Monday, April 15, 2013

Student Video #5: Unit S. Concept 7. #5




  • What is this video about? 
    This video goes over how to solve equations with half angles. With it, we incorporate the use of the half angle formulas. Moreover, this trigonometric concept allows us to further explore the utilization of the Unit Circle.

  • What does the viewer need to pay special attention to in order to understand the concept?
      You must be able to incorporate your knowledge from past units as well, not just simply Unit S. For example, in this particular problem I was able to identify that I would need to use the Pythagorean Identity for sin^2(x). For these problems as well as this unit, you need to able to focus on what you need to do in order to solve. If that means digging through your brain for formula or proofs from past units, than so be it.     

Thursday, April 11, 2013

(Unit S. Concept 4.) Half Angles Formula and it's similarity with (Unit R. Concept 1) Sum Formula

Solving the Sine, Cosine, & Tangent of 75* with Half Angle Formula



  • What does this picture show?
          This picture demonstrates how to solve for a half angle using the Half Angle Formula. Although this looks difficult and challenging, it really isn't--it is just plugging in and using simple algebraic skills to solve. Please understand that this is only one of the methods used to solve for the sin(75), cos(75), and tan(75); one of the other method is demonstrated in the picture below. I have done this problem step-by-step so that it is evident how to use the Half Angle Formula in relation to half angles of the Unit Circle.
  • What does the views need to pay attention to in order to understand the problem?
             The quadrant in which we determine whether sine and cosine will be a positive or negative value is dependent upon the 'u/2' NOT the 'u'. The 'u' value is twice as much as the 'u/2' value. In this case twice of 75* is 150*. Although 75* is not apart of the major angle values of the Unit Circle, 150* is. The cosine and sine used in this problem come from the ordered pair for 150*. Knowing these key parts of information along with the Half Angle Formula allow us to solve for the trig function values of half angles like 75* as shown in this example.


Solving the Sine, Cosine, & Tangent of 75* with Sum Formula




  • After having solved for the sin(75), cos(75), and tan(75) using these methods, how do you know that the answers are the same?
         Without even plugging anything into your calculator, you automatically know that although the answers look different between the sin(75), cos(75), and tan(75) because different methods were used, they are same. This is because the sin(75) can only be the sin(75), cos(75) can only be equal to cos(75), and tan(75) can only be the same as tan(75). This sounds very obvious, but many people assume that because the way the answers appear are different from one another that the they are different values, but they are not. Once plugged into your calculator, you will see for yourself that the approximate answers for all of them are the same. This is what the picture below shows. 



Tuesday, April 9, 2013

Student Video #4: Unit S. Concept 3 - #5





  • What is this video about? 
     This trigonometric video lesson goes over how to solve problems by using the     power-reducing formulas.


  • What does the viewer need to pay special attention to in order to understand the concept?    
      Please understand that we substitute the power-reducing formula to the entire given trig function, not just the (x). Moreover, note that for your final answer, you want the highest power to be 1.

***DISCLAIMER: These problems look a lot scarier than they are. With practice, they become super fast, easy, and even fun! :D

Wednesday, March 27, 2013

Student Problem #10: Unit R. Concept 3


*What is this problem about?

This problem in particular depicts how to use the sum formula in solving the trig of an inverse trig functions. Although this may sound big and scary, it really is NOT! As you can see from the above problem, the inverse trigonometric function is inside the brackets (thats the inverse trig part if that wasn't obvious enough.) The regular trig part is what is outside the problem. Now that you can distinguish each part, it doesn't sound so scary now, right? 

*What does the viewer need to pay close attention to in order to do the problem correctly?

Please understand that the way I got my coordinate pairs was through my referencing the Unit Circle. Yes, the Unit Circle is being used again! Moreover, for tangent, I realize that I could have used the ordered pair of (-1,0) as well because 0/-1 is also 0, however, I don't need to. I am picking the values closest to quadrant I. This has absolutely nothing to do with some sort of mathematical process, its just the way we solve these kinds of problems for my Math Analysis class, however, feel free to solve using that as well. :)


Tuesday, March 26, 2013

Student Problem #9: Unit R. Concept 2

What is this problem about?
This photograph shows how to the sum and difference formulas (look at Unit R. Concept 1 for reference) are integrated in the use of solving for right triangles.
What does the viewer need to pay attention to in order to do the problem correctly? 
Be aware that these triangles do lie in different quadrants! Because of this, the problem is still solved the same way as you normally would if both triangles were in the same quadrants, however, note whether sin, cosecant, cosine, secant, tangent, or cotangent are positive or negative [this depends on which quadrant we refer to.]  Please note that triangle drawn on the left is the 'u' and the triangle drawn on the right is the 'v' [my penmanship wasn't all that great here]. Be sure that you chose the correct formula for what is being asked. I can't stress that enough. If you use the wrong formula, evidently you will get the wrong answer!


Monday, March 25, 2013

Student Problem #8 :Unit R. Concept 1

Sum Formula:

*What is the purpose of this picture?
This picture demonstrates the use of the sum formulas for sine, cosine, and tangent being put to use to find exact values for these trigonometric functions.
*What should you pay close attention to in order to understand the concept?
Understand that 450* is not a normal angle from the unit circle. Here, I have added two values FROM the Unit Circle [of the normal reference angles to 30*, 45*, 60*, 90*, etc.,] that amount to 450*, those being 300* and 150*. (* = degrees symbol)



Difference Formula:


*What is the purpose of this picture?
This picture depicts the use of the difference formulas for sine, cosine, and tangent being put to use to find exact values for these trigonometric functions.
*What should you pay close attention to in order to understand the concept?
Due to the fact the 450* is going on to a second revolution of the Unit Circle, I have to use 720* because its a large value and can subtract another value (270*) to make 450*.  >>>NOTE: 720* is exactly two entire revolutions of the Unit Circle and its reference angle is 360* thus the ordered pair is the same. 

***CHECK: check if you are correct by plugging in the sin(450), cos(450), and tan(450) into your calculator. 

>>>NOTE: because sine is 1, cosine is 0, and tangent is undefined, what does that tell you about this angle???

         ***It is a quadrant angle!!! The reference angle of 450* is 90* :)





Hope this helped! If you need any clarification or for me to further describe verbally leave a comment down below. If you would prefer a video instead, also leave a comment. :)

Tuesday, March 19, 2013

Pythagorean Identities {Part 1 of 3}

Pythagorean Identity #1

The example below depicts how this identity came about. Using an angle, in reference to the Unit Circle, you see how I prove that sin^2(x) + cos^2(x) is ALWAYS going to equal to one.

Pythagorean Identity #2

This is derived in reference to sin^2(
𝞡
) + cos^2(
𝞡
) = 1 and dividing it by cos^2𝞡

Pythagorean Identity #3

This pythagorean identity is also derived in  reference to sin^2(
𝞡
) + cos^2(
𝞡
) = 1 and, but instead you are dividing it by sin^2𝞡