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

The Picture below will show you a problem as well as how to find all trig functions using IDENTITIES (Reciprocal and/or Ration and/or Pythagorean). 







This problem shows that you can find all trig functions by using the identities that we learned about in Unit Q Concept 1. It shows that it is helpful to know these identities in order to have expanded knowledge that would help us solve problems that could become more difficult. It follows that it is a good practice to get to know the identities of trig functions out of the top of your head.
Please notice that we figure out the quadrant that will be used by checking the answers of the trig functions given, whether they are negative answers or positive answers. Also notice on the right how we figured out the answers of  each trig function, whether they were negative or positive, based on the quadrant that we found out will be used.





And please do not forget to to check out the SECOND WAY to find all the trig functions of this problem. Just click on:


                         Kathy's Awsume Second Way

WPP # 13-14: Unit P Concepts 6-7 Word problems using law of sines and law of cosines

TO SEE THE PICTURES SHOWING THE WORK MADE IN COLLABORATION WITH KATHY C. PLEASE CLICK ON: 
                                                                KATHY'S AH-MAZING X10 BLOGPOST

Hershey has stopped at a stoplight and noticed that his best friend, Marlene, is due west of him at the next stoplight, 30 feet away. Both are going to Hershey's Bakery. Hershey walks N 30* W to get there while Marlene goes N 72* E to get there. What is both of their distances to walk over to Hershey's Bakery?




 After having a nice, long conversation with Hershey, they decided to see each other again the next day. They both leave the bakery at the same time. Marlene, in hurry to get to her cousin's Quinceanera, is headed at a bearing of 315* and is traveling 50 MPH. Hershey on the other hand goes home at 30 MPH at a bearing of 078*. How far apart are they after two hours?

I/D #3: Unit Q Concept 1: Rational Reciprocal and Pythagorean Identities

1. Where does sin^2 x+cos^2 x=1 come from to begin with?

To begin with. you should know that an "identity" is "a proven fact that is always true". Therefore, we assume that a Pythagorean Identity, which is the equation seen above, is called such because that equation can definitely be proven as a fact that is always true. Remember that the Pythagorean Theorem using variables x, y, and r is x^2+y^2=r^2. Now doesn't this look familiar? Yes, we did get it from the a^2+b^2=c^2 equation we've learned before, but notice that just by changing the variables to x, y, and r gives us the exact equation we learned that is of a unit circle.

Now let us play with this equation a little. Who knows what amazing discovery we will come across. If we wanted to set the Pythagorean Theorem (which we just found out is similar to the unit circle equation) equal to 1 as shown in the Pythagorean Identity above, what would we do to it? Oh, of course! Divide everything by r^2 because r^2/r^2 is equal to 1. We are left with x^2/r^2+y^2/r^2=1. You may think, what kind of mess is this? Well, do not fear, Math For Cheese Buckets, like you and me, is here to make it clear! First of all, you should know that the equation can be re-written as (x/r)^2+(y/r)^2=1. 

Moreover, the ratio of cosine is (x/r) and the ration of sine is (y/r). Wait, we've seen these before. Yes, a ton of times in the last 2 units, but wait a sec, we just saw them right now, in the re-re-written equation of the Pythagorean Theorem. We can look at the equation in the previous paragraph and switch the (y/r)^2 with sine^2 x and the (x/r)^2 with cosine^2 x. Guess that we did come across a discovery. We just derived the Pythagorean Identity from the Pythagorean Theorem. 

To show that this identity is true, we will choose one of the "Magic 3" ordered pairs from the unit circle.

• 30 degrees= (radical 3 over 2, 1/2)
• 45 degrees= (radical 2 over 2, radical 2 over 2)
• 60 degrees= (1/2, radical 3 over 2

I will show 30 degrees and 45 degrees (know that 30 degrees is similar to 60 degrees, just switched).

• 30 degrees: (radical 3 over 2)^2 + (1/2)^2 = 1. You cancel out the radical with the squared and end with just the 3 on the top of the first fraction. You square the 2 and end up with a 4 at the bottom of the first fraction, so you have 3/4. For the second fraction you square the 1 and still end with a 1 in the top of that second fraction, and you square the 2 to end up with a 4, so you have 1/4. (3/4)+(1/4)=1.
• 45 degrees: (radical 2 over 2)^2 + (radical 2 over 2)^2 = 1. You can cancel out the radical with the squared for the top of both fractions to end up with a 2 on top of both fractions. Then, square the 2 at the bottom of both fractions and end up with a 4 at the bottom of both fractions: (2/4)+(2/4)=1.

2. Show and explain how to derive the two remaining Pythagorean Identities.

sine^2 x + cosine^2 x = 1. You divide the equation by cosine^2 x first. We know that sine x divided by cosine x equals tangent x by the Ratio Identities. So, we can "power up" through our understanding that it will be the same thing if we squared the sine and cosine we can square the tangent. The first part of the equation becomes tangent^2 x. Obviously, cosine^2 divided by itself is equal to 1. The second part of the equation is 1. On the right side of the equation 1 divided by cosine^2 x will become secant^2 x, because we know by the Ratio Identities that 1 divided by cosine x is secant x, therefore we can again "power up" to show that secant x will become secant^2 x. Our final equation: Tangent^2 x + 1 = secant^2 x.

The picture below can give you a visual. Please ignore the bottom part. Just pay attention up until the equation that is squared.





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Now we divide the equation,  the one at the beginning of the top paragraph, by sine^2 x. Obviously, sine^2 x divided by itself is 1. Through the Ratio Identities we know that cosine x divided by sine x equals cotangent x. Again, we can "power up" to show that cosine^2 x divided by sine^2 x equals cotangent^2 x. On the right side of the equation we show that 1 divided by sine x equals co-secant x through the Ratio Identities, so once again we "power up" to know that 1 divided by sine^2 x will now equal co-secant^2. Our final equation: 1 + Tangent^2 x = Co-secant^2 x.

The picture below will give you the equations I am deriving the above work from.







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BQ #1: Unit P Concept 1-5: Law of Sines, Law of Cosines, Area formulas

#3: Law of Sines
The Law of Sines in a triangle that is SSA, or side-side-angle, meaning we know 2 sides and one angle of the triangle, is an ambiguous case because we can have either one, none, or two answers to the triangle. This is because we only know one definite angle. When we knew at least two definite angles, like in the cases of ASA and AAS, we were able to figure out the third definite angle without wondering if there could be another possible way to draw the triangle. With LIMITED information, it is possible that we can create two different triangles by forming a "bridge" with the side that is given to us for the angle that is not given to us. 

After figuring out the value of the angle using the Law of Sines, we use its reference angle to figure out its other possible value. The reference angle is relevant because you can have the angle in the sine quadrant (quadrant II) besides the all quadrant (quadrant I), since it is less than 180 and could be in the boundaries of the Triangle Sum Theorem.  

You know there is either no triangle when you hit a "wall" in the beginning or only one triangle when you hit a "wall" after figuring out the first triangle. The "wall" could be something that is not possible like sine x being greater that 1 or less than negative 1 and the angles adding up to something greater than 180 degrees.

The picture below will show you a problem that will give you two answers.










#5 AREA FORMULAS


The pictures below will help you see how the area formulas of a triangle work and how by using either of them you will still get the same answers. This will show you that the law of sines works with working out a triangle that is not a special right triangle.

WPP #12: Unit O Concept 10: Angles of elevation and depression

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The Problem: 15anera: Sugar, oh Honey Honey.

Marlene's cousin is about to celebrate her traditional quinceanera. She put Marlene in charge of the cake design. Marlene has a friend that works in a pastry and she meets with him to ponder over the design. However, one major issue she has is that they need to figure out how tall the cake should be. For aesthetic reasons, Marlene wants the corner of the table (where the cake will be placed) 25 inches away from the the edge of the top cake. She wants the table edge of the table to be 30 inches from quinceanera doll's eyes at the top of the cake. She wants the angle of the doll's eyes to the corner of the table plus the angle of the corner of the table to the edge of the top cake to be around 60 degrees. If they decide to make the cake 4 feet tall and put a 4 inch tall doll on top, find the angle of depression and elevation to see if 4 feet is close to getting the ideal angle that Marlene is thinking of.

I/D #2: Unit O Concept 7-8: Using the 30-60-90 triangle and using the 45-45-90 angle

30-60-90
To derive the pattern for the 30-60-90 triangles from an equilateral triangle with a side length of 1 we first need to know how an equilateral triangle is labeled. We have the information that it has the side length of 1, so since it is an EQUIlateral(meaning it has EQUAL sides) all three sides of the triangle will be one. Also, since it is equilateral triangle it means that its angles must be equal the same. This is when we absorb past knowledge about triangles. All angles of a triangle must equal to 180 when added. Therefore, in an equilateral triangle the angles will equal 60 each (60 times 3 equals 180). We can create two 30-60-90 triangle from an equilateral triangle by cutting it in half down the middle. We cut it in half down the middle first of all because we cut one of the 60 degrees angles of the equilateral into two 30 degrees, and second of all cutting a line straight down the middle will create two right angles, meaning they are 90 degrees (and now we have two 30-60-90).The lengths that the two 30 degrees angles are reflecting become 1/2 since we cut the equilateral triangle in half and its bottom side length of 1 is split to two halves. We see that the lengths reflected by the 90 degrees angles, the hypotenuse, are still.We do the Pythagorean Theorem, in relation to the special RIGHT triangles we have just formed, to get the side length of the line cutting straight down the middle. When we complete the Pythagorean Theorem we see that length is equal to radical 3 divided by 2. To get whole numbers for all our sides we multiply all the side values by 2, and the side reflected by the 60 degrees becomes radical 3, the side reflected by the 30 degrees becomes 1 and the side reflected by the 90 degrees becomes 2. We place the variable n in front of all these values to show that the pattern could be expanded. In other words, the length of 1 could be any other length, example 7, and the formula will still work.






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45-45-90
We have a square with side lengths of 1. Since it is a square, all the four side lengths are equal to 1 and all the four angles will be right angles, meaning equal to 90 degrees. We cut the square diagonally to make two special right triangles of 45-45-90. (Two 90 degrees angles in the square are split to make four 45 degrees angles).Both sides reflected by both 45 degrees angles are still  length of 1 since we did not cut the square's side lengths and instead cut it through the middle diagonally.Now, this requires us to instead find the one missing length of the diagonal line, which represents the hypotenuse in the two special right triangles we formed. Again, we utilize the Pythagorean Theorem to find this missing value. When we complete the Pythagorean Theorem we find that the value is equal to radical 2. We label all the side lengths with the variable n because we show that the value could be expanded and the lengths do not have to equal 1 in order for the rules to work.







http://blog.powerscore.com/Portals/156640/images/satblog-5.jpg

1. “Something I never noticed before about special right triangles is…that the 30-60-90 was derived from an equilateral triangle. Now all its measurements for its rule makes sense.
2. “Being able to derive these patterns myself aids in my learning because…I am able to apply my knowledge so that when I do not remember exactly how the rules are for special right triangles, at least now I am able to know how to find them.