SP #7: Unit Q Concept 2: Finding all trig functions when given one trig function and quadrant

This SP7 was made in collaboration with Adrie. Please visit the other awesome posts on her blog by going here

I/D3: Unit Q - Pythagorean Identities

1.) Where does sin^2x+cos^2x=1 come from to begin with? You should be referring to the Unit Circle ratios and the Pythagorean Theorem in your explanations. 

The Pythagorean Theorem is a^2+b^2=c^2. Since a, b and c are variables, we can replace them with any variable of our preference. But, according to the ratios, we use x, y and r, manipulating the theorem into x^2+y^2=r^2. We then have to divide the equation by r^2 to make it equal 1 and we would end off with (x/r)^2+(y/r)^2=1. Using our knowledge from unit O, we remember the trig ratio for cosines is x/r and y/r is the trig ratio for sines. If we simply substitute those variables for cosine and sine, the equation would be cos^2(x)+sin^2(x)=1. Proving this theorem to be an identity, we use one of the "Magic 3" ordered pairs from the Unit Circle and plug it in to our new equation. In example, the ordered pair for a 45 degree angle according to the unit circle is (radical2/2, radical2/2). Plugging it into the equation, we get (radical2)/2)^2 + (radical2)/2)^2, which ends up equaling 1, like every angle on the unit circle should.

2.) Show and explain how to derive the two remaining Pythagorean Identities from cos^2(x)+sin^2(x)=1. Be sure to show step by step. 

To show the left over 2 Pythagorean identities, we just have to manipulate the formula.

If we divide the equation by sin^2, the equation would be: cos^2(x)/sin^2(x)+sin^2(x)/sin(x)=1/sin^2(x). cos^2(x)/sin^2(x) would be substituted by cot^2(x). sin^2(x)/sin(x) would equal 1 since both top and bottom are the same. 1/sin^2(x) would be substituted by csc^2(x). In the end, the equation would end up being cot^2(x)+1=csc^2(x).

Now that we did sine, we do cosine. If we divide the equation by cos^2, the equation would be cos^2(x)/cos^2(x)+sin^2(x)/cos^2(x)=1/cos^2(x). cos^2(x)/cos^2(x) would equal 1 since top and bottom are the same. sin^2(x)/cos^2(x) would be substituted by the ratio tan^2(x). 1/cos^2(x) would be substituted by sec^2(x). In the end, the equation would end up being 1+tan^2(x)=sec^2(x).

INQUIRE ACTIVITY REFLECTION:

"The connections I see so far in Unit N, O, P, and Q so far are..." the use of the unit circle and the trig functions along with the ordered pairs from the unit circle.

"If I had to describe trigonometry in THREE words, they would be..." triangles, ratios and the unit-circle.

WPP #13 & 14: Unit P Concept 6 & 7 - ONE POST

This WPP 13-14 was made in collaboration with Jorge M. Please visit his other awesome posts by clicking here.

Law of Sines-

By the way, that's Adrie (not tyra) going to Farcos ice cream shop.
http://thejasminebrand.com/wp-content/uploads/2013/11/tyra-banks-drake-disneyland-the-jasmine-brand.jpg

Adrie and Drake are on a romantic picnic date together. When thier date is over, the decide to go get ice cream at Farco's icecream shop. But before that, Adrie and Drake play with a frisbee. Drake throws the frisbee 49 feet away to the East of him and Adrie runs after the frisbee. When she gets there, Drake yells out, "FIRST ONE TO FARCOS IS MY NUMBER ONE FAN!" And they both ran. Drake started off at N29*E and Adrie started off at N24*W. Who will get there first? (round to the nearest hundreths)

Law of Cosines-

Drake being attacked by fans.
http://www.popsugar.com/photo-gallery/31330666/Drake-surrounded-fans-during-his-VMAs-performance

The concert just finished and Drake and Adrie had planned hang out at Farco's Ice-Cream Shop. However, they want to keep it on the down low to avoid the crazy crowd of fans. From the stadium where the concert was performed, Drake's and Adrie's course to Farco's shop is at a bearing of 265 degrees and 1850 ft away. The crazy fan crowd heard a rumor that Drake was going to his hotel, which is at a bearing of 115 degrees. After the crowd walked 1780 ft towards the hotel, Farco, the owner of the ice-cream shop, tweeted that Drake and Adrie were in his shop and gave away their location! If the crowd change their route to the shop, how far are they? (Round to the nearest tenth)

BQ# 1: Unit P

1. Law of Sines - Why do we need it?  How is it derived from what we already know?  The derivation must be shown either in a video or in multiple sequential pictures and should include descriptions and information beyond what you can find in the SSS.

We need the law of sines to solve for triangles that are not 90* triangles. With the triangle having sides a, b and c, and angles A, B and C, C can be cut down to make 2 90* triangles. Using opposite over hypotenuse, the trig functions are rearranged. When everything is then equal to height, and everything is simplified, the law of sines is done.

4. Area formulas - How is the “area of an oblique” triangle derived?  How does it relate to the area formula that you are familiar with?

The area is derived from the original a=(b)(h) but the change is that instead of all 3 sides given, only 2 sides are given, with 1 angle (SAS). There are 3 given formulas we can use, depending on what information is given to us, but it is just a matter of manipulating the formula to get what we need.

Either: Area = 1 ab sin C 2

Or: Area = 1 bc sin A 2

Or: Area = 1 ca sin B 2

REFRENCES:

http://math.ucsd.edu/~wgarner/math4c/derivations/trigidentities/lawofsines.htm

http://www.mathsisfun.com/algebra/trig-area-triangle-without-right-angle.html

Unit O Concept 10- Solving Elevation and Depression Word Problems

I am having the time of my life at Drake's concert. The stage is exactly 5 feet away from me. The angle of elevation from me, to him is 30*. How many feet do I have to jump to make it on stage?

Photo taken by Adrie

In the picture above, I used cos35=5. I then multiplied by x, then divided by cos35. I plugged it into my calculator and got 5.019 feet.

After the concert, I stood outside Drake's tour bus to take a picture. When he came out, a million things ran across my head, but most importantly, I wondered how far he would have to lean in to kiss me. With Drake being 11 inches taller than me, and the angle of elevation being 21*, how far does he have to lean in to give me a kiss?

Photo photoshopped by Adrie

In the picture above, I used sin21. I multiplied x out and then divided by sin21. After plugging 11/sin21, I got 30.695 inches.

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

INQUIRY ACTIVITY SUMMARY

In order to get a 30-60-90 triangle, you ut an equatlateral triangle straight through the middle. Assumimg the legs in an equalateral triangle equal 1, when it is cut in half b equals 1/2. Then we must use the pythagorean theorem to solve for c. As seen above, we end up with radical 3/2 for b. Since b and c are both under 2, we multiply all sides by 2. Since not all triangles will have the side length of 1, we also mutliply everything by n.

In order to get a 45-45-90 triangle, you cut a  square diagonaly in half. In a 45-45-90 triangle, 2 sides are congruent. All sides are equal to 1. Using the pythagorean therom, seen above, c is equal to radical 2. Since not all triangles will have the side length of 1, we also mutliply everything by n.

INQUIRY ACTIVITY REFLECTION

“Something I never noticed before about special right triangles is…” I didn't know from just a square, we could get a 45-45-90 degree triangle which then we use the pythagorean theorem to figure out c.

“Being able to derive these patterns myself aids in my learning because…” if I become a cheesebucket and forget "c" then I could simply solve using the pythagorean theorem.