BQ# 5: Unit T Concepts 1-3

Why do sine and cosine NOT have asymptotes, but the other four trig graphs do? Use unit circle ratios to explain.  
       Sine and cosine do not asymptotes because asymptotes occur when the denominator of the ratios is 0 (undefined). The other trig graphs can be divided by 0 and it can be undefined while sine and cosine can't be divided by 0. They can only be divided by 1 because 1 is their restriction on the Unit Circle and on the Unit Circle itself it goes from 1 to -1 on both axis. They both can't be divided by 0 because it's not undefined but no solution. The other four especially tangent and cotangent have no restrictions and if their ratios equal undefined then it's their asymptote.

BQ#2: Unit T Concept Intro

A)   When is comes down to sine and cosine, their periods are are 2pi and it's like that due to their similarity to the Unit Circle.

Sine in the Unit circle is positive in the first and second quadrant and negative in the third and fourth quadrant. If we start from zero and do one complete rotation to get it to be positive again, then then it was a 360 degree or 2pi rotation. So when we stretch Unit Circle into a line, the cyclical will be stretched out too because the first two quadrants will be above the x-axis and the last two will be below the x-axis. 

The same thing goes for cosine since quadrant I and IV are positive and quadrant II and III are negative. We need to start at 0 degrees and as we move around in the Unit Circle well reach the positive when we get to 360 degrees. When it reaches 360 degrees the it made one complete rotation. If the stretch out the Unit Circle to a straight line for cosine then we will start above the x-axis since quad. I is positive and for quads II and III it will be below the x-axis. As we approach quad IV to make a cyclical then it go back up above the x-axis.

For tangent it's different because it's only half the trip to get a cyclical. Quadrants I is positive and quadrant II is a negative so we already have our cyclical. On the Unit Circle, it will start at 0 degrees and when we reach a positive again it will be at 180 degrees and the radian value is pi.

B)    Sine and cosine only have an amplitude of 1 because they have a restriction of 1 and -1. If we use any number and use it in their ratio, it will not work except those numbers less than 1. We only use the 1 in the Unit Circle then sine and cosine will work because it is in their restriction,  if it wasn't then it will be undefined and will not work.

Reflrction #1- Unit Q: Verifying Trig Identities

1. What it means is that when we are given a problem we try to use all three forms of identities to solve one side to see if it matches the other untouchable side. In other words we are verifying to see it we can get the same answer.
2. For me to solve them more easily, I try to get them to equal sine or cosine. I mostly try to see if any part of the problem can be made into a ratio or reciprocal identity and if they can be multiplied. If they are then it would be even better if they are written as fractions so I can cancel things out.
3. The first thing I try to see is if the problem is a simplifying or verifying and then I can try to use the conjugate if it's needed. It doesn't matter if there is because I can substitute in an identity or if it's a fraction combine them. If the problem comes out to have a monomial then it's even better because I can separate them and hopefully it turns to an identity.

SP#7: Unit Q Concept 2: Finding all trig. function

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

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

INQUIRY ACTIVITY SUMMARY:
     Pythagorean identities come from both the Unit Circle and the Pythagorean Theorem. First to clarify, an identity is "an equation that is true no matter what values are chosen". The Pythagorean Theorem is seen as an identity because no matter what two values you have and when you solve it it will give you the third value. If you check your work by using the found value and a value given, the answer will be that same given value you did not use. To get the Pythagorean identity, sin^2(theta) + cos^2(theta) = 1, I will work it out using the Unit Circle and the Pythagorean Theorem and I will show and explain it in the following pictures:

First I'll show it by using the Unit Circle using the first quadrant since sine, cosine, tangent, co-secant, secant, and cotangent are all positives. If we plot a triangle in the quadrant we can already see the connection between the two. Th height will be "y" since it follows the y-axis, "x" will be base since it's on the x-axis, and the hypotenuse will be "r" because the hypotenuse is the radius of the circle. When it gets plugged in to the Pythagorean Theorem "a" will be "y", "b" will be "x" and "r" will be "c".

When I use "x,y and r" in the Pythagorean Theorem it will make sense if it's written out like y^2 + x^2 = 6^2 because when we try to make it equal one then it will make sense. If I take the sine of angle A it will be y/r, the cosine of it will be x/r, and "r" will stay as "r". Now I can show how (y/r)^2 + (x/r)^2 = r^2 can equal one but there is a faster and easier way to get one. If I write out "opp" as opposite, "adj" for adjacent, and "hyp" for hypotenuse and then in Pythagorean form, it's opp^2 + adj^2 = hyp^2. If I divide both side by hyp^2 then the hyp^2 will be 1 and it will be opp^2/ hyp^2 + adj^2/ hyp^2 = 1. If I pug it in to the other equation I said, opp^2/ hyp^2 is the cosine of angle A and adj^2/ hyp^2 is the sine of angle A. So cos^2(theta) + sin^2(theta) = 1 is one of three Pythagorean identities.

To find the second one, which is the identity with secant and tangent, I will use cos^2(theta) + sin^2(theta) = 1 and divide both sides by cos^2. The cosines will divide to be 1, sin^2/cos^2 is the Ratio identity of tangent(theta), and 1/cos^2 is the Reciprocal identity of sec(theta). So we end up with 1 + tan^2(theta) = sec^2(theta).

    

To find the third Pythagorean identity that has cotangent and co-secant, it will be just like the second one except that I will divide it by sine^2. If I divide both side by sin^2 to both sides then cos^2/ sin^2 will be the Ratio identity of cotangent, sin^2/ sin^2 will divide to 1, and 1/ sin^2 is the Reciprocal identity of csc(theta).

INQUIRY ACTIVITY REFLECTION: 
1) The connections that I see between Unit N, O, P, and Q so far are the angles that can be found in the quadrants of the Unit Circle and how the triangles that are made in the quadrants can be used with the Law of Sine And the Law of Cosine to find any missing angle or side length.
2) If I had to describe trigonometry in THREE words, they will be hard, understandable, and progressive.

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

This WPP 13 & 14 was made in collaboration with Leo Escutia. Please visit the other awesome post on their blog by going here.
  

     One day Robin decided to go fishing but out at sea. He knew that his long time friend was fisher too and he had a boat. So his friend Gabriel took his to a good fishing spot off the coast. Hours passed but when they decided to head back, the engine failed. They radioed for help and two Coast Guard station responded back. They needed their exact location but the boat was acting weird and couldn't give the exact distance. Station B is due south of Station A and they are 100 miles apart. From what they can tell, the boat is S 45 degrees W of Station A, and N 65 degrees W of Station B. How far is each station to the boat?

Here we have it drawn to show how the triangle is suppose to look like. This a ASA problem since we give two angles and one side in between them.  Since we give two angles, the third missing side will be70 degrees.

We used the Law of Sine to solve for side "a". Angle C, which is where the boat is, will be our bridge to solve for side "a' and side "b". Sin 70/ 100 = Sin 45/ a, we cross multiply to give us a(sin 70) = 100(sin 450). We divide sin 70 to both sides to cancel sin 70 on one side so that it will be a = 100(sin 45)/ sin 70. Station A is 75.2 miles away.

Like I said, sin 70/ 100 is our bridge and it will be used to fine side "b". So sin 70/100 = sin 65/b, we cross multiply to give us b(sin 70) = 100(sin 65). We divide sin 70 to both sides so that sin 70 cancels on one side so that we can have b = 100(sin 65)/ sin 70. Station B is 96.4 miles away from the boat.

     Once they got rescued and taken back to Station A they are taken back home. They both leave the station at the same time, they diverge an angle of 100 degrees. If Robin is 4 miles away from the station and Gabriel is 5.5 miles away too, then how far away are they from each other?

With it's picture and work done together, we will use the Law of Cosine to find the missing side. It will be a^2 = 4^2 + 5.5^2 - 2(4)(5.5) cos 100. We just plug that in to our calculator and they are 7.3 miles apart.

BQ #1: Unit P Concepts 1 and 4: Law of Sines and Area of Obliques

Concept 1: LAW OF SINES
     We need the Law of Sines when are working on a triangle that is not a right triangle. When we have that we use the Law of Sines to solve it but only if the the triangle is an AAS or an ASA. How to derive the Law of Sines will be shown in these following pictures:

If we are given a triangle the A,B, and C as their angles and a, b, and c as their sides we can just split the triangle in half to form two triangles.

 

Now if we just focus on angle A, angle C, side a, side c and the height, h, I can show you how they can equal each other. If we take the sine of angle A, it will be h/c but since this is a part of a triangle we don't know what h is. So c will be multiplied to both side and it will give us h = c(sin of angle A). The same thing will happen if we do the sine of C, except it's h/a and we multiply a to both side and we'll get h= a(sine of angle C). We can get the height in two way but in the end the height will be the same for the both of them. Since c(sine of angle A) = a(sine of angle C), and we cross multiply, then ( sine of angle A)/a =  (sine of angle C/c.

If we stay with the same angles and side but with a different height then I can show how sine of B can be the same as the other two.

In this version of the triangle we'll use the sine of angle B and it is k/c but we don't know what k is so we multiply c to both side and we'll get k= c(sine of angle B ). On the other triangle it's the sine of angle C and that's k/ b. Just like the other one we multiply b to both side and get k= b(sine of angle C). But if  c(sine of angle B ) = k and b(sine of angle C)= k then they are both the same. Which mean that if we cross multiply them then (sine of angle B )/b = (sine of angle C)/c and if that is true then (sine of angle A)/a will be the same and all three will work and we'll get the same answer.

Concept 4: AREA OF OBLIQUES.
     The "area of obliques" is derived from the area of a triangle which is A = 1/2 bh. The base is b and the height is h in a right triangle. But if we don't have a right triangle then it will almost be the same but that depends on the side and angle given to us. If we have angle A, side b, and side c given to us then the area formula will be A = 1/2 b(c(sine of angle A). The formula can be rewritten to work with the two other angle. The following picture will show which formula will work for that problem.