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

This SP7 was made in collaboration with Adrie. Please visit the other awesome posts on her blog by going here. 



What is this problem about?
This problem will go over the steps on how to all trig functions when given one trig function and a quadrant by using identities. We will also try to fund the trig functions using SOH-CAH-TOA to show that both methods can give us the same answers.


Identity Work (Unit Q Concept 2)

SOH-CAH-TOA Work (Unit O Concept 5)

Make sure you pay attention to the rationalization, because we can never have a radical as a denominator. Along with that, make sure to remember your identities, so you know why we plug the values in those numbers. Also, keep in mind that 

Pay close attention to the relationships because in the end, SOh-CAH-TOA and trig function identities both give us the same answer to all trig functions.

I/D #3: Unit Q: Pythagorean Identities

1. Where does sin²x + cos²x  = 1 come from?

http://i1.ytimg.com/vi/o-fAx_96lgw/maxresdefault.jpg

Well, let's start by defining this as the main Pythagorean Identity. So what is an identity? They are proven facts and formulas that are always true, so we can go ahead and manipulate the expressions and equations. The equation that we can manipulate (not redefine) is the Pythagorean Theorem because it  is also an identity that will help us help us derive sin^2x+cos^2x=1.
So let's back up a little. The Pythagorean Theorem is a^2+b^2=c^2. However, when referring to a unit circle, the Pythagorean Theorem is written as x^2+y^2= r^2.

http://www.contracosta.edu/legacycontent/math/Pythagoras.htm

This is how the Pythagorean Theorem is drawn on the unit circle, with r=1. So how do we get r^2 to equal 1? We simply do divide it all by r^2 (both sides)

x^2 + y^2 =r^2
                               >>>  (x/r)^2 + (y/r)^2 = 1    We cancelled off r^2
r^2     r^2   r^2

Now, take a closer look at the fraction and what do you see? Anything familiar, perhaps the ratio for cosine and sine on the unit circle?? Yes? Yes!
Based on one of previous units, we found out that the ratio for cosine is x/r and the ratio for sine is y/r. Notice how they are written the same way, except that they are both squared. So we can just replace the ratios by the trig function name. Therefore, we get: sin^2x+cos^2x=1
This helps prove the reason why the Pythagorean Theorem is an identity since it can be manipulates, but the facts and formula is always the same even though a little rearranged.
Let's actually prove that this works though. Since we are talking about the unit circle, we can use one of the "Magic 3" pairs from the Unit  Circle. Let's use 45 because it has different x and y values.
The ordered pair is :(√2/2)>>>> Substitute them in the equation:  (√2/2)² + (√2/2)² = 1. (√2/2)²
When we simplify it, we get (1/2) + (1/2)=1, so this helps us prove that then identity is true!

Now we can proceed to find the next two Pythagorean Identities:
We can first find the tangent derivation by dividing the main identity by cos^2/x.
sin^2x   +   cos^2x    =    1
                                                                >>  tan²x + 1 = sec²x.                         The cos^2x cancel to equal 1
cos^2/x       cos^2/x     cos^2/x                                Then we find the rest by using the ratio/reciprocal identities


It is very important to remember the ratio identities in order to make your life easier and know where we get our substitutions!

http://www.sosmath.com/CBB/viewtopic.php?t=41908

Now we can continue to find cotangent derivation by dividing everything by sin^2x
sin^2x   +   cos^2x    =    1
                                                                >>  cot ²x + 1 = csc²x                The sin^2x will cancel to equal 1
sin^2x         sin^2x       sin^2x                                 Then we find the rest by using the ratio/reciprocal identities


Inquiry Activity Reflection:
1. The connections that I see between Units N, O, P, and Q so far are that the unit circle and the trig functions will continue to reappear and help define most of the trig functions and anything that relates to it even though equations are always being rearranged. Also, I learn the significance of sin and cos because all trig functions eventually relate back the them. It is almost like the universal trig function. 
2. If I had to describe trigonometry in THREE words, they would be... interconnected, confusing,and mind-blowing. 

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 Vanessa Morales, by visiting their blog here.  Also be sure to check out the other awesome posts on their blog

BQ#1: Unit P Concepts 1 and 4: Law of Sines, Area of an Oblique Triangle, and their Derivations

http://htmartin.myweb.uga.edu/6190/quickreview3a.html

1. The Law of Sines AAS or ASA
When we use trig function, we usually use them only with right triangles. All we have to remember if SOH -CAH- TOA or the Pythagorean Theorem. What if it is not a right triangle? Well, we can still use the trig function, to determine the Law of Sines. This way we can determine for non-right triangles.

Here it how we derive the Law of Sines:

Add cahttp://www.mathamazement.com/Lessons/Pre-Calculus/06_Additional-Topics-in-Trigonometry/law-of-sines.htmlption

This is a non-right triangle, but we can simply make it a right triangle by drawing a perpendicular line from <B.  Now we have two right triangles that we can work with.

http://www.mathamazement.com/Lessons/Pre-Calculus/06_Additional-Topics-in-Trigonometry/law-of-sines.html

When we drew the perpendicular line, we got two right triangles. The perpendicular line can be called "h".  Let's look at the relationships that the sides and angles have based on the trig function. When we do this we could get SinA=h/a and SinC=h/a (opposite.adjacent). Both equations have fractions, so to get ride of the fractions, we can , multiply by the denominator. Once we do that we get cSin A=h and aSin C=h. Since both equal "h" we can use the transitive property to get c Sin A= a Sin C. Lastly, divide both sides by "ac" to cancel off the c in one equation and a in the other.

Now we have Sin A/a= Sin C/c. However, the equation also includes SinB/b. Why don't we have it. Well, that's because we cut <B with a perpendicular. However, if we can drawn the perpendicular line from another angle, such as <A, then we could get Sin B/b.
We follow the same process and we get Sin C= h/b and Sin B=h/c.  Use the transitive property and divide by "bc" and we can also get Sin B=b. Let's put it all together and we get.

                    

http://www.clausentech.com/lchs/dclausen/algebra2/formulas/Ch12/Chapter12Formulas.htm

Check this out for more clarification on the Law of Sines and see an example of the Law of Sines.

Also, learn more about the Law of Sines and the different solutions: two solutions, one solution, or no solutions. Click here!


4. Area of an oblique triangle

According to Mrs. Kirch, "the are of an oblique (all sides different lengths) triangle is one-half of the product of two sides and the sine of their included angle."

http://www.compuhigh.com/demo/triglesson07.html

Similar to the Law of Sines, we can draw a perpendicular line "h" to make the triangle into an oblique triangle. You may know that the formula for a triangle if A=1/2bh where b represents the b and h is the height.
Based on the triangle above and concept 1, we know that Sin A= h/c and Sin B=h/c and Sin C=h/a.
When we multiply the denominators, we get cSinA=h or aSin C=h. Since we already know "h" then we can easily plug them into the original area formula of A=1/2bh. In other words, we leave the equation as is and substitute "h" for cSinA, cSinB, and bSinC

When you do this, we get:
A= 1/2 b(cSinA)                       A= 1/2 a( cSinB)                       A=1/2a (bSincC)


References:
http://www.youtube.com/watch?v=bDPRWJdVzfs
http://www.lessonpaths.com/learn/i/unit-p-supplementary-resources-and-help/ambiguous-case
http://www.compuhigh.com/demo/triglesson07.html
http://htmartin.myweb.uga.edu/6190/quickreview3a.html

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

Directions: Read the word problem and answer the question.Show all work and steps clearly.

This picture will go over the solution to this word problem.Make sure to try it on your own first before looking at this picture.


Part A:
Scott went to San Francisco to sight see what his friends called the best city in the world! Of course, he wanted to intake the beauty of the city by standing from one of the great buildings. He was approximately 45 ft. from one of the buildings. He measured the elevation to the building to be 25 degrees. He wanted to know how many feet above the ground his view would be. How tall was the building?



Note:  "X" is the missing side that we need, the height of the building (also the opposite side). The distance between Scott and the building is 45 ft, which is adjacent to the angle (elevation). Remember to use tan and not the inverse since we are not looking for an angle, but a side instead.

Part B:
Once he was at the top of the building, he spotted his best friend Anna who happened to be at the ice cream shop eating ice cream. Suddenly, he got a crave for ice cream and wanted to accompany her. The angle of depression from the top of the building to Anna was 29 degrees. How far apart was the building that Scott was standing on to the ice cream shop that Anna was at? (Hint: Use "Part A" to find the length of the building. This will help you complete the ratio.)


Note: Since we referred back to the previous part and found out that "x" (height) was 20.98 ft. we can make that the adjacent side to the ratio. The missing distance between both building is represented by the variable "x", which also tangent. Again, we only use tan and not the inverse since we are only looking for the distance and not the angle.

I/D #2: Unit O - How can we derive the patterns for our special rights triangle?

HEADING FOR THIS SECTION: INQUIRY ACTIVITY SUMMARY Please click here to watch video


HEADING FOR THIS SECTION: INQUIRY ACTIVITY REFLECTION

Something I never noticed before about special right triangles is how they all connect one way or another. I knew that triangles all add up to 180 degrees, but I had forgotten that if you divide the triangle by half, you get a right triangle and then a 30-60-90. Of course, the 45-45-90 also derives from another shape, but this time it is the square. I never noticed that the triangle is simply a ratio that will always remain the same even as the triangles expand.

Being able to derive these patterns myself aids in my learning because I can always rewrite the pattern in case I happen to forget. It would only really take me knowing the Pythagorean theorem, which I hope to never forget. Finding out the pattern is actually really simple and now I have the ability to easily figure it out simply by finding the angles measurements of a triangle and not by memorizing the sides, but by actually understanding the factors of triangles.