BQ #5: Unit T Concept 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 graphs will never have asymptotes and reason being is their unit circle ratio. The ratio for sine is y/r and for cosine it is x/r. In the unit circle we know that r will always equal 1. And we know that asymptotes are there when it is undefined which means there is a 0 as the denominator. But for sine and cosine according to the ratio it will always be over 1 keeping it from being undefined. All the other graphs may have asymptotes because their ratios are either over x or y. And it is possible for the x and y values to be 0. When any ratio is over 0 it becomes undefined creating an asymptote.

BQ #2 Unit T

How do the trig graphs relate to the unit circle?

The trig graphs relate to the unit circle with the use of the quadrants in the unit circle.For example sine is positive in quadrants 1 and 2 so the graph will be positive through pi since the end of the second quadrant is 180. As soon as the graph passes pi it becomes negative because that's where in quadrant 3 sine becomes negative. The graph will stay negative through quadrant 4, 2pi, because of the unit circle. Once the graph gets to 2 pi the graph as well as the unit circle start over. Since the pattern of this will begin again, the first whole time the graph goes through the unit circle is called a period. This is how the graph of a trig function comes from the unit circle.

Why is period for sine and cosine 2pi whereas period for tangent and contingent is pi?
This is so because sine and cosine have to go through the whole unit circle before it repeats itself again and to go around the whole unit circle is 360 degrees, or 2pi. Tangent and cotangent is only pi because in the unit circle their quadrants are positive negative positive negative. So within the first two quadrants tangent has gone through a period because after that it will already start repeating itself. The end of quadrant 2 would be 180 or pi. Leaving tangent and cotangent pi to become a period.

 How does the fact that sine and cosine have amplitudes of one (and the other trig functions don’t have amplitudes) relate to what we know about the Unit Circle?
Sine and cosine are the only trig functions that have amplitudes because these are the only ones that have restrictions. In the unit circle sine and cosine can't be greater than 1 our less than -1. All the other trig functions do not have restrictions therefore they do not contain amplitudes.

Reflection #1 - Unit Q

24. What does it actually mean to verify a trig identity?
        
      To verify a trig function is to make sure the equation is true by having both sides equal to each other. In order to do this we must not touch the right side of the equal sign. We must use identities to change the equation and find something to cancel out. Verifying a trig function is making both sides of equal sign look identical. 

25.What tips and tricks have you found helpful?

     A trick I have found helpful is looking for the cosine and sines to change into. Doing this makes it easier to change it into another function. It is also helpful to label the steps you are doing because it will help know where you are in the problem. And re-watch videos that you don't understand. Things make more sense when it's explained the second time. 

26.Explain your thought process and steps you take in verifying a trig identity. 

     My thought process looks for a trig function that can be changed in to either cosine or sine. If there is no possible way to do this then I will look for any other identity to replace the original function. If the function is complicated I will seek to split them apart such as (1/tanxsinx • secx/1). In this we split the original equation of secx/tanxsinx. After looking to split the functions I will see if there is any possible way to cancel anything out. This will lead to verifying the trig function. 

I/D3: Unit Q - Pythagorean Identities

Pythagorean Identities

We know that an identity is always true. So this makes the Pythagorean Theorem an identity, with a^2+b^2=c^2. We replace a,b, and c with x,y, and r making x^2+y^2=r^2. We need to make the theorem equal to 1. We do this by dividing everything by r. Our equation is now (x/r)^2+(y/r)^2=1. From our unit circle we know the ratio of cosine is x/r which means we can substitute cos for x/r. The ratio for sine from the unit circle is y/r and now we can plug in sin for y/r. Plugging both of these in gets us cos^2x+sin^2x=1 which is our main identity.

To derive the identity with secant and tangent from cos^2x+sin^2x=1 we must divide everything by cos^2. We now have sin^2x/cos^2x=1/cos^2x. From the Ratio identity we know tanx=sinx/cosx therefore we are able to plug in tan^2x for sin^2x/cos^2x. We are able to do this because the ratio identities are allowed to power up. As for the other side we know from the Recipricol identity that secx=1/cosx. The Reciprical identity is also allowed to power up, the only one not allowed to is the pythagorean identity. So now we are left with the final identity of 1+tan^2x=sec^2x.

To derive the identity with cosecant and cotangent we must divide everything by sin^2x. Giving us cos^2x/sin^2x+1=1/sin^2x. In the ratio identity cotx=cosx/sinx and even though the problem is squared the ratio identities are allowed to be powered up allowing it to be cot^2x to equal cos^2x/sin^2x. The reciprical identity says that cscx=1/sinx. This identity is also allowed to be powered up, so it equals csc^2x=1/sin^2x. The final equation is cot^2x+1=csc^2x. 

   INQUIRY ACTIVITY REFLECTION

The connections I see between units N, O, P, and Q are that they all use the trig functions in some way. They are also the same because it is all derived from the unit circle.

If I had to describe trigonometry in three words, they would be, simple once learned.

WPP #13 & 14: Unit P Concept 6 & 7

WPP 13 & 14 Unit P Concept 6&7

This WPP was made in collaboration with Daniel. Please visit the other awesome posts on their blog by going here.

Blake and Steve both see a ball. They get 3 miles apart east and west of each other to race to the ball. Blake runs at a bearing of 050* at 20 mph for 15 min to get to the ball. How far away is Steve away from the ball?

Law of Cosines

Once Blake gets to the ball he throws it at a bearing of N 25 W with a distance of .8 miles. When it lands Steve looks at the at a bearing of N 70 W. How far apart are Blake and Steve? 

BQ#1 - Unit P

BQ#1 - Unit P 

1. Law of Sines - Why do we need it?  How is it derived from what we already know?  

Here we have the non-right triangle and once we make a line straight down from B we get two right triangles and could use what we know with the trig functions.

To derive the Law of Sines we know that to get A we need to use Sine and get the opposite/hypotenuse giving us h/c. We multiply both sides by c to get h by itself. And now have cSinA=h. We must also do the same to get C. Taking the same steps as A, we will get aSinC=h. Since both equal h, this means they both equal each other, cSinA=aSinC. We now need to get SinA and SinC by themselves, so we divide both sides by ac. And c will cancel out on one side and a will cancel out on the other leaving us with SinA/a=SinC/c

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   4. Area formulas - How is the “area of an oblique” triangle derived?  How does it relate to the area formula that you are familiar with?
   
   We derive this formula because we first know the original equation is A=1/2bh. We dont know the h so to find this we must use the trig functions. We know that the SinC=h/a and to get h by itself we multiply each side by a. We now replace h with aSinC giving us A=1/2b(aSinC).
   
   
   This relates to the original area formula because we are using the exact same formula except replacing h because in order to find h we must use the trig function. And instead of just finding h and plugging it in, we plug in the trig function to find h in the original equation.

WPP #12: Unit O Concept 10

WPP #12: Unit O Concept 10

David is about to jog a a hill. The hill has a 25 degree elevations and is 46 feet high. What is the length he'd run to the top?

David gets to the top of the hill and takes a break. As he looks he sees the other side of the hill is much steeper with an angle depression of 32 degrees, he also sees a lone tree that is about 92 feet on the trail down. What is the length of the horizontal line from top of hill to the tree?