SP 7: Unit Q: Concept 2: Finding all trig functions when given one trig function and quadrant

Please see my SP7 made in collaboration with Eriq, by visiting their blog post here.  Also, be sure to check out our other collaborative blog post as well as his awesome individual ones! (:

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

http://www.softmath.com/tutorials-3/relations/articles_imgs/5385/fundam5.gif

There are the reciprocal identities that I will constantly be referring to.

http://calculustricks.com/wp-content/uploads/2011/03/Trigonometric-Identities-3.jpg

These are the ratio identities that I will constantly be referring to.


1. Identities are proven facts or formulas that are always true. The Pythagorean Identity sin²x+cos²x=1 comes from using the Pythagorean theorem.  The Pythagorean theorem, instead of using a²+b²=c², we will replace the letters with x²+y²=r², which is the same exact thing when placed on a coordinate plane. Then we want the right side to equal 1 so we divide both sides by r² and we get(x/r)²+(y/r)²=1. When looking at what we now have, we can make to changes which we learned previously. First, we can change (x/r)² to cos² (cosine) because the equation for cosine on the unit circle is x/r. Also, we can change (y/r)² to sin² (sine) since that is the equation for sine on the unit circle. Now, to make sure that this is really identity, we must apply the unit circle to this to get the Let's plug in an angle of 45* to this identity. First, we find cos of 45 and sin of 45 which turn out to be √2/2. So we plug it into the Pythagorean Identity that we found. So it is going to be (√2/2)²+(√2/2)²=1. When we distribute the power and add both of them together, our answer becomes 4/4 which equal 1. Remember to put x or theta because that means that it can be any angle. Now we have one of our Pythagorean Identity which is cos²x+sin²x=q.

©Kelsea

©Kelsea

2. There are two remaining Pythagorean Identities, 1+tan²x=sec²x and 1+cot²x=csc²x. To obtain the first equation I mentioned, first divide both sides by cos²x. By doing so we get cos²x/cos²x+sin²x/cos²x=1/cos²x. Now when we look at our ratio identities and our reciprocal identities, we can see that some of the fractions we have created are equal to some sort of single identity: sin²x/cos²x=tan²x and 1/cos²x=sec²x. So we can replace the fractions with these single identities to get the equation 1=tan²x=sec²x.  We now have one last Pythagorean Identity to find. To do so, we divide by sin²x. Now our identity will be cos²x/sin²x+sin²x/sin²x=1/sin²x. Again, the fractions correspond with some of the ratio and reciprocal identities. cos²x/sin²x=cot²x and 1/sin²x=csc²x. Now replacing the fraction identity with a single identity, we get the last Pythagorean Identity which is cot²x+1=csc²x.

©Kelsea

References:

http://www.softmath.com/tutorials-3/relations/articles_imgs/5385/fundam5.gif

http://calculustricks.com/wp-content/uploads/2011/03/Trigonometric-Identities-3.jpg

WPP#13-14: Unit P: Concept 6 and 7

This WPP 13-14 was made in collaboration with Eriq on his blogspot! Please visit his other awesome posts by going here.

http://www.kevinandamanda.com/whatsnew/travel/sweet-tooth-fairy-bakery.html

The Problem: Bakery Meet Up

While on her way walking around town, Kelsea notices her friend Eriq across the street, about 20 feet away. He notices her and waves back. They both text each other, to avoid walking, and look ahead and see a bakery in which they both decide to meet up. Kelsea is N30*E and Eriq is N25*W. 

a) what is the distance from Kelsea to the bakeshop?

b) what is the distance from Eriq to the bakeshop?

The Solution:

Copyright: Kelsea and Eriq

The Problem: Leaving

After having a few sweets and catching up on each other's lives, their rides arrive. They bid each other goodbye and leave at the same time, traveling on a course with an angle of 150* between them. If Kelsea, on the right, travels 30 mph and Eriq, on the left, travels 40 mph, how far apart are they after two hours have passed?

The Solution:

Copyright: Kelsea and Eriq

BQ#1:UnitP: Concept 1-4:Non-Right Tiangle Equations

1. Law Of Sines
For non-right triangles we need an equation that will help us find the angles and sides. We can't use the Pythagorean theorem because that only applies to right triangles. So observing a non-right triangle, we can try and find an equation to help us. First of all, let's try and make that non-right triangle into two right triangles by taking angle B and putting a straight line through it.

(http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm)

Above, we can see how the line that cuts in between angle B has created two right triangles. Also, the line is now being called h. Now, we use Sine to find and equation with h in it.
In order to do so, we use Sin A=h/c and for the other side we use Sin C=h/a. Then we cross multiply to get cSinA=h and aSinc=h. Now, here's where it gets a little tricky.

(http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm)

(http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm)

(http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm)

In the above picture, we can see how we already did the first two steps. Now to put it simply, all we had to do was equal both equations to each other since they both equal h. Then we divide both by the side by the side that it has. So we divide by ac. Then one we do that we get the equation SinA/a=SinC/c. Now this is the equation for sines. Since we have this, we can safely assume that SinB/b also equals the others. And if you are skeptical, we can do the exact same thing but instead just cut a straight line through angle A

(http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm)

We did the same thing we did at the beginning but the line cuts through angle A. Now we solve the exact same way but for angle B. We did the same thing by cutting a straight line but only to a different angle. Then let's call it k. It could be h but we already used it.

(http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm)

(http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm)

(http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm)

 we solved the exact same way but for angle B and got the same and similar answer. So now we know that not only do they all equal each other but also that this works for all triangles

4. Now let's try and find the area. As we know, for right triangles, the area equation is A=1/2bh. Now of course' this is for right triangles. Now, to find one for non-right triangles, we simply replace h with

aSinC. Now you're probably wondering how that is so. It's simple. Remember when we were finding the origins of the law of Sines each one was equal to h? Well instead of using the equation area for right triangles we simply replace the h with the sin that we got when fixing the Law of Sine.

(http://pcb2s11.blogspot.com/2011/06/chapter-6-summary.html)

References:
http://pcb2s11.blogspot.com/2011/06/chapter-6-summary.htmlhttp://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm

WPP#12: Unit O Concept 10

http://cdn.snowboardermag.com/files/2009/10/IMG_4157.jpg

The Problem:Lauren has a week off from work and decides to go snowboarding at the mountains and stay at a resort. When she arrives, she is admiring the snow white mountain at ground level at the snowboarding resort "Mountain High." 

a) Lauren measures the angle of elevation to the top of the mountain to be 60*. If, at this point, she is 60 feet away from the mountain, what is the height of the mountain?

b) Lauren reaches the top of the mountain and see's her friends waving at her and encouraging her to ride down the mountain. She now see's the flag where she is supposed to snowboard down and she measures the angle of elevation to the flag to be 20*. How long is the path to the flag? [keep in mind to use the height of the mountain]

The Solution: a) Draw out the mountain and draw a person some distance away from it. Then draw a line from the top of the mountain to the persons feet and then draw a line connecting the person to the bottom of the mountain. That line will be 60. The angle we just made will be 60*. Now label the height of the mountain x and that is the one we will be solving for. Now we are going to be using tangent to find the height of the mountain. The equation should be tan(60)=x/60. multiply both sides by 60 and the answer should be x=103.9 ft. 

b) Draw the mountain again but this time put a person on top of the mountain and a flag some distance from the mountain. Connect the flag to the person and that will be our x. Make a flat line across the top of the mountain and then connect the flag to that line. That will create a height which will be the height of the mountain, being 103.9. Now, using sin, we will find the distance from the person to the flag. The equation should be sin(20)=103.9/x. Multiply both sides by x and then divide both sides by sin(20). The answer should be 303.8 ft. 

*anything highlighted are important to solving the problem*

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

Inquiry Activity Summary

[Property of Kelsea]

1. A 30-60-90 triangle is derived from an equilateral which is cut in half. An equilateral is a triangle where all sides are equal as well as all angles. All the angles in the equilateral triangle are 60* and all the sides are 1. When we cut the triangle in half, we get different angles and sides. half of the triangle is now made up of angles which are 90*, 60* and 30*. The 60* is the one that was not affected by the cut. The 30* is the one at the top where the angle was cut in half, which makes sense since half of 60* is 30*. The 90* comes from the bottom where the side is perfectly perpendicular to the bottom side. The sides are now different as well. The hypotenuse, the slanted on, is not affected by the cut so is remains 1. The bottom one, which is cut in half becomes 1/2. However, there is no given value for the side that has been cut. In order to find that side, we use the Pythagorean Theorem [a^2 + b^2 = c^2]. The hypotenuse is c, the bottom side is a, and w solve for b. Once we plug it all in and solve for b, the answer should be b=√3/2. Now the sides should all be as follows: a=1/2, b=√3/2, and c=1. Now, since nobody is very fond of fractions, we can multiply all the sides by 2 to eliminate the fractions. Now the sides should be a=1, b=√3, and c=2. Now we are almost done! Now we must add an n to each side number because although the angles will remain, the sides are subject to change. That means n can be any number and the angles will remain the same. So finally, one last time, lets list what the angles are: a= n, b=n√3 [n should not be behind it], and c=2n. Our 30-60-90 triangle is now derived and conquered!

[Property of Kelsea]

[Property of Kelsea]

[Property of Kelsea]

[Property of Kelsea]

2. A 45-45-90 triangle is derived when a square is cut diagonally. A square is a shape where all four sides are equal [in this case we will set them equal to 1] and all sides are 90*. When cut diagonally, the two angles change, and one side changes. Since it is cut diagonally, the two angles affected will become 45* which makes sense because half of 90 is 45. The other angle remains 90* since it was not affected by the cut and it is perfectly perpendicular to the bottom side. Two of the sides were not affected, the ones that are not slanted, so they remain 1. However, one was, the hypotenuse. In order to find the hypotenuse, we must once again use the Pythagorean Theorem [a^2 + b^2 = c^2] to find the hypotenuse. Hypotenuse will be c, the bottom side will be a, and the other side will be b. Now we plug our numbers into the equation and solve for c. The answer should b √2. Now all our angles should be as follows: a=1, b=1, c=√2. Yay! Wait ... we are forgetting something that we did in the 30-60-90 triangle. What is it? Oh Right! Our lovely buddy n. Add n to all the sides because, as stated before, the angles can remain the same but the sides are subject to change! So now, our sides should be a=n, b=n [It is 1n but since any number multiplied by 1 is that number, we do not need the 1], and c=n√2 [remember, no n behind the radical!]. Congratulations! The angles are now derived and we now know how to derive the 45-45-90 triangle! Good Job!

[Property of Kelsea]

[Property of Kelsea]

[Property of Kelsea]

Inquiry Activity Reflection

1. "Something I never noticed before about special right triangles is ..."  that the sides are found using the Pythagorean Theorem equation. [Hypotenuse for 45-45-90 and the b for the 30-60-90]
2. "Being able to derive these patterns myself aids in my learning because..." if I ever forget what the side is, I can derive it using the what I learned from this activity, which is to cut a square diagonally for a 45-45-90 or an equilateral triangle for a 30-60-90 and then use the Pythagorean Theorem to find the angle that is missing.