SP #7: Unit Q Concept 2: Finding All Trig Functions When Given One Trig Function and Quadrant (Using Identities)

Given this problem:

Find the missing trig functions.

Using Identities

Using SOHCAHTOA
The first thing that you need to do for this problem is find the hypotenuse. Do this by using the Pythagorean Theorem. You should get rad34. 
We know that:
Tanx= -5/3
Secx= -\|34/3

If tan & sec are negative the. We know we are in the 2nd quadrant. To find sin, cos, csc, cot, we use the ratios. Go back to unit N if you need to review the ratios. If you need help look at the picture for reference.

A viewer needs to pay special attention to the two different methods of solving these trigonometric functions. Even though it uses both identities and the old SOHCAHTOA method, the answers still turn out the same. This shows that one can use either identities or SOHCAHTOA to solve for different trig functions, because they both work for the same problem.

This SP was made in collaboration with Bryan Arciniega.

I/D3: Unit Q - Pythagorean Identities

Inquiry Activity Summary:
1. Where does sin^2x+cos^2x=1 come from?
This identity is derived from the Pythagorean theorem, and that is why it is known as a Pythagorean identity. Drawing from our knowledge about the unit circle, the Pythagorean theorem using x, y, and r would be: x^2+y^2=r^2. From here, we would perform an operation that makes the Pythagorean Theorem equal to 1, which is part of the identity. This would then result in: (x/r)^2+(y/r)^2=1. A pattern that we notice is that the unit circle trig ratio for cosine is x/r, while the unit circle trig ratio for sine is y/r. As a result, we can plug these two trig functions into our equation, and thus we derive the Pythagorean identity.
2. Show and explain how to derive the two remaining Pythagorean identities from sin^2x+cos^2x=1.
To derive the two remaining Pythagorean identities, we can perform similar operations:

• For the first derivation, we divide the equation by cos^2x. This will leave us with several functions that can be converted using trig identities. Cos^2x/cos^2x cancels out as 1, sin^2x/cos^2x is equivalent to tan^2x (as seen in a ratio identity), and 1/cos^2x is equivalent to sec^2x (as seen in a reciprocal identity). This leaves us with 1+tan^2x=sec^2x.

• For the second derivation, we divide the equation this time by sin^2x. This will leave us with several functions that can be converted using trig identities. Cos^2x/sin^2 is equivalent to cot^2x (as seen in a ratio identity), Sin^2x/sin^2x cancels out as 1, and 1/sin^2x is equivalent to csc^2x (as seen in a reciprocal identity). This leaves us with cot^2x+1=csc^2x.

Inquiry Activity Reflection:
The connections that I see between Units N, O, P, and Q so far are that they all deal with triangles and angles in some way. We are learning how to solve different kinds of triangles, from right triangles to acute and obtuse ones that we needed to apply the Law of Sines and Cosines to. We are using the trigonometric functions derived from triangles (sin, cos, tan, etc.) and using them as identities or to find angles in the unit circle. Another connection that can be seen between these four units is that most of them deal with the unit circle--for example, we just derived the Pythagorean identities using trig functions from the unit circle and can prove them using points from it.
If I had to describe trigonometry in three words, they would be complicated, challenging, and rewarding to learn. It is complicated because there are many derivations that we must learn and, more importantly, understand if we want to grasp a concept. Everything that we learn is built on something else, and it is important to understand these things before we move on to something new. Also, trigonometry is challenging because there are many formulas, many identities to memorize in order to succeed. Memorization is the biggest and most challenging part, because once we know the material well, we will be able to do well on the tests. Finally, trigonometry is almost rewarding to learn because I enjoy the feeling of success that comes after solving a triangle or simplifying a trig function using identities. It may be difficult at first, but once I know the concepts well, I am very satisfied with learning them.

WPP 13-14: Unit P Concepts 6-7: Application of the Law of Sines and the Law of Cosines

The Ultimate Crossover Finale: Part 1

http://25.media.tumblr.com/5927df0165a2a3c16b1299503df2d3be/tumblr_mhb0d5qkFy1rmt8bgo1_500.jpg

Bryan and Sarah are on the same team when playing capture the flag. However, whoever gets to the enemy's flag first will win all the glory for their team. They are 47 yards apart on an east-west line. Between them in the distance, they see the enemy's flag mounted on a pedestal. Bryan's bearing to the flag is N 23 degrees E, while Sarah's bearing to the flag is N 36 degrees W. Find each person's distance from the flag. Who is closer? (Round your answers to the nearest hundreth.)

The Ultimate Crossover Finale: Part 2

http://designlap.com/eye-catching-beach-wallpapers/

Bryan and Sarah are now working as lifeguards. Bryan and Sarah's life guard post are 550 feet apart. Sarah spots a person in need at 213°. Being a great mathematician, Sarah calculates she is 700 feet away from this person. How far is Bryan from the person in need?                                         This WPP was made in collaboration with Bryan Arciniega.

BQ# 1: Unit P Concepts 1-5

1. Law of Sines: The law of sines is important in finding the values for non-right triangles. By setting up a proportion, one can find the value of a side or an angle in a triangle if we have all the necessary information. The law of sines works best when we have an angle-angle-side, angle-side-angle, or even an side-side-angle triangle.

5. Area Formulas: All three formulas to find the area of the triangle--the "traditional" formula, area of an oblique triangle, and Heron's formula--will result in the same value: 4.2 units squared.

WPP 12: Unit O Concept 10: Finding Angles of Elevation and Depression

http://www.timeanddate.com/holidays/un/international-mountain-day

a) Percy is going on a hike with his friends up a mountain. His friend, Annabeth (who is a total genius), measures that the angle of elevation to the top of the mountain to be 72 degrees. Assuming that she, Percy, and their group of friends are standing 205 feet away from the base of the mountain, how tall is the mountain? (Round to the nearest foot)

b) Percy and his friends reach the peak of the mountain a few hours later. They are going to hike down to a lodge at the base of the mountain from the other side, which is less steep. It has an angle of depression of 41 degrees. Since they are ____ ft high up on the mountain, how long is the trail they will hike this time?

Solution:

I/D1: Unit O - Derive the Special Right Triangles

Inquiry Activity Summary
1. To derive the 30-60-90 triangle from a equilateral triangle with all side lengths of 1, we first have to cut the triangle in half down the middle, perpendicular to the base side. The angles that will have been formed in the new triangle (either half of the divide) are 30, 60, and 90 degrees. The 90 degrees is formed from the perpendicular cut and the base of the triangle, the 60 degrees is the untouched angle of the new triangle, while the 30 degrees angle has been the angle cut in half where the divide had begun at (half of 60 degrees is 30 degrees).
For the side lengths, the hypotenuse is still one, since that was the original length of a side of the equilateral triangle. The 30 degree side is 1/2, since the triangle had been cut in half from the base and we are deriving half of the original side length of 1. Finally, the 60 degree side can be derived using the Pythagorean theorem. We must solve for 'b' once we plug in the numbers for 'a' and 'c' (1/2 and 1, respectively, corresponding to their angles). When we simplify and get the 'b' variable on its own, b^2 will be equal to 3/4. As a result, the value of b would be radical 3 over 2. Finally, we must multiply all the side lengths by 2 to simplify them, since we do not want fractions. The new side lengths would be, in ascending order: 1, radical 3, and 2. The last step is to substitute the values with the 'n' variable where n=1. This shows that these values can be applied to any multiples of themselves. Now, the new side lengths would be, in ascending order, 'n' to the 30 degree side, n radical 3 to the 60 degree side, and 2n to the 90 degree side.

2. To derive the 45-45-90 triangle from a square with all side lengths of 1, we first have to cut the square in half diagonally, from one corner to another. The angles that will have been formed in the new triangle (either half of the divide) are two 45 degree angles. This is because the two 90 degree angles that have been cut in half are both now 45 degrees, since half of 90 degrees is 45 degrees. The third and last angle is 90 degrees, which is the corner of the former square that had been untouched. These angles fulfill the 180 degree rule for all triangles.
For the side lengths, the two legs of the 45-45-90 triangle are both one, since they were from the sides of the square. If we plug this into the Pythagorean theorem, we see that c, or the hypotenuse, is radical 2. This is because c^2=1+1=2, so we take the radical of it. So, our three side lengths would be, in ascending order, 1, 1, and radical 2. Then, we substitute the values with the variable 'n', where n=1. This shows that these values can be applied to any multiples of themselves. Now the new side lengths would be 'n', 'n', and n radical 2.

Inquiry Activity Reflection
Something I never noticed before about special right triangles is that by using the Pythagorean theorem, we can derive the sides of both of the special right triangles. I can understand how the values came to be in the special right triangles, such as how the 60 degree side of the 30-60-90 triangle was originally radical 3 over 2, until we simplified the sides of the triangle further. It is amazing how the Pythagorean triangle can be applied to so many things!
Being able to derive these patterns myself aids in my learning because if I ever forget the rules of the 45-45-90 or 30-60-90 triangle, I will be able to derive them from the square and the equilateral triangle with the sides lengths of 1. I will know how to use the Pythagorean theorem to determine the sides and then substitute the values for the variable n, so I know that any other multiples of those values can be used. It broadens my understanding of the special right triangles even further because now I have a better grasp of where they come from and what different math concepts could be applied in deriving the special right triangles.