Identities Test Correction and Unit Reflection

I seriously struggled with this unit. I never quite got it. I can do things with formulas much better in math than proofs.

5) tan^2x/secx-1=secx+1 - Prove

1) sin^x/cos^2x/                                               1) Put the left side of the equation in terms of sine and cosine
1/cosx-1

2) sin^2x/cos^2x/                                                                                       2) Change 1 to (cosx/cosx)
1/cosx-(cosx/cosx)

3)sin^2x/cos^2x/                                 3) Combine the numerators because there is a common denominator
1-cosx/cosx

4) sin^2x/cox^2x*cosx/1-cosx                                                                 4) Multiply by the reciprocal

5) sin^2x/                                                                                                    5) Multiplication done out
cosx(1-cosx)

6) 1-cos^2x/                                                         6) Change sin^2x to 1-cos^2x using Pythagorean identity 
cosx(1-cosx)

7) (1+cosx)(1-cosx)/                                                                                      7) Factor out 1-cos^2x
(cosx)(1-cosx)

8) 1+cosx/                                                                                                           8)Cancel out 1-cosx
cosx

9) 1/cosx+cosx/cosx                                                                                    9) Separate the numerator

secx+1





8) 1/cotx+tanx=sinxcosx   -Prove

1) 1/((cosx/sinx)+(sinx/cosx))                                                     1) Put in terms of sine and cosine

2) 1/ ((cos^2x/(sinxcosx))+sin^2x/(sinxcosx))                           2) Make common denominators

3) 1/ (cos^2x+sin^2x)/(sinxcosx)                                              3) Combine numerators

4) 1/1/(sinxcosx)                                                                       4) Simplify

sinxcosx

Vectors Test Corrections and Unit Reflection

I understood this unit fairly well. The component method made more sense to me than the law of cosines method, though I could use the law of cosines when I had to. Sometimes with the component method I wasn't sure which angle to solve for. On the test, the problem I struggled with (number 2) was hard for me because I couldn't figure out how to draw the picture. The pictures are important for me.

2a) Use simple trig to calculate the speed of the current. You know the angle is 34 degrees and the hypotenuse is 4 km/hr. You need to calculate the opposite side, so you use sine.
sin34=x/4
x=2.24 km/hr.

2c) Here, you know that the adjacent side to the angle you need is 4 km/hr and the opposite side is 2.24. In order to find the angle, you can use tangent.
tan2.24/4=
29.25 degrees

Parametric equations test corrections

Overall I felt that I understood this unit more than most. The book problems were a lot easier than the problems given in class, such as the Ferris wheel problem and the basketball problem. One thing I could not get was how to get the right window when graphing on the calculator.

Test corrections:
2a) I wrote my equations as x=350t, y=t and x=475(t-1.5)+250, y=t, but my y values should have equaled 1 and 2.

3f) the 45 degree angle would get the most distance (44.7 yd). I got this part but I did not state that it landed 5.3 yd from the end zone.

4) when writing the components, in the vertical component I wrote 70 twice (70cos0+70sin30), but it should have been 90sin30.

Inverse Trig Functions Test Corrections

5a)
My equation was y=3sin18(x+10).
I ignored the vertical shift, which is supposed to be +4.
I also only calculated half the period. The period should be 40, making it 3sin9, instead of 18.
I also made it +10 when it should be -10.

2b)
Cos-1(sin90)
Cos-1(1)
the only time this happens is at 0 degrees because the Cos indicates only principle values. I included all values.

I am pretty good at memorizing the unit circle values and evaluating compound functions. I can also graph equations fairly well. At the time of the unit, I struggled with providing an equation given a graph. After the subsequent units working with trig equations, I am able to do this more easily.

Identities and Trig Equations Test Corrections

6b)
5sinx^2-3=0
sin^2x=3/5
(I messed up here; when I took the square root of both sides I didn't take the +/- square root)
sinx=+/-.77

6c)
cos2x+sinx=1
1-sin^2x+sinx-1=0
sinx(sinx+1)=0   (I messed up on the factoring here, I made it sinx(x+1))
sinx=0            sinx=-1
0,180               270


I found this unit to be difficult. I am much better at algebra than at "puzzles." If there are equations that need to be memorized and applied to different situations, I will succeed. If there is no concrete way to do it, I have a difficult time. This is not so for me in many other areas of life. In English, in history, in music, in sports...I can think "outside of the box." Math and science are weak for me, so if there is not, basically, step by step instructions I can know in my head and follow, I struggle.

Simple Harmonic Motion Test Corrections

I got one point off for calculating the vertical shift incorrectly for number 12. I should have averaged the max and min to get 59.1. In this case, the average for February should have been 36.67 degrees. On the last problem, I lost one point because I didn't assume that February 15 was the middle of the month. I should have subtracted 1/30 of 36.67 from 36.67, in which case the temperature on that day should have been, according to the model, 35.45 degrees.

I enjoyed this unit because it has many real life applications. Many of the word problems we did could be applied to real life. I need things to be more concrete in math like this section was. It was very algebraically based, which I am good at.

Graphs of Trig Functions Test Corrections

5c) I originally had the equation of this graph as 3sin15x-8. The correct answer is 3sin30x+9. My answer was wrong because when calculating the period, I only calculated half of a period. My vertical shift was wrong because I counted wrong.

5d) I originally had the equation of this graph as 2sec1/2(x-2). I don't know how to find the equation of secant graphs, so I got this pretty much completely wrong. The correct answer is -3csc(x/2). For this problem, it is easier to use cosecant. The amplitude of -3 can be found by plugging in points. The b value is 1/2 because the period is 4pi, or 2 times the normal period of 2pi.

I struggled with this section. I could find the graphs of sine and cosine fairly well, but tangent, cotangent, secant and cosecant were difficult. I never really learned algebraically how to find these graphs. I didn't really get how to plug in points to find an equation.