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

Please see our SP #7, made in collaboration with Nga Mai, by visiting her blog here. Also be sure to check out the other awesomely awesome posts on her blog.

I/D 3 Unit Q- Pythagorean Identities

1. sin^2x and cos^2x=1 comes from the Pythagorean theorem of x^2+y^2=r^2. We know that x/r equals cosine and y/r equals sine from the unit circle ratios as we should know that our trigonometric functions of cosine is adjacent over hypotenuse and sine is opposite over hypotenuse. Since in the unit circle we considered a side opposite of an angle to be opposite of the angle that is protruding from (0,0) in all cases, we can deduce that the adjacent value equals x and that the opposite value equals y. If we take the Pythagorean theorem and solved to make r^2 1 by dividing the whole formula by r^2, we get what we wanted which is (x/r)^2+(y/r)^2=1. As we already went over, x/r is cosine and y/r is sine, and since the equation is already squared, that is where we get our equation of sin^2x and cos^2x=1. The x value that accompanies the formula can be substituted for theta if we never knew it.

2. In order to derive the identity with Secant and Tangent, we start off with our standard formula of sin^2x+cos^2x=1. We want to get secant, so we divide the whole equation by cos^2x in order to get that one the right side. Once we have 1/cos^2x on the right side, we see that same ratio is secant itself, just used twice, so we substitute 1/cos^2x for secx. On the left side of the equation, we also see sin/cos being the identity for tan, so we substitute tan^2x for that since sin/cos was squared as well. Cos^2x/cos^2x is 1 because they are the same number, just one over the other. We then obtain our final formula of tan^2x+1=sec^2x.

3. Just like the last problem, we start off with our formula of sin^2x and cos^2x=1. We want to get cosecane and cotangent this time, so we divide the equation by sin^2x this time. We get sin^2x/sin^2x+cos^2x/sin^2=1/sin^2x. For the right side, we can substitute that for csc^2x because we know that one of our trig identities is that csc equals 1 over sin. We can sub sin^2x/sin^2x for 1 since they equal each other. We substitute cos^2x/sin^2x for cotangent because the identity of cotangent is cos/sin, or x/y. After we substitute everything, we get 1+cot^2x=csc^2x as our final equation.

Inquiry Activity Reflection

1. THE CONNECTION I SEE BETWEEN UNITS N, O,P, AND Q SO FAR is that they all utilize the unit circle as well as the Pythagorean theorem to solve a multitude of problems. 

2. IF I HAD TO DESCRIBE TRIGONOMETRY IN THREE WORDS, THEY WOULD BE sine, cosine, and tangent.

WPP #13&14 Unit P Concepts 6&7

This WPP was made in collaboration with a handsome fellow named Mason Nguyen. Please visit the other awesome posts on his blog at Masonnperiod1.blogspot.com.
1. Law of Sines
       Hannah and her friend Miley are going on a trip to Los Angeles. However, they are at different locations, being 100 miles apart. Hannah flies there in a helicopter at a degree of N 29* E. Miley drives there in a BMW at a degree of N 40* W. They finally arrive at Los Angeles and give each other patty cakes. What was the distance it took for Miley to get to Los Angeles?

 In the above picture, we see the set-up for the problem. Hannah and Miley are 100 miles from each other so thus that is our side value. We draw the north-south line to show our N 29* E angle and our N 40* W angle. x stands for the value which we want to find, which is the distance from Miley's starting point to L.A.

 We find our angles by subtracting the corresponding values we gotten from 90*. We get 61* and 50*. Since we know two values of our angles, we can easily solve for the third by subtracting the two angles from 180*. Therefore, 180-61-50=69. 69* is our third angle. Now that we have a "bridge," we can use the law of sines.

We set up the ratios as shown above. We then cross multiply to eliminate the fractions. We then try to get x alone by dividing by sin69. We plug in the right side to our calculator and thus get 93.68 as our answer.

2. Law of Cosines
        Hannah has appointments in two different places. She can't be in two places at once, or CAN SHE? She asks Justin Timberlake if she can borrow his cloning machine. She then clones herself and orders her Hannah duplicate to go due east to Bakersfield. Hannah herself goes northeast to Riverside. The angle they create is 046*. The Hannah clone travels 25 miles to Bakersfield. Hannah travels 50 miles to Riverside. How far are the two apart now?

 We label everything accordingly. Since the problem said for the clone to go due easy, we know that Bakersfield is directly east of the starting line. The clone traveled 25 miles while Hannah traveled 50, so we label those. We also label the angle as 46*. Now that we have SAS, we can use the law of cosines.

We plug in our numbers into the formula as so. Let us substitute the variable a for x so that we can use b and c for the other sides. The formula is thus, to put it variably, a^2=b^2+c^2-2(b)(c)cosA. We then get our value when we plug it all into the calculator (do it all in one step) and get our next value. We have to square the value due to a being squared, so then we get our answer of 37.26 (miles).

BQ #1- Unit P: Law of Cosines and Area of an Oblique Triangle

3. Law of Cosines- We need Law of Cosines because it can solve for any angle and side of any triangle. As long as we have the right information (three sides or a side angle side), we can easily solve for any value as it is derived from the distance formula. Since we know that on a triangle, angle C is at (b,0) and angle B is at (cCosA, cSinA) since b is the distance from angle B to between angle A and angle C (Since you draw a perpendicular line down angle B) we can derive the (b,0). Also, since c is the distance from angle C to between angle A and angle B, we can also derive the latter point.

                      https://www.youtube.com/watch?v=lPP-pABvwdA

In the above video, we see the method to deriving the Law of Cosines. We see our desired formula of a^2=b^2+c^2-2bccosA. However, how do we derive that formula? First of all, we draw a triangle (as shown in the video, not a right triangle) and label the corners appropriately as A,B, and C. We then label their opposing sides as a,b, and c. We then draw a perpendicular line down angle C as demonstrated in the video to form two right triangles. We now have to cut the c side into two different values. Label the left triangle as "x", and since that is x, then the other side's value must be c-x because c is the total side value while x is one portion of it. Now that we know all the side values (as variables) we can use the Pythagorean Theorem. For the left triangle, we have x^2+h^2=b^2 (remember that h is the height we made from cutting the triangle in 2). For the right triangle we have (c-x)^2+h^2=a^. We have to foil this, so after we do that, we get c^2-2cx+x^2+h^2=a^2
            Where do we go from this?  Since x^2+h^2 is present in both the left and right triangles, and since the left triangle says that it equals b^2, then we can substitute that in for the right triangle formula to make c^2-2cx+b^2=a^2. We can move b^2 to the front and have b^2+c^2-2cx=a^2. As we reverse the equation to make it more like our original, desired formula, we have a^2=b^2+c^2-2cx. The x value in that formula does not correspond to what we want, so how do we get it? We get it by using our first triangle and taking the cos of angle A to get cosA=x/b. We solve for x and get x=bcosA. We substitute that in for our formula and get the desired formula, thus deriving the Law of Cosines.

4. Area of an Oblique Triangle- As we already know in an ABC labeled triangle, the sin of angle C is h/a because like the last problem, we draw a perpendicular line down the middle angle to make 2 right triangles. Since sinC=h/a, then that also means that h=asinC when we solve for h. Now when we substitute our common area equation for h, we get A=1/2b(asinC), since t'he regular area equation is A=1/2bh. This relates to the area equation because we use both base and height as it does except that we go a step beyond by using our laws of sin to derive our formula.


       https://www.youtube.com/watch?v=gMnBMJwpyc0

In the above video, we see how an oblique triangle is derived. This formula is very simple. It is A=1/2bcsinA. Of course, there are variations, so it can also be 1/2acsinB or 1/2 absinC. When you know the SAS value of the triangle, then you can easily plug in the corresponding values and then plug into your calculator the equation. For example, in the video, you merely plug in 1/2 x 12 x 8 x sin(135) to get your answer.

References:
http://www.mathsisfun.com/algebra/trig-cosine-law.html
https://www.youtube.com/watch?v=gMnBMJwpyc0
https://www.youtube.com/watch?v=lPP-pABvwdA

WPP #12 Unit O Concept 10: Elevation and Depression

         1.)  Hannah realizes that she wants to glide down a building into a soft bush. She climbs a 500 ft building, reaches the top, and gazes down at the bush at an angle of 25 degrees 10'. If Hannah was to glide perfectly horizontally, then suddenly fall onto the bush, what would be the distance she traveled horizontally.

THE PROBLEM

In order to solve this problem, first we need to calculate the value of 25 degrees 10'. We accomplish this by placing 10 over 60 (because there are 60 seconds in a minute) to get 10/60 which simplifies to 1/6 which, in decimal form, is .17. After we get the angle value, we can utilize trig functions. In the below picture, I use tan.  Tan is y/x so we use 500/x. We don't want variables as the denominator so we multiply both sides by x. We have to get x by itself so we divide both sides by tan 25.17. Our answer is 1063.99 which rounds to 1064 ft.

                                                                       The Solution

          2.) Hannah, after landing on the bush, has just discovered that she is terrified of falling. She deduces that what she loves is parkour. She loves to jump from buildings to buildings. She climbs the same building as the last problem, but wants to jump to the top of a 1000 ft building. She looks at the top of that building at an angle of 32 degrees 6' and reassures herself that she is the queen of pop. There is no way that she can fail this measly task of jumping 500 feet. How far is the building she is currently on from the building she wants to jump to?

THE PROBLEM

 To solve this problem, we must again find the angle value. We take 6 and divide it by 60 to get 1/10 which is essentially .1 which we add to 32 of course. To find our opposite side, just take the side value of the big building and subtract it by the height of the smaller building. Once we have found that out, we can use a trig function. I am a fan of tan, so I used tan once again. It is the same old song and dance as we multiply both sides by x and then divide by tan 32.1 to get our answer of 797.068 which is simplified to 797.07 feet.

THE SOLUTION

I/D #2 Unit O: Deriving Patterns For Special Right Triangles

                                    Inquiry Activity Summary:
1. 30-60-90

Step 1.) I drew the equilateral triangle and labeled all sides 1 as well as labeling the degrees as 1. Because it is an equilateral triangle, all sides are equal. If all sides are equal as each other, then it makes sense that the angles are equal to each other as well. The angles also add up to 180 degrees as well, a golden rule for triangles.

 Step 2.) I splitted the triangle in two by drawing a line down the middle, creating two 30-60-90 triangles. By drawing a line down the middle, I separate the 1 value on the bottom into two 1/2 values. Also, the 60 degrees at the top becomes two 30 degree angles.

 Step 3.) In order to simplify things and make it a little easier, we will be using one of the triangles for this portion of the derivation.The only thing we are adding here is "b" to the opposing side of angle 60.

 Step 4.) We are going to be using good ol' classic Pythagorean Theorem to solve for "b." In step 1, we use the formula of a^2+b^2=c^2. In step 2, we plug in the values. In step 3, we square the values to get the resulting answers. We then, in step 4, subtract 1/4 to both sides to get b^2 by itself. We get an answer of b^2=3/4, but we don't want the exponent. We then square root both sides in step 5. In step 6, we get our answer as depicted below.

 Step 4.) We label the "b" value we solved for on the triangle. However, we have two fractions for the triangle. We want no fractions and thus we multiply every side by two to cancel out the fractions. We then receive a hypotenuse of 2.

 Step 5.) Why do we use "n"? Well, we need "n" in order to solve problems."n" allows us to find values of all sides of a 30-60-90 triangle since if we have one value of a side of a 30-60-90 triangle, we basically have all the side values. It is all a matter of solving for the other 2 hidden values. Of course, in this example we do not need to use "n" to find the other sides because they were already given. However, it is crucial to understand that without "n," it would be more difficult to identify all sides. Knowing just the "n" value is enough to completely solve a special right triangle. The sides of a 30-60-90 triangle regarding n are n, n radical 3, and 2n.

2. 45-45-90                                                                                                                                             

Step 1.) I drew a beautiful square as shown below. The sides are labeled 1 and right angles shown.                                                                                                                                            

 Step 2.) I diagonally splitted the square to turn it in two 45-45-90 triangles. The angles are cut into two 45 degree angles and now we can derive the pattern.

 Step 3.) Let us solve for one of the triangles. Ignore the radical 2=c. We are going to find "c" so we can find all sides of the triangle and consequently find the same value of "c" for the other triangle.

 Step 4.) Since we know 2 of the values of the sides, we are going to be using the theorem of Pythagorean. We input the values into their proper variables, with both 1 values going into a and b, respectively. We then square the 1's to get a value of....ONE! We then add the 1's together to get 2 and square it to get radical 2. That will be our value for c.

 Step 5.)We then place our glorious value of radical 2 smack daddy in the middle and we have just found the hypotenuse for TWO triangles! It's just like hitting two stones with one bird.

 Step 6.) But wait, we need to put the n's in our special right triangle as well to, as said before, indicate to create a sort of formula to solve any special right triangle problems. By using n, we can easily solve for the hypotenuse or for a side. The hypotenuse is n radical 2 and the side values for n are n and n. We just hav eto multiply n by radical 2 to get n radical 2 and divide n by radical 2 to get n.

Inquiry Activity Reflection:

1. Something I never noticed before about special right triangles is that they are very easy to solve once you break up the steps to solve them one by one.

2. Being able to derive these triangles myself aids in my learning because now I can apply these lessons to my special right triangles as I now know the pattern for solving both 30-60-90 triangles and 45-45-90 triangles.

I/D#1: Unit N: Concept 7: Unit Circle and Special Right Triangles.

Inquiry Activity Summary:
      In this activity regarding unit circles, we labeled 3 special right triangles according to the Special Right Triangle Rule which will tie in to our derivation of the unit circle.

1. In this picture, we see a 30 degree triangle. To go by this problem step, our numbers in circles are chronological, which means they are in order of what to solve for first. For number 1, we label the triangles according to the aforementioned rule, so our hypotenuse value is 2x, our x value is x radical 3, and our y value is x. We got these values by looking them up on google and applying them here in this problem. For step number 2, we first follow instructions by setting the hypotenuse's value to 1, then divide the value of every other side by the value of the hypotenuse, which is in this case, 2x. Therefore, the y value becomes 1/2, and the x value becomes radical 3/2. We now see the relations of the unit circle to this activity as the values of the sides are identical to the ordered pairs in the unit circle. Step 3,4, and 5 has us simply labeling the r,x, and y value, which are respectively shown in the picture. Step 6 has us draw the coordinate grid of this triangle, and this is where we see how this triangle is literally identical to what it is on the unit circle.

In this picture, the 45 degree triangle is represented. We will go step by step just like the last picture. For step 1, the labeling of this triangle is thanks to the internet, and the hypotenuse value is x radical 2 with the x and y values being x. How this is so is because the 45 degree triangle has two sides which are equal due to this triangle being an angle that is in the middle of 90 degrees. Step 2 has us again set the hypotenuse to 1. We then divide the x and y values by the hypotenuse value, which is x radical 2. We therefore get radical 2/2 for the x and y value because they are the same values. Steps 3,4, 5 are shown as the labeling of the sides and step 6 has the graph of this triangle, and as we can see, the angle is representative of the unit circle.

In this picture, we have a 60 degree triangle, which is basically the same as the 30 degree triangle but the x values and the y values are reversed. The labeling is the same, so step 1 remains unchanged, except that the labeling for x and y are of course different from the first picture in that they are now respectives of each other now. Step 2 has the hypotenuse as 2x, the x value as  1/2, and the y value as radical 3 over 2. We derived these numbers from the work shown in the picture. The steps 3,4, and 5 are shown above clearly. Step 6 is where we draw the triangle. In this example, the x and y swap off in order to get that 60 degree shape. 

4. This activity assisted me in deriving the unit circle by teaching me 3 of the most important degrees I need to know for this unit. The x and y values are identical to those on the unit circle and by recognizing these special right triangles I can label the unit circle accordingly and solve real problems with ease.

5. The quadrant the triangles in this activity lies on is the 1st quadrant. If I placed the triangles in quadrant two, then the x value will be negative. If I placed them in quadrant 3, then both values would be negative. If I placed them in quadrant 4, then the y value would be negative. In this picture we see the 3 triangles being placed in different quadrants. The 30 degree triangle has its x value as a negative just like we mentioned. The 45 degree triangle has both ordered pair values being negative. The 60 degree triangle has only its y value being negative.Another change that occurred was how the overall angle measurements of each angle has changed. the 30 degree triangle is a reference angle for sure, but is 150 degrees from the initial axis. The 45 degree triangle is 225 degrees, and the 60 degree triangle is 300 degrees.

Inquiry Activity Reflection:

1. The coolest thing I learned from this activity was how this related to the unit circle I did last year and how easier this activity made my life easier by teaching me the core basics of the unit circle. 

2. This activity will help me in this unit because I will be able to derive the unit circle with greater ease by knowing the most important degree properties: 30, 45, and 60. 

3. Something I never realized before about special right triangles and the unit circle is how easy it can be after you do this activity of how to solve the properties of the 30, 45, and 60 degree triangles.