DEV Project

1. Verify the identity.

Sin(x)(cos(x)+1)/cos(x)=(sin(x)*csc(x))(tan(x)+sin(x))

-First, distribute sin(x) to cos(x)+1.

(cos(x)*sin(x)+sin(x))/cos(x)=((x)*csc(x))(tan(x)+(x))

-Separate the addition being done to (Cos(x)*Sin(x)+Sin(x))/Cos(x).

(Cos(x)Sin(x))/Cos(x)+Sin(x))/Cos(x)=(Sin(x)*Csc(x))*(Tan(x)+Sin(x))

-Then multiply (Sin(x)*Csc(x)) together and see that they multiply to equal 1.

(Cos(x)Sin(x))/Cos(x))+(Sin(x)/Cos(x))=Tan(x)+Sin(x)

- In the (Cos(x)Sin(x))/Cos(x))+(Sin(x)/Cos(x)) section, have the first cosines divide out.

Sin(x)+(Sin(x)/Cos(x))=Tan(x)+Sin(x)

-Turn the (Sin(x)/Cos(x)) into Tan(x).

Sin(x)+Tan(x)=Tan(x)+Sin(x).


2.The time on an analog clock is 7:54. Find the angle made from the minute hand, clockwise to the hour hand.

-From the 11 to the 7th hour on the clock there are 8 ∏/6 's, which is just 8∏/6.

-From the minute hand to the 11 there is  1/5th of a cycle. To turn the 1/5 into radians you have to multiply it by ∏/6 to get ∏/30.

-From the hour hand to the 7 there are 9/10ths of a cycle and when multiplied by ∏/6 it equals 9∏/60.

-To get the total distance between the minute hand, clockwise to the hour hand we need to add all of these distances together which would be (8∏/6)+(2∏/60)+(9∏/60)= 91∏/60

-91∏/60 is the final answer.

3. Convert 76∏/12 to degrees and find its co-terminal angles.

In order to first convert this angle to degrees, multiply it by 180/∏ since there are 180 degrees in 1∏.

(76∏/12)*(180/∏)

The pies make a 1, and you can instead multiply 76*15 to make it easier since 180/12=15.

76*15=1140 degrees.

Now that we have converted the radian measure to degrees, we now will find the coterminal angles of this measure on the unit circle. Since there are 360 degrees in the unit circle, we both add and subtract 360 from 1140 to find the angles that are the same on the unit circle, but have different degrees
because they are in different cycles.

1140+360=1500 degrees
1140-360=780 degrees

You have now found two coterminal angles of 1140 degrees.

4. The highest point of a ferris wheel is 471 feet off of the ground. The loading point is 13 feet off of the ground. It takes 3 minutes and 30 seconds to make one revolution. Find an equation that represents height as a function of time in seconds.

In order to write the equation, we will need to find the amplitude, the vertical shift, and the b value.

To find the amplitude/wavelength, we need to take how high the bottom of the ferris wheel is off the ground and subract it from the highest point of the ferris wheel.

471-13=457 ft.

This will give us the height of the ferris wheel. Then, since the amplitude is half the height of the ferris wheel, we divide the height of the ferris wheel by 2.

457/2=228.5 ft.
amplitude=228.5 ft.

Now we need to find the vertical shift, which is how high the point of origin is off of the ground. Since the origin is half of the ferris wheel added to how high the bottom of the ferris wheel is off the ground, we will just plug in the numbers.

amplitude/half of ferris wheel: 228.5 ft.
height of bottom of ferris wheel off the ground/ loading point: 123 ft.
228.5+123=241.5ft
vertical shift/k value= 241.5 ft.

Finally, we need to find the b value. The equation to find the b value is 2∏/period, and the period in this case is 3 minutes and 30 seconds, which is the time of one revolution. First, we need to convert minutes into seconds.

3 minutes and 30 seconds= 210 seconds

2∏/ 210= ∏/105

Now that we have all of the necessary components, it's time to write the equation that represents height as a function of time.

y=aCos(b+h)+k

Now, just plug in the numbers.

y=241.5-228.5Cos∏/105t


5. Find the tan(105 degrees)

First, you have to find the individual tangents of two known angles on the unit circle that add to 105. For this problem, 60+45 is the easiest way. Use the tangent addition identity equation and plug in the numbers.

tan(60+45)= (tan60+tan45)/(1-tan60*tan45)

Next, replace the tan60 and tan45 with the actual values, calculated using Y/X.

tan(60+45)= (√3+1)/(1-√3*1)

Then, rationalize the denominator by using the special form of 1.

=(√3+1)/(1-√3)*(1+√3)/(1+√3)

=(3+1+2√3)/(1-3)

After that, finish solving the equation.

=(4+2√3)/(-2)

We can factor out a 2 in the numerator, since 2 goes into both 4 and 2√3.

= 2(2+√3)/(-2)

The 2's then divide out, leaving a negative 1 in the numerator. This gives us the final answer of:

-(2+√3)


6. Simplify the identity.

sec^5(x)+1/(cosx+sin^24x)+5/(cos^2x)

=1/(cos^5x)+1/(cosx+sin^24x)+5/(cos^2x)

First, we need to get common denominators. Unfortunately in this problem, the common denominator ends up being a combination of all 3 denominators, which would be (cos^5x)(cosx+sin^24x)(cos^2x). So to find this, you have to multiply the numerator and denominator by the terms that aren't there to get all 3 in the denominator, which in turn also multiplies those terms by the numerator. It gets a little messy.

So the new equation is:

(cosx+sin^24x)(cos^2x)/(cos^5x)(cosx+sin^24x)(cos^2x) + (cos^2x)(cos^5x)/(cos^5x)(cosx+sin^24x)(cos^2x) + 5(cos^x+sin^24x)(cos^5x)/(cos^5x)(cosx+sin^24x)(cos^2x)

Super messy, but now we can add the numerators together. But, before this, we have to distribute the terms in the numerators to each other to get the terms outside of the parentheses.

(cosx+sin^24x)(cos^2x) = cos^3x+sin^24x*cosx

(cos^2x)(cos^5x) = cos^7x

5(cosx+sin^24x)=(5cosx+5sin^24x)(cos^3x)=5cos^4x+5sin^24x*cos^3x

So, now we have terms to add together without parentheses.

(cos^3x+sin^24x+cosx+cos^7x+5cos^4x+5sin^24x*cos^3x)/(cos^5x)(cosx+sin^24x)(cos^2x)

Notice there are both (cos^2x) and (cos^5x) in the denominator, which can be combined to become (cos^7x).

This new (cos^7x) can be divided with the (cos^3x) in the numerator, becoming (cos^4x) in the
denominator, since both are being multiplied.

So, the final simplified equation becomes: (really messy, but about as far as you could easily go)

(cos^3x+sin^24x+cosx+cos^7x+5cos^4x+5sin^24x)/(cos^4x)(cosx+sin^24x)


Ryan's reflection: We chose these problems for the D.E.V. project because we believe they best represent the units we've learned this trimester. We chose a few problems, such as #6 and the ferris wheel to challenge ourselves and to really see how well we understood everything. I think that working through these problems with Tyler and Edmund made me understand the content more than ever before. It helped with prep for the exam and my future calculus endeavors. We tried to use problems that were a little bit of everything from the trimester. Not only will it help me with math related problems in the future, it also helps to gain experience working through problems with a group of people, which is almost guaranteed to happen sometime in life. What I learned is that even when you challenge yourself and the problems seem impossible, time is the most valuable resource. If you take time to work through extremely difficult problems, you can achieve the correct answer. This was my 2nd DEV and overall another great experience for me.

Tyler's reflection: Creating problems for the D.E.V. project has made me look back at earlier tests and homework assignments, giving me a chance to look back at how to do certain problems to get me ready for the upcoming exam. When working through some of the identity problems, I had to figure out how to start from the end and change it into the beginning of the problem which helped me figure out different ways to verify identities.We chose these problems because it took a little section out of every unit we worked with this trimester and used problems that accurately represented our knowledge of these units.These problems accurately provide an overview of my best mathematical understanding of what I've learned so far by how complex we tried to make them while still being able to work with. What I learned from this project is that i knew more about these units than i really
thought and that this will really help my understanding when I go to take the exam.

Edmunds reflection: Our entire group learned a lot from this trimester's dev project. We came up with problems from all three units we have learned, and made sure the problems we picked reflected the main ideas of each unit. For example, we did a convert radians to degrees problem as well as a clock problem. We did  these problems because the conversion helps you to better understand the unit circle while the clock problem requires previous understanding of the unit circle to complete. One of the problems we took from unit 2 was a Ferris wheel problem, because in order to write the equation you would need to have an understanding of trigonometric equations. We picked a verifying identity problem for unit three because it combined all previous problems we had to solve in that unit. I learned a lot from this project, because this tri I attempted to explain the problems like I was actually teaching, as opposed to just writing out steps. My mathematical understanding of many concepts improved because of this project.