SP# 7: Unit Q Concept 2: Find all Trig Functions when given one trig function and quadrant (using identities and SOH CAH TOA)

Please see my SP7, made in collaboration with Ivan L, by visiting their blog here.  Also be sure to check out the other awesome posts on their blog

I/D #3 Unit Q: Pythagorean Identities

Inquiry Activity Summary:

1. Where does sinx^2+cosx^2=1 come from?

In earlier units we learned about the three basic trig functions which are sin, cosine, and tangent. Sine which means x/r so the y is the side that rises and r is the hypotenuse. R replaces the c^2 in the Pythagorean theorem. So it would really be a^2+b^2=R^2. Cosine we know is y/r   . We know that when looking at the problem above that it really is x/r^2+y/r^2=1. Using the unit circle to help prove that this statement is true is fairly easy we use the 45-45-90 triangle. Which is square root of 2/2, square root of 2/2. When squaring these you get one half plus one half equals one. Which in this case is an identity which means it is a true proven statement. 

2. Show and explain how to derive the two remaining Pythagorean Identities from sin^2x+cos^2x=1

To derive the two remaining identities all you have to do is divide by either cosine or sine. We know that if we divide by the same function we will get one so that is how we get the one on the right side. We will also notice that if we divide by a trig function like sin/cos we simplify that to x/r /y/r we multiply the reciprocal of the denominator to top and bottom to get the new one which is tangent. You do the  same thing for the other identity. 

Inquiry Reflection Activity 

1. The connection I see between Units N, O, and P are that we still have to use the unit circle in some we and that we still use the trig functions even more. 

2. If I had to describe trigonometry in 3 words, they would be challenging, depressing, frustrating. 

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

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

BQ#1:Unit P: Concept 1 and 4: Law of Sines and Area Formulas

1. Law of Sines 
We need the law of sines to help determine the missing sides of an oblique triangle. We can also use it to determine the sides of a right triangle. Sin is opposite/ hypotenuse. When we have a problem we will always be given an angle we use one of those angles to help us determine one of the missing sides. The law of sines comes from trig. Sine is only used for right triangles but with non right triangles we can use them too. When we use some we have to make sure the question gives us AAS or ASA 

4. Area Formulas 

Always remember that the area of a right triangle will always be 1/2 bh. B is the base and h is the height. We will not always be given the height so we have to find the height using sine. You use the same method of finding the area of a right triangle for finding the one of an oblique triangle. You split the oblique triangle in half or wherever you can so you can determine the height. Use an angle given to you to determine one of the sides. You keep going until you find the height and use the same equation. It will always relate to the area of a triangle formula because we have to use it in order to get what we are looking for. 

WPP#12: Unit O: Concept 10: Solving angle of depression and and elevation

Picture:

http://nassau.happeningmag.com/wp-content/uploads/2013/01/boat-show.jpg

Problem: Julian decided to take a boat ride on his brand new yacht. He takes it out for a spin and notices a cliff upcoming ahead. He wonders, if the angle of elevation to the top of the cliff is 52.53 degrees,  and the base of the cliff is 1236 feet away from the boat, how high is the cliff? (Remember round to the nearest foot)

Ivan was on top of a lighthouse observing Julian on his brand new yacht. He looked around and noticed at the bottom of the cliff where he was positioned, that there was a boiling crab. He found a parachute and thought, " I wonder how long will be the path I glide if the angle of depression is thirty degrees?"  Keep in mind he is already 300 feet above ground. 

The Solution: 

a)

B)

I/D #2: Unit O: How can we derive the patterns for our special right triangles?

Inquiry Activity Summary:

For this activity we had to derive the special right triangles not from the unit circle though. We had to derive the two special kinds ,45-45-90 and 30-60-90 right triangles. Each are completely different we had to find out what n was and why n couldn't just be a number. 

How can we derive the 45-45-90 triangle from an square with a side length of 1? 

We know that if we are given a square with side lengths of 1 we will have two right triangles if we split the square in half from the corners. For a 45-45-90 triangle we know the base and height are the same,1. If we do the Pythagorean theorem we will see that the hypotenuse will equal radical 2. N is there to be any given value, without n the triangle sides cannot be altered to match the initial constants, it is multiplied by the initial constants. For example if n were to equal 2 then the base and height are 2. The hypotenuse will equal 2 radical 2. 

How can we derive the 30-60-90 triangle from an equilateral triangle with a side length of 1?

To derive the pattern for the 30-60-90 triangles we have to cut an equilateral triangle in half straight down the middle. Each side length of the triangle is one so when we split the triangle in half the base turns into 1/2. We also split the triangle in  half to get 30 degrees as one of the angles. Then we will have a 30-60-90 triangle. Then we notice that we do not have the height of the triangle, we have to use the Pythagorean Theorem. Once you get the constants they can be altered to get rid of the ugly fractions by multiplying by two ( look at picture). The constants are the same because they are all proportional and everything was multiplied equally. 

Inquiry Activity Reflection: 

 Something I never noticed before about special right triangles is how we have to tweak equilateral triangles and squares to get the sides for n. 

Being able to derive these patterns myself aids in my learning because now I can refer to this square or triangle if I ever forget what the sides of the triangle equal.