SP #7: Unit Q Concept 2: Finding all trig functions values when given one trig function and quadrant (using identities)

Please see my SP 7, made in collaboration with Sol S., by visiting their blog here. Also be sure to check out the other awesome posts on their blog!

I/D#3: Unit Q- Pythagorean Identities

INQUIRY ACTIVITY SUMMARY:

• First of all, we must make the connection that the Pythagorean Theorem is an identity. An identity is "a proven fact or formula that is always true". So in the PY, a^2+b^2=c^2, it's a proven statement. On a graph, instead of using "a,b,c" we use "x,y,z" which we can refer back when we were graphing special right triangles within the Unit Circle. So how do we get from x^2+y^2=r^2 to the pythagorean identity of sin2x+cos2x=1? We simply just divide r^2 on both sides to the x^2 and y^2. So now we have (x/r)^2+(y/r)^2=1. This is the moment where things just connect together and we can all turn on our lightbulbs! We see the ratio x/r and y/r and we know the trig functions that go with it. So when we plug that in we get sine from (y/r) and cosine from (x/r)! Since it's squared, we get the PI of sin2x+cos2x=1.
  
• How can we be sure that it actually works? We can use one of the Magic 3 pairs from the Unit Circle! Below is an example using 60 degrees:
  

• Here are the remaining Pythagorean Identities are derived from sin2x+cos2x=1
  
  

• ID w/ secant and tangent
  When we derive secant and tangent from our base Pythagorean Idenity, sin2x+cos2x=1, there are two things we know from memorizing the chart! We know that to get tangent, we would need sin2x/cos2x. Also, to get secant, we know the recipricol identity will be 1/cos2x. So how do we get this? Simply divide cos2x from everything and we are left with our Pythagorean ID of tan2x+1=sec2x!
  
  
  
• ID w/ cosecant and cotangent
  Similar to the previous problem, there are something we already know from the chart! To get cosecant and cotangent, we must divide sin2x from everything. From here, we know that cos2x/sin2x is the ratio identity for tan2x. As for 1/sin2x, we know that is the recipricol identity for csc2x. After we plug in the identities to the corresponding one, we get our Pythagorean ID of 1+cot2x=csc2x.

INQUIRY ACTIVITY REFLECTION:

• The connections that I see between Units N, O, P, and Q so far are that the Unit Circle from Unit O will always be our base/foundation. From it, we can use trig functions that we found from the special right triangles in the circle. The SRT also used the Pythagorean Theorem! So from Unit Q back to N, it all connects in the way that the identities are made up from ratio, recipricol, and pythagorean.
• If I had to describe trigonometry in THREE words, they would be tricky, complicated, "oOoOoOoOh" (when I understand the concept)!

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

*This WPP #13-14 was made in collaboration with Anthony S. Please visit his blog for more awesome posts by clicking here

Introduction: The Bakery Collaboration

(http://rainydaydesserts.files.wordpress.com/2013/03/macarons.jpeg)

(http://sallysbakingaddiction.com/2012/08/21/soft-andes-mint-chocolate-chunk-cookies/)

The Problem:

Law of Sines 

1.) Hannah decides that she wants to pair up with Juana, the owner of Juana's bakery, to sell her macarons! While Hannah walks towards the bakery, she sees that Juana is across from her in parking lot at a distance of 24 feet. Juana is N26*W from the bakery while Hannah is N43*E from the bakery. What is the distance bewtween Juanna and the bakery?

Law of Cosines 

2.) The two meet up at Starbucks to talk then decide to head over to the bakery. Both leave from the same point. Juana heads 027° driving at 32 mph while Hannah heads 146° driving 46 mph. If they drive for 2 hours, what is the distance between them? 

The Solution:

Solution to #1

Solution to #2

BQ #1: Unit P Concepts 2 & 4: Law of Sines and Area of an Oblique Triangle

2. Law of Sines

Why is SSA ambiguous?

• SSA is has three possible solutions: one, none, or two. When solving a problem for this triangle, you need to solve it as if you're going to get two possible solutions. The only way you know there's only one or no solution is when you "hit a wall". A wall can either be when sine is less than -1 or more than 1. Remember that the trig function sine should be between -1 and 1 to be possible. Another wall can be if the angle of your triagle adds up to be greater than 180 degrees. We know that a triangle can only be 180 degrees so if two angles already add up to that amount, we will hit a wall. It is possible that when we find the two possible value of an angle (one acute in the first quadrant; one obtuse in the second quadrant), one can still work for a possible of one solution!

Two possible triangles-

When given the triangle, we match each angle to its corresponding side and we can indicate the "magical pair" and "bridge". The "magical pair" is the one with both given angles and side while the bridge is the one with either a given side or and angle. We use the law of sines using the pairs of B and C to find the value of angle B.

In order to make sure a wall was not hit, the Triangle Sum Theroem can be used. After finding the possible values of B, we add them to the given angle of C and make sure it does not go over 180 degrees. Afterwards, we can find the degree of angle A and A'. Afterwards we use the same maigcal pair and use the Law of Sines but with the unknown variable of side a. 

Our answer results with two possible triangles, one acute and the other obtuse.

One Possible Triangle-

As usual, we find the magical pairs and bridge. Then we use the Law of Sines to find the value of angle C. However, only one of the angles work in this problem because we have hit a wall. The value of C' is 159 and with that added to 84, it exceeds 180 degrees. So now we know that angle won't work for a triangle but we must continue to find the rest of the other triangle.

Using the Triangle Sum Theroem again, we find the value of angle B. Then we use the Law of Sines to find side b.

Our answer results in one possible triangle.

No Possible Triangle-

In this problem, when we find sin C, we result with 1.96 Remember that sine can only be -1< sine Ø < 1. Since sine is over 1, there is no solution and this results in no triangle.

4. Area Formulas

How is the “area of an oblique” triangle derived? 

• The area of an oblique triangle is derived from the area of a triangle formula.

Since we know the area of a triangle is A=1/2bh, we can make a substitution for h that will help us find the area of an oblique triangle. If we are looking for the angle C, we know that the sine formula will be h/a sine it's opposite over hypotenuse. After we multiple a to both sides, we get h alone and this is where "a sin C" can be subbed into h from the regular area of a triangle. From there on, we can find the area of an oblique triangle by plugging in the two sides and the sine of the included angle. 

It can also look like these options depending on what sides and angle is given:

(http://wps.prenhall.com/esm_blitzer_algtrig_2/13/3560/911520.cw/index.html)

How does it relate to the area formula that you are familiar with?

• This relates to the area of a triangle (A=1/2bh) because it is derived from it. There is both the base and height multipled together and divided by two, however, the oblique formula also has the sine of an angle multiplied.

Works Cited-

1. http://wps.prenhall.com/esm_blitzer_algtrig_2/13/3560/911520.cw/index.html

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

Introduction: Hannah's Hike

 
(http://www.destination360.com/north-america/us/vermont/vermont-hiking)

The Problem: Hannah's Hike

After all the bow business and shopping, Hannah takes time to appreciate nature by going on a hiking trip with her friends. She is determined to hike up the cliff and be able to look down the side of the cliff to see the ledge where most people jumped off after such a long hike! (Don't worry it's safe even though it's 300 feet!)

1) Before they start their hike, Hannah wants to know the height of the cliff. The pole where she is standing from is 600 feet away from the cliff itself. She knows from the base of the pole to the base of the flagpole, there is an angle of elevation of 45 degrees. What is the total height of the cliff? (Round to the nearest foot!)

2) After reaching the top of the cliff, Hannah lies down next to the flag pole and looks down towards the tip of the ledge. She estimates the angle of depression from where she is lying to the tip of the ledge to be 21 degrees. If the ledge is 300 feet higher from the ocean floor, what is the distance she is looking at from the base of the flagpole to the tip of the ledge? (Round to the nearest foot!)

The Solution: Hannah's Hike

(This is a visual of what Hannah's hike looks like)

First we need to find the height of the cliff (we'll need this piece of evidence for problem #2). We know we must use trig functions to be able to find the missing side we're looking for. Remember that in an angle of elevation, there will be a horizontal line then the angle will go upwards.

Next, we can use the height we found earlier and other clues to find the side labeled "opp." to help us solve the whole problem. Again, we use trig functions to find the desired side. Remember that in an angle of depression, there is a horizontal line but the angle will go downwards.

I/D #2: Unit O- Deriving the SRTs

INQUIRY ACTIVITY SUMMARY:

1.) 30-60-90

(Deriving a 30-60-90 triangle from an equilateral triangle)

In a 30-60-90 triangle, we know that it is derived from an equilateral triangle. In an equilateral triangle, all the angles add up to 180, 60 at each angle. Since we know that the sides are the same in an equilateral, we can label them all the same. We are given 1 as our side length in this activity. Once we split the triangle in half, we get a 30-60-90 triangle. The value of a is 1/2 since it was split from 1 and c (the hypotenuse) stays the same as 1. To solve for the last side, b, we must apply the Pythagorean Theroem.

(finding the missing vairable, b using the Pythagorean Theorem)

We know that the Pythagorean Theorem is a^2+b^2=c^2. If we plug in our known variables, which will be a and c, we can find the missing variable, b. After we plug it in and solve, we are left with b=(rad)3/2 after finding the squareroot of (rad)3/4.

(Triangle after all sides are found)

Once we figure out the side length for each side, we are left with the above picture. Dealing with fractions can be such a hassle so it's easier to just multiply everything by 2 and get a non-fraction answer for each side. We are left with a=1, b=(rad)3, and c=2.

(final 30-60-90 triangle derived from an equilateral triangle)

When we are using SRT's it doesn't necesarily always have to have a hypotenuse of 1. By putting the variable n where 1 would be, it allows us to plug in any given value and keep the ratio at the same time. So in the above picture, we are left with the variables n(rad)3, 2n, and n. When given a value for n, we can easily plug it in and figure out the value of each side.

2.) 45-45-90

(Deriving a 45-45-90 triangle from a square)

When deriving a 45-45-90 triangle from a square with a side length of one, we know that the sides will be 1 all around. We know that in a square, all the angles are 90 because the total degree of a square is Once we split the square in half, our angles will be 45-45-90. Our a and b value will be 1 and to solve for side, c (the hypotenuse), we can apply the Pythagorean Theorem.

(Using Pythagorean Theorem to solve for missing variable, c)

To find the missing sides, c, we plug in our known variables into the Pythagorean Theorem. We end up with c=(rad)2. 

(45-45-90 triangle with all the sides found)

After we find the missing piece, our 45-45-90 triangle will end up like this. Unlike the 30-60-90 triangle where we had to deal with fractions, we can just leave the numbers alone since they're whole numbers.

(final 45-45-90 triangle derived from a square)

After finding all the sides, we can do the same thing as in the previous trianlgle and plug n into where the 1's are. We are left with n(rad)2, n and n. The n is used to plug in any given value while keeping the ratio, which represents the relationshp between the three sides of a special right triangle.

INQUIRY ACTIVITY REFLECTION

1. Something I never noticed before about special right triangles is how it derived from a square (for a 45-45-90 triangle) or an equilateral triangle (for a 30-60-90 triangle).

2. Being able to depreive these patterns myself aids in my learning because now I can derive it on my own and I know where the values came to be.