SP: 7 Unit Q Concept 2

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PROBLEM: tanx= -7/2
sinx>0

The picture above shows various steps on evaluating all the different trig functions when you are only given 2. Follow the steps I have given to understand where each trig function is derived from.

The picture above shows how you can solve the different trig functions when using SOHCAHTOA as the trig ratios.


In order to understand this concept, the viewer must able to recognize all the different trig functions that coexist within one another. Apart from this, the viewer must be able to understand the trig ratios of each trig function so that you can know how the triangle will be pictured as.

I/D #3: Unit Q- Pythagorean Identities

1. The equation sin2x+cos2x=1 is derived from the Pythagorean Theorem. The Pythagorean Theorem, as we know, is x^2+ y^2= r^2 when it is compared to the unit circle. To derive the equation given, we must follow the steps given on the picture below. We must first divide by r^2. This will allow us to get certain trig functions. These trig functions will be able to be arranged so that we end up having cos^2 + sin^2=1.

2. To derive the next two pythagorean identities, we must use the first pythagorean identity to get the next two. For the first one, we must divide everything by sin^2. This will create new trig ratios which will soon be simplified to make the new equation cot^2+ 1= csc^2.

To get the next pythagorean identity, we have to start off using the first pythagorean identity that we derived. From here, we must divide the entire equation by cos^2. This will again make new trig functions that will be able to simplified into the new equation, 1+ tan^2= sec^2.

Inquiry Activity Reflection

1. The connections that I see between units N,O,P,& Q so far are that they all basically rely on the Pythagorean Theorem in some way. They all also fall under the concept of the unit circle and the "magic three" ordered pairs.

2. If I had to describe trigonometry in three words, they would be tricky, mind-blowing, and interesting.

WPP 13&14: Unit P Concepts 6 and 7

This WPP13-14 was made in collaboration with Vanessa Ceja.  Please visit the other awesome posts on her blog by going here.

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BQ #1: Unit P Concepts 1 & 5: Law of Sines and Area of Oblique Triangles

1. Law of Sines

Why do we need it? 

We use the law of sines because it helps us evaluate the side lengths or angles for triangles, in this unit mainly oblique triangles. It is a simple process of evaluating these missing sides by making to proportions based on what we know and what we need. The proportions you will need will be based on the law of sines formula given to the right.





How is derived from what we already know? 

First, we know that triangle ABC is oblique making it have different different side and angle lengths. Since it oblique, we are able to cut the triangle in a section where it makes two right triangles. (as shown on the top picture) This procedure will help us evaluate the height of the triangle as well.

Since we know that Sine is opposite/hypotenuse, we can use both angles A and C to make two trig functions. The two trig functions that came from this triangle are shown on the second picture. (The very top one)

The next step we must take is to cross multiply which makes h remain alone while we have cSinA and aSinC as the two trig functions. We want to have the Angle value the same as its corresponding side so we must divide both functions by ac. This will make the first equation get rid of the "c's" while the other will get rid of the "a's".

Finally, we get what we were looking for which is SinA/a = SinC/c









5. Area Formulas

Do all three area formulas work? 

As we can see on the pictures below, we notice that all three equations are basically the same the same answers. They all have a 12 unit area which proves that all three equations work to find the area of oblique triangles.

WPP 12: Unit O Concept 10: Angle of elevation and depression

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ID#2: Unit O Concepts 7-8: How can we derive the patterns for our special right triangles?

Inquiry Activity Summary:

In the activity we did during class, we were supposed to derive the special right triangles making all the sides equal to one. To derive a 45-45-90 triangle, we were given a square with each side equal to 1. From there, we had to divide the square into two parts so that we can have a 45-45-90 triangle and then find the value of the hypotenuse. To derive a 30-60-90 triangle, we were given an equilateral triangle with side lengths equal to 1. We had to again cut the triangle in two parts to get the triangle degrees of 30-60-90. Finally,we found the values of the side missing.


1. When deriving the equilateral triangle, you must cut the triangle in half so that one side can remain as 60 degrees, one can be 90 degrees, and the other one can be 30 degrees since it is half of the triangle. We know that all sides of the equilateral are equal to 1. Since it is cut in half, the bottom side is equal to 1/2. (as shown on the bottom left) The hypotenuse is equal to 1, so we know what a & c are equal to. So to find the value of the height, we must use the pythagorean theorem. My work is shown on the bottom right which gives us the value of the height which is rad 3/2. We know that 

 a 30-60-90 triangle has side values of n, n rad 3, and 2n. The variable "n" is a constant. In this case, our n is 1/2 which gives us these answers but the value of n can vary depending on the problem they give you. But, they all end up falling under the same guidelines. 

2. When deriving a 45-45-90 triangle by an equilateral triangle we must first cut the square diagonally. It must be diagonally because it will create two triangles and also making two 45 degree sides. It can't be cut in the middle because it will make a rectangle. We know that two sides are equal to 1 so we must use the pythagorean theorem, as shown below, to find the value of the hypotenuse. We use "n" to represent our sides because it is a constant. This means that any number will follow the same guidelines for each side we must know. 

Inquiry Activity Reflection: 

1. Something I never noticed before about special right triangles is that the value of the sides are directly connected to equilateral triangles and squares. The value of their sides make sense when you work out the problem to its individual parts.

2. Being able to derive these patterns myself aids in my learning because I finally understand where the values of the sides for special right triangles come from and I am able to see how things are directly connected to one another.