Sp7 Unit Q Concept 2

My amazing partner Ana posted this SP on her blog here

I/D3: Unit Q - Pythagorean Identities

INQUIRY ACTIVITY SUMMARY:  

1)What is an identity? Why is the Pythagorean Theorem an identity.? 

An identity is a proven fact or formula which is always true. The Pythagorean theorem is an identity in that whatever numbers you plug in you're able to get the correct numbers.

b) What is the Pythagorean Theorem using x y and r?

Since our original theorem used a^2+b^2=c^2 With triangles being inside our unit circle we can replace a with x, b with y and c with r which results in x^2+y^2=r^2

c)

Here we divide by r to set our equation equal to one

d) What is the ratio for cosine on the unit circle?

x/r

e) What is the ratio for sine on the unit circle?

y/r

f, g and h)

What do you notice about part c in relation to parts D and e? What can you conclude?

Looking at our very top equation in this picture we take our value of the ratio for cosine and sine of the unit circle and plug it in and what we get is that our equation consists of is taken from the unit circle. Its also derived from our new transformed formula. Our sinx^2+cos^2x=1 is referred to as an identity because whatever number you plug in will result in one respectively. Using 45 45 degrees I plugged the numbers into our equation and got 1. 

2) Perform a single operation fairly to derive the identity with secant and tangent. 

Here we take our equation and divide it by cos in which results the first part results in one and our second part results in tan (because we memorized it) and our 1 over cos is a reciprocal which results in it equal to sec.

b ) Perform a single operation fairly to derive the identity with Cosecant and Cotangent.

Here we divide by the value of sin in which results it in being one for our first part. Our second part it results in being cot, and our third part which is a reciprocal leads it to being csc.

INQUIRY ACTIVITY REFLECTION:

1)The connections I see between units N, O, P, and Q are how we use triangles within the unit circle, and the use and value of trig functions such as y/r= sin and etc and how the values can be taken the root and lead into the unit circle.

2) If I had to describe trigonometry in three words, they would be : headaches are present.

WPP#13-14: Unit P Concepts 6-7: Solving law of Sin and Cosine word Problems: The Skydiving Pizzeria Evening

This WPP was made with the amazing Ana who is fabulous so its on her blog so check the wpp here

BQ#1: Unit:P Concept 3 Law of Cosines SSS or SAS

3. Law of Cosines - 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 inshould include descriptions and information beyond what you can find in the SSS.


 This video derives the law of cosin starting off by starting off with the use of the trig value functions such as sin= opp/hyp We start off by cutting or triangle and dividing line is labeled as h. Since SinC=h/a we multiply a on both side and get h=asinc. He then speaks of how the sides are achieved. Using the Pythagorean theorem A^2+B^2=C^2 we take the squared values of the legs of a triangle and set it equal to our hypotenuse. Leading into identities sinC and cosC are equal to one.


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 of an oblique triangle is taking from the original area equation of a triangle. We take the SinC=h/a we multiply each side by the value of a and get h=asinC. We then take our our original area of a triangle formula and replace the h with our h value in which results in a=1/2absinC. We can do this with our a and b values. For example if we use  Sin A= h/c multiple each side by c and get h=csinA and plug it into our orignial equation of 1/2bh replacing the h. So our new equation would be sinA=1/2bcsinA.


 Reference:
http://www.youtube.com/watch?v=_DR0BfWh5Jk&list=WLy-6lkQIkegNnF6iwmkqxSLjHvc-2mXVZ

http://www.youtube.com/watch?v=pyftYzmOwr4

WPP#10 Unit O Concept 10: Elevation and Depression

 




 


A) Read the problem and solve for what is asked. Show all works and steps clearly. 
Nancy is about to go skiing, she measure the angle of elevation to the top of the slope to be 80. She is 30 feet away from the slope, what is the height of the slope? 

  





  

  

 B) What is the angle of depression 
 Once Nancy reaches the bottom of the slope, she estimates the angle of depression from the top to the end to be 40. Nancy knows shes 140 feet higher then the base of the course. How long is the path?

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

Inquiry activity summary:

 
1. 30-60-90 We begin by taking a regular triangle which is 60 in each of its corners because the sum of a triangle is 180 degrees. We split a line down the triangle in order to achieve our special right triangle which is a 30 60 90 triangle. When we split the triangle we make a 90 degree angle and split the 60 in half which changes the value to 30 and our 60 in the corner remains. Since we split the triangle in half our one becomes 1/2 for each side because 1divided by 2 is one half.  Our base of our triangle is labeled as A while the side is labeled B and the hypotenuse as C.We then take Pythagorean theorem which is a^2+b^2=c^2. Our base is labeled as A, our side is labeled as B, and our hypotenuse is labeled as C. We then plug in and square our values.  The product results in b= radical 3 over 4 in which we square the 4 and leave the radical 3. This value of radical 3 over 2 is across from our 60 degree angle. Since we have fractions, and a majority of people despise fractions we multiply each value by 2 so radical 3 over 2 becomes simply radical 3 across from our 60 degree angle. Our 30 degree angle which was once 1/2 becomes simply one. Our hypotenuse becomes then 2.  When our values vary we use a variable like n for extended values so we can use them for the numerous amount of numbers.
 

 



 
2. We are given a square with equal sides of one which has the angles of 90 degrees in 4 spots. Taking our square, we split  it diagonally so we can get our 45 45 90 triangle. With two 90 degree angles cut in half  we get the result of 2 45 degree angles and have a 90 degree angle remaining.We start off using the Pythagorean theorem  with a^2+b^2=c^2.  Our A is labeled at the base our B is the side and our hypotenuse is C. We plug in our value of one which results in c equals radical 2 which is our hypotenuse. We put N on the side just in case our numbers change.

Inquiry activity reflection:
1. Something I never noticed before about right triangles is how you can derive them from other shapes like a whole square or a whole triangle.
2. Being able to derive these patterns myself aid in my learning because I can use our knowledge of shapes and break down and use for knowledge later on.