Unit Q - Concept 2 - Finding All Trig Functions with Given & Identities

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

Things to notice when going through the Pythagorean/Ratio/Reciprocal identities and with right triangles is that these both result in the same answers even though the methods aren't exactly the same thing. In the end both methods are used to find the ratio for their corresponding trig functions and so that is the relationship between Unit Q Concept 2 and Unit O Concept 5. It is also important to note that where the triangle lies in the quadrant determines which trig functions are positive or negative and that so always be sure to check the quadrants. Once again, thanks for checking out Derik Vierya and my blog!

I/D 3 Unit Q - Concept 1 Pythagorean Identities

Inquiry Activity Summary

1. An identity is proven facts and formulas that are always true. The Pythagorean Theorem is an identity because it is a proven fact and formula that is always true. When presented with the variables x,y, and r, the Pythagorean Theorem is x^2+y^2=r^2. For example, when you want r to =1 then you would use the Pythagorean theorem (x/r)^2 + (y/r)^2 = 1. The ratio for cosine is (x/y) like what you saw in the Pythagorean theorem above and the ratio for sine is (y/r) like the one from above. With this you notice that the sin and cos ratios equal to 1 and there is a relationship between the Pythagorean theorem and the ratio for trig functions. From this we can conclude that cos^2theta + sin^2theta = 1 This is a Pythagorean identity because it is just a rearrangement of the Pythagorean theorem with trig functions that always works. The example below shows how this identity is true. 

2. Our first Pythagorean identity has secant and tangent. To get our other two Pythagorean identities, we have to divide by either cosine or sine because they are basically the golden ratios that give us the rest of the trig functions and are the easiest to deal with. Basically, with sin or cosine, you can form a trig function from them to make cosecant, secant, tangent, and cotangent. With the 2 examples below, this shows us that our pythagorean identity is truly an identity. 

Inquiry Activity Reflection 
1. "The connections that I see between Units N, O, P, and Q so far are.." trig functions being used for explaining almost every concept and the unit circle being referenced back to as well to explain the triangles and the uses. 

2. " If i had to describe trigonometry in THREE words, they would be..." Too many trigs. 

WPP#13 & 14: Unit P Concept 6-7: Applications with Law of Sines & Cosines

Please see my WPP13-14, made in collaboration with Derik Vieyra, by visiting their blog here. Also be sure to check out the other awesome posts on their blog. 

BQ# 1 Unit P Concept 1 & 4 Law of Sines Derivation and Area of an Oblique Triangle

1. Law of Sines - Law of Sines is important because life isn't so perfect that it will always give right triangles. Normal trig functions that we use for special right triangles don't apply to non-right triangles. Therefore we use the law of Sines to solve non-right triangles. Essentially, what we do is set a perpendicular from angle B to create two right triangles and label the line as H. From here we can determine the relationships and use the transitive property
c Sin A = a Sin C to solve for what is needed. From this, we can solve for any side or angle depending on it opposite angle or side. This video below will help you fully understand the concept of Law of Sines.

 

4. Area Formulas - Sometimes you might not be able to get the value of H from the law of Sines. Not to fear because the area of an oblique triangle is very similar to the area of a triangle (A=1/2bh) but since we don't know H, we use (A=1/2b(asinc). This will help us find the value of H so we can use either law of sines or cosines to determine the rest of the triangle if needed. Remember, "The area of an oblique (all sides different lengths!) triangle is one half of the product of two sides and the sine of their included angle." This video below will help you understand this situation better.




Sources
Law of Sines Deriviation
Area of an Oblique Triangle Deriviation

Unit O - WPP 12- Trig Functions with Elevation and Depression

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Things to look out for when doing problems like these is to remember what you are looking for. Like any other concept in this SSS packet, we have to remember which of the trig functions we are using in order to be able to get the correct answer to our problem. Remember to make sure that your answer is a reasonable answer and not some absurd number. Thanks for looking at my post!

I/D2: Unit O - Derive the SRTs

Inquiry Activity Summary 



 1. To start off, we are given an equilateral triangle with a side length of 1 and are asked to derive a 30-60-90 triangle from it. Following the image above, we first cut a straight line down from the center of the equilateral triangle. Since equilateral triangles are 60* on each corner, the line cut results in a 30* at the very top, 90* at the bottom of the cut line, and a 60* at either the left or right corner since it is left untouched to form a 30* special right triangle. . The height is derived from using the Pythagorean formula (a^2+b^2=c^2) and the image above shows how it is done. Since our answer from the Pythagorean results in fractions, we want to make the answer easier to remember so we multiply by 2 to each side length to result in √3 for 60*, n for 30*, and 2n for the 90*. Since the original side lengths are 1, we can infer that we only have to substitute different values and we should result in the same ratio as the original side lengths of 1 triangle. This is where we substitute in a variable like (n) to represent a value that can be used. 

  2. Next, we are given a square with the side lengths and asked for a 45-45-90 triangle. To do this, we simply cut the square in half with a diagonal line. We do this because each corner of the square is 90* and if we were to cut a side in half, it would result in a 45* so that is where we get our special right triangle. To get our hypotenuse, we simply use the Pythagorean formula like above and after plugging everything we can, our C (hypotenuse) ends up as with the ratio as 1 for the 45* and √2 for our 90*. Since the side lengths are 1, we can infer that using different side lengths will still result in the same ratio so we just add an (n) to substitute as our new value so our ratio becomes n,n,n√2.

Inquiry Activity Reflection 

1. "Something I never noticed before about special right triangles is.." there is actually a ratio that we can use to solve for the sides rather than just memorizing something like a unit circle and its value. 

2.  being "Being able to derive these patterns myself aids in my learning because.."able to understand these concepts is much better for me and my test grade than just barely getting the gist of it.