SP #7: Unit Q Concept 2: Finding All Trig Functions (using identities and SOH CAH TOA)

   In this problem, it is given that tan(x)= 2/3 and sec(x) < 0. It is also given to us we are going to use Quadrant III and the values to start our problem. Before we start solving, it is better to decide first if the sign is going to be negative or positive based on what is given to us. It can only work in ONE quadrant and only ONE of these quadrants can work (in this case it is tangent or Quadrant III).

  Reading from left to right, this is one way to find the missing pieces accordingly. When solving for cotangent, you can use the reciprocal identity. Now that you know what cotangent is, you can use the Pythagorean Identity. Plug in the value of cotangent, add, and take the square root. From there, you will find your value for cosecant. Because you know the value for cosecant, you can use it to find the value for sine. Thus, you can use the reciprocal identity. Remember, you must rationalize in order to get rid of the square root on the denominator. Now, if you look back to see what trig you are missing, you find that you still need to solve for secant and cosine. Since tangent was given to us, we can use the Pythagorean identity to find secant. Once we find the value for secant, we can use the reciprocal identity to find cosine.

  We can use our knowledge of SOH CAH TOA to find/check our values. This picture demonstrates the ratio and steps to get the value. Remember, "r" does NOT equal to 1. If you look back at the problem, it is given to us that "r", the radius, is 2 radical 13. 

I/D #3: Unit Q Concept 1: Using Fundamental Identities to Simplify or Verify Expressions

INQUIRY SUMMARY ACTIVITY

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

     To start off, an "identity" is proven facts and formulas that are ALWAYS true. The Pythagorean Theorem is an identity because it has been proven that in a triangle, for instance, a^2 + b^2= c^2. For this unit, the Pythagorean Theorem can be consist of x,y, and r similar to that of a, b, and c according to the Unit Circle in quadrant I.

   Because a^2 + b^2=1, we divide r^2 on both sides. We can use our knowledge from the Unit Circle to determine what is (x/r)^2 and (y/r)^2. From there, we have found our first Pythagorean Identities. 

  Cos^2x + sin^2x=1 is referred to as a Pythagorean Identity because we used the Pythagorean Theorem to find it. Basically, we substituted variables and connect that to what we have learned. 

   To be sure if this Pythagorean Identity is true, we can use the angles (30, 45, or 60 degrees) from the Unit Circle and substitute those values. This picture shows an example of using the 60 degrees. 

2. How to derive the two remaining Pythagorean Identities from sin^2x + cos^2x=1

   The picture above demonstrates the various steps of how to get tan^2x + 1 = sec^2x. First, you will have to divide both sides by cos^2x. Then, with your memory of the ratio identities, reciprocal identities or Pythagorean Identies, you would substitute it into the equation. The identity with Secant and Tagent is tan^2x + 1 = sec^2x.

   The picture above shows the various steps of how to get 1 + cot^2x = csc^2x. You divide sin^2x on both sides. Then, using your memory, plug in what you know. At last, you should be able to find the identity for Cosecant and Cotangent. 

INQUIRY ACTIVITY REFLECTION

1.THE CONNECTIONS I SEE WITH UNIT N, O, P AND Q SO FAR ARE ratios that we use again from the Unit Circle and how we use quadrant I to know what variables to plug into the equation that we are trying to find.

2. IF I HAD TO DESCRIBE TRIGONOMETRY IN THREE WORDS, THEY WOULD BE difficult but fun.

WPP #13 and 14: Unit P Concept 6 and 7: Law of Sines and Law of Cosines

This WPP #13 and 14 was made in collaboration with Ashley V. and Sarahi L. Please visit the awesome blog post here!

BQ #1: Unit P Concept 3 and 4: Law of Sines and Area of an Oblique Triangle

Law of Sines:

1. Why do we need it? How is it derived from what we already know?

    The Law of Sines is used for non-right triangles. We cannot use the Pythagorean Theorem for non-right triangles. The Law of Sines can be used when it comes to non-right triangles with AAS and ASA. Remember, just like sine on the Unit Circle, the Law of Sines cannot be greater than 1 or less than 1. To make sure where the Law of Sines is derived from, draw a non right triangle.

   To make a right triangle, draw a perpendicular line from the top of the triangle to the bottom and name that "h." Now that you have two triangles, you can use SOH CAH TOA. When doing using the soh cah toa, you found that sinA= h/c and sinC= h/a. To get rid of "h", you multiply both sides by its denominator. The final answer is cSinA= aSinC. To get the Sine angle alone, divide both sides by "a" and "c".

Area of an Oblique Triangle:

1. How is the "area of an oblique" triangle derived?

   The area of an oblique triangle is derived when you substitute sinC=h/a, which is h=asinC, into the regular area equation of a triangle, A=(1/2)BH. By doing that, you will get 1/2b(asinC). Remember, we are still using the trig equations or SOH CAH TOA to get "h", the perpendicular line.

How does it relate to the area formula you are familiar with?

   This relates to the area formula that we are familiar with because you are simply using sin, cos, or tan to find "h" which is the perpendicular line that we "dropped" since these triangles are not right triangles. The area of an oblique triangle needs to contain two sides and an included angle. Therefore, it has to be SAS in order to use the area of an oblique triangle.

References:
Mrs. Kirch's SSS packet

WPP #12: Unit O Concept 10: Solving Angle of elevation and depression word problems

Problem: It is that time of the year again... for Superbowl Puppies! The husky, on its paw ready to compete the game, is about to jump on a stool. The angle of elevation to the top of the stool is 33 degrees and 8 minutes. a) If the base of the stool is 12 feet from the puppy, what is the height of the stool (to the nearest foot)? In the next round, the same puppy has to slide down the mini puppy slide from where he is standing (the stool). The puppy is standing on top of the slide looking down at the ground. The vertical distance from the ground to the puppy's eyes are 6 more feet than the height of the stool. The angle of depression is 54 degrees and 16 minutes. b) How long is the slide that the puppy is going to slide on(to the nearest foot)?


                                     

http://www.visualphotos.com/photo/1x8325432/siberian_husky_puppies_in_bucket_AFA-Z-01502.jpg

I/D #2: Unit O Concept 7-8: How can we derive the patterns for special right triangles?

  The activity of how to derive the patterns for special right triangles includes a square and an equilateral triangle. Based on the angles of these shapes, you can easily define what type of special right triangle it is, whether it is a 45,45,90 triangle or a 30, 60, 90 triangle. Each side length will be equal to one. We will used what we have learned about special right triangles and relate it to different formulas (Pythagorean Theorem) or concepts (Unit Circle) that we know. 

INQUIRY ACTIVITY SUMMARY

1. 30-60-90 triangle

  This is an equilateral triangle that has a sum of 180 degrees. Since all angles and sides are equal, each angle has to be 60 degrees. To get a special right triangle out of an equilateral triangle, you will need to split the triangle in half vertically. Side A will have an angle of 30 degrees, side B with 60 degrees, and side C with 90 degrees. According to the 30-60-90 special triangle, each side is equal to n, n radical 3 and 2n accordingly. Remember, "n" is a variable. For instance, if a triangle has a side length of one more than one, the ratio for the triangles will be the same. The variable "n" tells us the the approximate length of the triangle.

  The side length of the triangle is one. Equilateral triangles all have the same length and angles. Thus, each side for this triangle is one. Since the special right triangle only occurs when the equilateral triangle is in half, then the side A will have to be equal to 1/2. As a fact, 90 degrees equals to 2n. 1/2 (which is n) multiply by 2 will give you 1 as your hypotheses length. To find side B, you can simply use the Pythagorean Theorem. Side B will then eventually equal to radical 3 over 2. Another way to solve for side B is to plug 1/2 (which is n) to n radical 3. This step will also give you the same side length for B.

2. 45-45-90 triangle

  It is given to us that the square has a side length of one. A square has all equal sides and angles. Since a square has a sum of 360 degrees, each angle is 90 degrees. To get a special right triangle out of a square, you must cut the square in half diagonally. The special right triangle will be 45-45-90. Side A and B are equal to one (a given). In a 45-45-90 triangle, it is n, n, and n radical 2 accordingly. Since one is your "n", you can simply plug in 1 to n radical 2, which will give you radical 2 for the hypotheses. 

  You can also use the Pythagorean Theorem to find side C. Side A and B are both equal to one. As you solve for c^2, you get radical 2. Radical 2 cannot be simplified any further, so radical 2 is the side length for your hypotenuse. Remember, "n" is only a variable. You could have also plug in your side A or "n" to find the length of the hypotenuse. The variable "n" is a ratio for the special right triangle whether the side length is one or more than one. Thus, it is like a constant. The variable gives the special right triangle an approximate length. 

INQUIRY ACTIVITY REFLECTION

1. SOMETHING I NEVER NOTICED BEFORE ABOUT SPECIAL RIGHT TRIANGLES IS how we can derive these triangles to fully understand why the variable "n" is so important. Once you know how to apply "n" and the Pythagorean Theorem, it becomes a lot easier to find the side length.

2. BEING ABLE TO DERIVE THESE PATTERNS MYSELF AIDS IN MY LEARNING BECAUSE I can fully understand how a special right triangle functions and use this knowledge to apply to concept 7 and 8. I can also fully recognize where these special right triangles come from. I will forever remember the variable "n" and the patterns that these triangles contributed.