SP#7: Unit Q Concept 2: Finding All Trig Function Values Using Identities

 

 

This SP was made in collaboration with Leslie E. Please visit the other awesome posts on their blog by going here

 

Identities

 

 

SOH CAH TOA

The viewer must need to understand what identities are and how to use them for this problem. The viewer needs to be aware of the substituting that goes on while trying to find all trig functions. The viewer must be familiar with SOH CAH TOA in order to do the problem the other way. They need to pay attention to the ratios when solving it with SOH CAH TOA.

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

INQUIRY ACTIVITY SUMMARY

1. The Pythagorean Theorem is an identity because it is a proven fact that is always true. The Pythagoreom Theorem is a^2+b^2=c^2, but in the Unit Circle, these letters are mentioned as x, y and r, making a slight shift to the original theorem. With these letters, the Pythagorean Theorem is now x^2+y^2=r^2. In order to make the Pythagorean Theorem equal to one, we must divide both sides of this equation by r^2, as seen in the picture below.

 

Then, going back to the Unit Circle,  the ratio for cosine was x/r and the ratio for sine was y/r. These ratios can then be plugged into (x/r)^2+(y/r)^2=1 in order to get sin^2x+cos^2x=1. This is referred to as a Pythagorean Identity because it is just simply the Pythagorean Theorem rearranged in a different way. To show that this is identity is true, we can use one of the "Magic 3" ordered pairs from the Unit Circle. If we have a 45 degree angle, we know that the ordered pair is (radical2/2, radical2/2). When this is plugged into the equation, it will be radical2/2^2+radical2/2^=1. This is true because when radical2/2 is squared, it results to being 1/2 and then when it is added to the other 1/2, it is one. Therefore, the identity is true.

2.To derive the identity with Secant and Tangent, cos^2x must be divided by both sides in the equation of sin^2x+cos^2x=1. You will get:

 

In the unit circle, sine has a ratio of y/r and cosine has a ratio of x/r. When the ratios are divided,with sine being the numerator and cosine being the denominator, you get y/x, which is the ratio for tangent. The two cosines will then cancel and result in being 1. 1/(x/r)^2 (the ratio of cosine) is equal to sec^2. This is how you get tan^2x+1=sec^2x.

To derive the identity with Cosecant and Cotangent, we must now divide sin^2x+cos^2x=1 by sin^2x.

 

The sines will cancel, leaving us with 1. The ratio of x/y equals cotangent (refer to picture for further explanation). 1/sin^2x equals csc^2x because of reciprocal identities. This ends up being 1+cot^2x=csc^2x.   

INQUIRY ACTIVITY REFLECTION

 

1. "The connections that I see between units N, O, P, and Q so far are..." that they all somehow connect to the Unit Circle and also that they somehow involve triangles as well.

 

2. "If I had to describe trigonometry in three words, they would be..." complicated, intricate, and time-consuming. 

WPP #13 & 14: Unit P Concept 6 & 7

Please see my WPP 13-14, made in collaboration with Leslie E,  by visiting their blog here. Also be sure to check out the other awesome posts on their blog.

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

 

1. Law of Sines
We need the Law of Sines to help us find sides or angles that are not necessarily right triangles. Since the Pythagorean Theorem can only be used for right triangles, the Law of Sines helps us find the sides for triangles that are not right triangles.

 

http://etc.usf.edu/clipart/36700/36740/tri19_36740_lg.gif

To derive the Law of Sines, we first need to draw an imaginary line down angle B.




http://etc.usf.edu/clipart/36700/36738/tri17_36738_lg.gif



This imaginary line can be labeled h, as the picture above shows. This allows for two right triangles to be formed. We can now use SOH CAH TOA to help us figure out the rest (just SOH in this case)

Sin A=h/c

Sin C=h/a

Since both of these have h as a common variable, we can simplify to get c sin A=a sinC
Then, divide by ac and then you will get sinA/a=sin C=c.
(this applies to angle B too, if a perpendicular line was drawn from either angle A or C.)

You finally get:

 

 http://www.clarku.edu/~djoyce/trig/lawofsines.jpg



4. Area Formulas
The area of an oblique triangle is derived from the area formula which is A=1/2bh

http://www.compuhigh.com/demo/lesson07_files/oblique.gif

 

To find the height, a verical line is drawn and labeled h. The trig functions, which are sinA=h/c, sinB=h/b and sinC=h/a can be simplified by their denominators in order to make them equal to h. Then, this is plugged into the area of a triangle. H is then replaced with either asinC or csinA. Finally, you end up with the area of an oblique triangle, which is:

 

https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgZrTRmOJrogy3QUkgKj4h6OGuoWqoGscJ0J0Vx9qiKdr2IOoRiD67ngtI6YPhH06z0jxjEKdDKrUYCPBh8lVllMYlDLUvRHJ3G3KjA0Mm_OFJp-z-5Au4VdDeRht56fsINaXJ00Sijqu4/s400/hi.bmp

 

The area of an oblique triangle related to the area I am familiar with because it is what was used to derive the area of an oblique triangle. The area of a triangle is essential in finding the area of an oblique triangle because you need to plug the height into the familiar area equation in order to derive the area of an oblique triangle. 

 

 

 

WPP#12: Unit O Concept 10: Solving Angles of Elevation and Depression Word Problems

http://image.yaymicro.com/rz_1210x1210/0/9d9/baby-boy-climbing-on-bookcase-9d9831.jpg

 

The Problem

A.) Sasha is walking to a shelf to get the flour she needs to bake a recipe. She is 124 ft. away from the shelf and the angle of elevation is 24 degrees. She wants to know the height of the shelf so that she will know how high she has to reach for the flour. What is the height of the shelf?

B.) Sasha needs to get on top of the shelf in order to get her flour. When she is on top, she realizes that she had to climb 55.2 ft. to get there. She spots a flour sack on the floor and notices that she did not have to go through the trouble of climbing to get a flour sack. She then decides that she wants to jump on the flour sack that is on the ground to avoid getting hurt by the floor. The angle of depression to the flour sack is 42 degrees. How long will her fall be?

The Solution

 

A.)

 

 

B.)

 

 



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

INQUIRY ACTIVITY SUMMARY

 

In this activity, we were given a square that told us to derive the pattern for 45-45-90 triangles. We were  only given the side length, which was 1. We were also given an equilateral triangle and we were asked to derive the pattern for 30-60-90 triangles. The information we were given was that the side of the equilateral triangle was 1. This activity is meant for us to understand why special right triangles have these patterns.

 

30-60-90 Triangle

To derive a 30-60-90 Triangle, we first need to label the given information, which is that the side lengths of the equilateral triangle are 1. Then, we need to split the equilateral triangle in half. This will create two right triangles. Since the side lengths of the equilateral triangle are one, the two right triangles now have a side that is 1/2. We then need to solve the missing side using the Pythagorean Theorem. We know that one side is 1/2 and the hypotenuse is 1. We then need to square both sides. 1/2 squared is 1/4 and 1 squared is 1. Then, we need to subtract 1/4 from both sides. We will get 3/4, and finally we need to take the square root of it. The top stays as radical 3, but the bottom is 2. The final side ends up being radical 3/2. Lastly, we need to multiply the sides by two to get rid of the fractions. We will end up with radical 3, 1, and 2. A 30-60-90 Triangle has"n" in its pattern simply because "n" is a variable that represents any number. If any number were placed in to replace "n", the pattern would still work. (refer to pictures below for steps)

 

 

 

 



45-45-90 Triangle

To derive a 45-45-90 Triangle, we first need to label the given information, which is that the side lengths of the square are 1. Then, we need to split the square into two right triangles by cutting the square diagonally. We know that the lengths of both legs of the triangle are 1, so we need to use the Pythagorean Theorem to solve for the hypotenuse. We need to square 1 twice, and then add it. Since 1 squared is simply 1, 1+1 equals 2. Then, we need to take the square root of two, so the hypotenuse ends up being radical 2. The side lengths are now 1,1,radical 2. If 'n' is multiplied to this, it ends up being n, n, n radical 2. The relationship between the sides is the same, so n applies to any number because it is just a constant. " n" basically just represents a number that can be substituted in (refer to pictures below for steps).

 

 

 

 

 

 

 

 

 

 



 

 

INQUIRY ACTIVITY REFLECTION

 

1. "Something I never noticed before about special right triangles is" that the Pythagorean Theorem helps derive the patterns for special right triangles.
 

2. "Being able to derive these problems myself aids in my learning because" now I know why these patterns are there, so I do not have to memorize things anymore and can use this logical reasoning instead to help me solve problems.