaboration with Miguel G. Please visit the other awesome posts on their blog by going here.

In this post we will be finding the six different trigonometric ratios using two different routes. The first is for this current concept-using the identities- and the second for a concept in unit o- using the triangle-, however your answers will be the same exact thing. Be careful to pay attention as we walk you through the problem step by step and each identity. It is simple and easy to understand once you get a walk through. Here we go.
 
Unit Q Concept 2

 

 

 

 

Unit O Concept 5

As you can tell they are the same for both of the concepts. The answers are the same, the path to that answer is different. Both ways will work though.

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

Inquiry Activity Summary:

1. Where does sin^2x+cos^2x=1 come from?
A) First off, an identity is a proven fact or formula that is always true. The Pythagorean Theorem is an identity because it will always work, there is no number that cannot be plugged into that equation.
B) Using the letters x, y, and r we can create the Pythagorean Theorem. The Pythagorean Theorem comes from these letters. We can use x as one side, y as the other side, and r as the hypotenuse-remember the Pythagorean Theorem is for triangles- and you square all of them. Your result is the equation of x^2+y^2=r^2. These are terms we have derived from the unit circle and that is why we use them so that they can be associated with the trigonometric ratios.
C) Now we want the theorem equal to one so that we can create the identity. In order to do this we have to divide each letter by r^2. This leaves us with (x^2/r^2)+(y^2/r^2)=1. With the law of exponents we can make this a little bit easier to see and understand by making it a fraction squared for both. This leaves us with (x/r)^2+(y/r)^2=1. This is easier to understand and read.
D) Now we know that the ratio for cosine in the unit circle is (x/r). Remember that this comes from our triangles and the values they represent.
E) We also know that the ratio for sine is (y/r) which also comes from our triangles and the unit circle.
F) If you pay attention you will see that the ratios in d and e match the ratios we have in c. Using the substitution property we can replace our fractions with the functions and we have our new identity. We have cos^2(x)+sin^2(x)=1. This is what our identity is.
G) It is known as a Pythagorean Theorem Identity because all we used was the Pythagorean Theorem (which is an identity also) to figure it out and substitute some terms. It is the only formula that we used though.
H) We can verify this by using one of the angles from the unit circle. We will use a 60 degree angle. The cosine is (1/2)^2+ the sine which is (radical3/2)^2 and that produces the answer we are looking for which is one. Overall it is a working and true formula therefore it is an identity. You can also make it work with the 30 degree or 45 degree angles.

2. Show and explain how to derive the other two Pythagorean Theorem Identities from the first..
A) To get the second identity we need to divide everything in the equation by cos^2(x). This then gives us three fractions that fit with our other identities. The first is (cos^2x)/(cos^2x) which gives us the answer 1. The second is (sin^2x)/(cos^2x) which is equal to the powered up form of the reciprocal identity of tanx=sinx/cosx [Powered Up means that it is in the second or third or fourth power for each of the pieces of the identity]. That gives us tan^2x. Then the third, on the other side of the equal sign, is 1/cos^2x which is the powered up form of the reciprocal identity of secx=1/cosx. That gives us sec^2x. The resulting equation is 1+tan^2x=sec^2x. It is as easy as that.
B) To get the third one you divide the first identity by sin^2x. That gives you three fractions. The first is (cos^2x)/(sin^2x) which is the powered up form of the ratio identity of cotx=cosx/sinx. That gives us the answer cot^2x. The second fraction is sin^2x/sin^2x which is just 1. The third one on the other side of the equal sign is 1/sin^2x which is the powered up form of the reciprocal identity of cscx=1/sinx. That gives us the answer of csc^2x. The final equation-place the one in the front just to make it match the second one- is 1+cot^2x=csc^2x.

Inquiry Activity Reflection:

1) The connections that I see between Units N,O,P, and Q so far are that they are all focus on using the trigonometric ratios of sine, cosine, tangent, cosecant, secant, and cotangent. They also all are associated with the Unit Circle. Yeah for the Unit Circle!!

2) If I had to describe trigonometry in three words they would be ratios, triangles, and identities.

WPP #13/14: Unit P Concepts 6-7: Law of Sines and Cosines

This WPP13-14 was made in collaboration with Miguel G.  Please visit the other awesome posts on their blog by going here.

Into the City: A Tale Of Two Cousins

 

Problem 1: Mr. Q was on his way back from his trip to Antarctica and was flying into the Los Angeles Airport (LAX). It just so happened that his cousin Jack the Pumpkin Seller (JPS) was also flying into LAX from the small farming town in the plains. They had talked about meeting up before they got on the planes. The one thing they did not plan was where they would meet up. When Mr. Q arrived at gate 4A he got off, collected his luggage, and went to stand and look out the window. When JPS got off at gate 12A he got his luggage and went to stand by the windows. From the point where Mr. Q was standing to where JPS was standing is three hunna yards. They are both looking out at the incoming plane wondering if it is the other cousin coming in. Mr. Q is looking at it from a bearing of North 47 degrees East and JPS is looking at the same plane at a bearing of North 56 degrees West. What is the distance that Mr. Q is from the plane? What about JPS?

 

 

 

Problem 2: Mr. Q and JPS finally found each other after making a simple phone call to one another. By this time it was 4:30pm and both of the cousins were starving. Mr. Q knows of a place called Mustache Mikes right down the street and around the corner. However, JPS used his phone and thought it was down the street and around the corner in the other direction. They decided that they would go their own routes and see who was right. Mr. Q headed there at a bearing of 290 degrees while JPS headed there at a bearing of 080 degrees. Mr. Q heads in his direction at a speed of 8mph for 30 minutes and JPS heads in his direction at a speed of 6mph for 30 minutes. It turns out that Mr. Q was right. How far is JPS going to have to travel to get to the restaurant? 

 

BQ#1: Unit P: Concept 3: Law of Cosines SSS or SAS and Concept 4: Area of an oblique triangle

iii. 3. Law of Cosines - We need it because it allows us to find the missing sides and angles of oblique triangles with side angle side (SAS) or side side side (SSS). It is derived by making two right triangles, using the trigonometric ratios, and the distance formula. It is helpful and simple to understand. We derive it from a triangle in the first quadrant because all the ratios are positive and the origin is (0,0) making it all easy to derive. It works for a triangle with any lengths and with any angles. It will always work as long as the angles are realistic. The pictures below will walk you through it step by step. 

 
iv. 4. Area formulas -It is derived by drawing a line down the middle of the triangle and that creates two right triangles. Then you apply the trigonometric ratio and substitute that in for h. Now we have the area formula for an oblique triangle that looks almost the same to the one we know and love (A=.5bh), but we have our replacement for h, what h equals, and get the new formula (A=.5b(asinC)). The pictures below will walk you through it step by step. 

WPP #12: Unit O Concept 10: Angles of elevation and depression.

REMINDER! YOU HAVE TO TRY THE PROBLEMS YOURSELF AND THEN LOOK AT THE PICTURES TO CHECK! HAVE INTEGRITY!
A) Mr. Q decided that he was going to visit Antarctica. While he was down there he decided that he was going to stand on this mound of snow to look up at a group of penguins on a cliff. This meant that he was looking up at the top of the cliff from 5 feet off the ground. The base of the cliff was 180 feet away from his precious mound of snow and the cliff was 103 feet tall. What is his angle of elevation from eye level of Mr. Q? (Round to the nearest hundredth)

B) After that Mr. Q continued his adventure through the Artic. Mr. Q decided that he was going to stand up on a hill and look down at a group of penguins by the ocean of ice. His angle of depression was estimated to be about 37*. The path down the hill to the spot by the icy sea is 240 feet. How tall is the cliff that he is standing on? (Round to the nearest hundredth)

I/D2: Unit O- How can we derive the patterns for our special right triangles?

Inquiry Activity Summary
 45-45-90

Above you can see that we have an equilateral triangle. The length of each side is 1. You divide the square in a diagonal you get two triangles. Since the angles for the triangle were each 90*(because you must remember that the angles of a square add up to 360*), you now have two 90* angles and four 45* angles. Looking at the bottom triangle (the pink one) we see that each of the sides remain 1. We will label these sides a and b. The side across from the 90* angle will be the unknown c. Now that we know two of the sides we can plug it into the Pythagorean Theorem. Yeah for the Pythagorean Theorem! This is a^2+b^2=c^2. If you look at the top picture you can see it worked out completely step by step. You plug your known values in (which are 1 for both a and b) and you get c equals radical 2. That is what your diagonal is. Now this is all great if we are always working with equilaterals, but the truth is that we are not always working with them. In order to make sure that this can work with any of them, we multiply each side by the variable n. "n" just represents an unknown value. When you multiply each of the sides (a, b, and c) by n you get the new values of n (for side a), n (for side b), and n radical 2 (side c). [Clarification is on the top photo.] The variable allows a consistent expansion with any triangle that does not have the value of 1. If you look at this triangle it has the angle measurements of 45-45-90. This creates a consistent pattern for your triangles with these angle measurements. The n, n, n radical 2 will now and forever be associated with this type of triangle.

30-60-90

This is an equilateral triangle. The sum of all the angles in a triangle is 180*. Each angle has a measurement of 60*.  You divide the triangle right down the middle so that your top angle becomes two 30*, your two bottom angles remain 60*, and the split creates two 90* angles. Since an equilateral triangle is equal on all sides, your two sides on the left and right will remain 1 and your bottom side will become two sets of 1/2 because you divided it into two. Now look at the highlighted triangle and label the side across from the 90* angle c and the one across from the 60* b and finally the side across from the 30* a. If you look at the values and the variables you just put in place you see that you have two known variables and one unknown. We are going to use the Pythagorean Theorem to solve for the unknown. The theorem is a^2+b^2=c^2. You know the values for a and c. Plug those in and in the end you get b=radical 3/2. [Look at the top picture for clarification and walk through.] These all work if we were always going to use a triangle that had these measurements. However, we will not always use a 1 as the value so we need a variable and we multiply each value by the n. Wait, we also need to get rid of the fractions because we want to keep the formula consistent with every triangle we go through and fractions are not consistent so we will be multiplying all three values by 2n. If you look at the picture above you will see this worked out. Your new values are n(a 30*), n radical 3 (b 60*), and 2n (c 90*).  Now that you have these new values with the variable n you can solve other triangles that are not set with values of 1. Then if we look at other triangles with these same angle measurements we see that these form a pattern that we can depend on for 30-60-90 triangles.

 

Inquiry Activity Reflection

1. Something I never noticed before about special right triangles is that you can derive both types or triangles from a set of equilateral shapes (the triangle and the square) and also that this is how the pattern of the triangles was established. I also realized what the variable n was there for now. It was a good refresher of things I forgot and things that I never really knew.

2. Being able to derive these patterns myself aids in my learning because if I ever forget how to do it I will be able to figure it out and go back to the very beginning and work it all out. It will help me fully understand and explain how to do some of the concepts in this unit (like concepts 7 and 8) and also allows me to understand why each side is what it is. It helps me not just memorize a pattern, but really understand it and make sense of it in my head and all of the work that I do. It will forever help me.