## Wednesday, March 26, 2014

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

Please see my SP#7, made in collaboration with Vivian Pham, by visiting her blog here. Also be sure to check out the other awesome posts on her blog.

## Wednesday, March 19, 2014

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

__Inquiry Activity Summary__

**Where does sin²x + cos²x = 1 come from to begin with?**
First off, in order to understand the concept, we must acknowledge that an identity is a proven fact and formula that is

*always*true. Therefore, the Pythagorean Theorem is an identity because it is a proven formula that is always true. The Pythagorean Theorem is usually demonstrated through the formula a^2+b^2=c^2. However, in reference to the unit circle, the Pythagorean Theorem is x^2+y^2=c^2. To get the theorem equal to 1, we divide both of the sides by r^2. The result is x^2/r^2+y^2/r^2=1. These can also be written as (x/r)^2+(y/r)^2=1. Now, it is in complete reference to the unit circle! Cosine for the unit circle is x/r and sine is y/r. Therefore, if you substitute the equation with cos^2θ and sin^2θ. The new equation, cos^2x+sin^x=1 is now referred as an Pythagorean Identity. We can prove this by plugging in a pair from the unit circle such as 60 degrees. The pair for 60 degrees is (1,2, radical 3 over 2). When plugging it into the formula, after simplifying, you get 1=1. IT IS TRUE. :D

**How to derive the two remaining Pythagorean Identities from sin2x+cos2x=1.**
This shows how to get tan^2x+1=sec^2x. First, you have to divide both sides by cos^2x to get tangent and secant. cos^2x/cos^2x cancels out to become 1. After, using your identities, you simplify the equation. Thus, with our memory of identities, we get the identity tan^2x+1=sec^2x.

This Pythagorean Identity consists of cotangent and cosecant. To get there, we divide by sin^2x for both sides. We know that sin^2x/sin^2x cancels out to become 1. We then use the identity for cos^2x/sin^2x which is cot^2x. Then, we know that 1/sin^2x is csc^2x. The final answer is 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 it uses the unit circle and trig functions. Also, it uses the Pythagorean Theorem as we have been using in triangles.

**2. “If I had to describe trigonometry in THREE words, they would be**crazy, mind-blowing, interesting.

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

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

## Sunday, March 16, 2014

### BQ# 1: Unit P Concept 1-5: Law of Sines and Area of an Oblique Triangle

Law of Sines:

The law of sines is needed when solving for triangles. To find angles and lengths in right triangles,we use the Pythagorean Theorem and normal trig functions. However, when we are faced with a triangles that are not RIGHT triangles, we result to using the law of sines, which we can find using ASA or AAS.

We start off with a triangle that is not a right triangle.

We draw a perpendicular line down from angle A, forming two right triangles. Label the line h (for height).

Since we have two right triangles, we can use trig functions. We have to find h, and we know the equations are sinA=h/c and sinC=h/a in right triangles. To get h by itself, we multiply the denominators by itself for both equations. So the result is csinA=h and asinC=h. Since they are both equation to h, you can make them equal to each other. So it will become csinA=asinC. Next, we divide the equation by the coefficients together, ac. The result is sinA/a=sinC/c.

Using the same concept, we can now find angle B. Let's draw a perpendicular line through C.

With the two new right triangles, the trig functions are asinB=h and bsinA=h. Set them next to each other by the transitive property. Next, we again divide the denominators multiplied together, ab, and we get sinB/b=sinA=a.

Area of an Oblique Triangle:

The area of a triangle is a=1/2bh. When we want to know the area of an triangle, but we don't know the height, this is when when use an oblique triangle. In order to, we have to have two sides and one angle. To find our height, we will use the trig functions. For example we know sinA=h/c. Multiply both sides by c and you get csinA=h. After, you can plug this into the area of a triangle formula, which results to a=1/2b(csinA). This is our new formula for the area of the oblique triangle. We can also find the other two angles as well using the same method.

__Why do we need it?__The law of sines is needed when solving for triangles. To find angles and lengths in right triangles,we use the Pythagorean Theorem and normal trig functions. However, when we are faced with a triangles that are not RIGHT triangles, we result to using the law of sines, which we can find using ASA or AAS.

__How is it derived from what we already know?__We start off with a triangle that is not a right triangle.

We draw a perpendicular line down from angle A, forming two right triangles. Label the line h (for height).

Since we have two right triangles, we can use trig functions. We have to find h, and we know the equations are sinA=h/c and sinC=h/a in right triangles. To get h by itself, we multiply the denominators by itself for both equations. So the result is csinA=h and asinC=h. Since they are both equation to h, you can make them equal to each other. So it will become csinA=asinC. Next, we divide the equation by the coefficients together, ac. The result is sinA/a=sinC/c.

Using the same concept, we can now find angle B. Let's draw a perpendicular line through C.

With the two new right triangles, the trig functions are asinB=h and bsinA=h. Set them next to each other by the transitive property. Next, we again divide the denominators multiplied together, ab, and we get sinB/b=sinA=a.

**The result: Law of Sines**http://precalculuswinch.wikispaces.com/The+law+of+sines(+Section+4.4) |

Area of an Oblique Triangle:

__How is the "area of an oblique triangle" derived?__http://www.compuhigh.com/demo/lesson07_files/oblique.gif |

Our final formula for area of an oblique triangle |

http://wps.prenhall.com/wps/media/objects/551/564474/StudyGuide/7c1h7_1.gif |

__How does it relate to the area formula that you are familiar with?__
We used the original area formula a=1/2bh to find the area of the oblique triangle. The only difference is that we had to find the height using trig function sine, instead of normally being given it. The rest of the area formula is the same.

http://precalculuswinch.wikispaces.com/The+law+of+sines(+Section+4.4)

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

__References:__http://www.compuhigh.com/demo/lesson07_files/oblique.gif

http://wps.prenhall.com/wps/media/objects/551/564474/StudyGuide/7c1h7_1.gif

(and beautiful triangles drawn by me!!! :D)

## Wednesday, March 5, 2014

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

http://www.novinite.com/view_news.php?id=131240 |

**The Problem:**

**Part A: Carrie is going to go to the store. She is 110 feet away from the store. The angle of elevation is 30 degrees. She wants to know the height of the building because she is very curious, what is it?**

Part B: Carrie decided to climb the store because she is crazy. She is on top of the store. She is 63.5 feet higher than a car she wants to jump on. She estimates the angle of depression from where she is now to the car to be 20 degrees. How big will her fall be if she decided to jump?

**The Solution:**

**Part A:**

**Part B:**

## Tuesday, March 4, 2014

### I/D2: Unit O - How can we derive the patterns for our special right triangles?

**Inquiry Activity Summary**

1. In order to derive a

__30-60-90__triangle from an equilateral triangle, we start off with the given information that since it is an equilateral triangle, all sides are the same, and in this case they are 1. Also, since the sides are all the same, this also means that the angles are the same as well. Therefore, having a triangle add up to 180, we can conclude that each angle is 60 degrees if you divide 180 degrees by the amount of sides, which is three. To form the 30-60-90 triangle, we simply split the triangle in half. Once we do that, there is now a 90 degree angle on the interior. Since we split the triangle in half, we split the 60 degrees angle in half as well, thus making it 30 degrees on each side. We now have two 30-60-90 triangles.

To find the sides, we know that the equilateral triangle has sides of 1. But, since we split the split the triangle in half, the side across the 30 degrees is also split in half. We know the angle across 90 degrees is still 1 because the splitting of the triangle didn't affect the side (now the hypotenuse). To find the angle across the 60 degrees with our newly made triangle, we use Pythagorean Theorem. Pythagorean theorem is a^2+b^2=c^2. The hypotenuse is c, and the angle across the 30 is a. We are looking for b, which turns out to be radical 3 over 2.

Lastly, we multiply each side by 2 because we want to get rid of the fractions. After multiplying by 2, our sides conclude to be a=1, b= radical 3, and c equaling to 2. The importance is that a pattern begins to form. Since a=1, we see that side c is twice of it. This creates the notion that a is equal to n, and side b is 2n. The third side concludes to be n radical 3 since 1 times radical 3 stays the same. We place

**n**to show that there is a pattern, that the numbers from the sides don't just magically appear by memorization. There is a relationship between each sides in relation to their angles and the n shows that, and you can solve it by using n.
2. In order to derive a

__45-45-90__triangle from a square, we know that a square adds up to 360 degrees and all sides are equal. We also know that a square has equal angles of 90 degrees.
To create a 45-45-90 triangle, we draw a diagonal through the square. Because each side was 90 degrees in the square, we see that two of the angles are split because of the diagonal. Therefore, each of the angles split is 45 degrees, forming a 45-45-90 triangle.

We know that each side from the 45 degrees remains 1 because the diagonal cut did not affect them. To find the hypotenuse, the diagonal cut, we use Pythagorean Theorem. We know that leg a is 1 and so is leg b, so we plug it in to 1^2+1^2=c^2. We then result to 2=c^2. To get our c we square root both sides and we get radical 2=c. Thus, we see that our sides is a=1, b=1, and c=radical 2.

This is when n comes to play. We see that n is equal to 1, therefore n takes place for a and b. Since the hypotenuse is radical 2 times 1, and we know that n is equal to 1 (which both sides a and b have), we can conclude that our legs are going to be n radical 2. N represents the relationship between the sides in the 45-45-90 triangle and no matter what numbers we are given, we can find the other sides using the relationship of N.

**N IS POWERFUL! :)**

__INQUIRY ACTIVITY REFLECTION__**“Something I never noticed before about special right triangles is…”**there is an actual reasoning behind why 30-60-90 and 45-45-90 have the sides that they do. Before I just assumed to know from memorization!**“Being able to derive these patterns myself aids in my learning because…”**IT GOES BACK TO THE UNIT CIRCLE AND IT HELPS ME UNDERSTAND THE UNIT CIRCLE SO MUCH BETTER NOW :)

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