assignment 2 query

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course Mth 164

IIAsst # 2

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Question: **** query ch. 5 # 102 f(x) = cos(x), f(a) =1/4, find f(-a), f(a) + f(a+2`pi) + f(a - 2 `pi)

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00:05:16

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Your solution:

Cos and sec are functions so f(-a) = f(a). f(-a)=¼. ¼ + (¼ + 2pi) +(¼ - 2pi)….the 2 pi’s cancel leaving the equation equal to ¾.

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Given Solution:

** the cosine is an even function, with f(-a) = f(a) so if f(a) = 1/4, f(-a) = 1/4.

The idea of periodicity is that f(a+2`pi) = f(a), and the same for f(a-2`pi). Since f(a) = 1/4, all these terms are 1/4 and f(a) + f(a+2`pi) + f(a - 2 `pi) = 1/4 + 1/4 + 1/4 = 3/4.

It is helpful to visualize the situation on a unit circle. If a is the angular position on the unit circle, then at angular position -a the x coordinate will be the same. So the cosine of a and of -a is the same.

Angular positions a + 2 `pi and a - 2 `pi put you at the same location on the circle, since 2 pi corresponds to one complete revolution. So any trigonometric function will be the same at a, a + 2 pi and a - 2 pi. **

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00:05:18

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Question: query (no summary needed) **** How does the circular model demonstrate the periodic nature of the trigonometric functions? Be specific.

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00:13:53

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Your solution:

Don’t clearly understand this question.

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Given Solution:

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Question:** the circular model demonstrates the periodic nature of the trigonometric functions because if you go all the way around the circle you end up at the same point, giving you the same values of the trigonometric functions, even though in going around an additional time the angle has changed by 2 `pi.

This is the case no matter how many times you go around.

Every time the angle changes by 2 `pi you find yourself at the same point with the same values. **

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00:13:54

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Self-critique (if necessary):

Not sure I even understand the answer let alone the question. Help me understand what you are asking, please.

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Question: **** How does the circular model demonstrate the even or odd nature of the sine and cosine functions? Be specific.

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00:16:35

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Your solution:

The coordinates stay the same, whether they are positive or negative depends on which side of the axis they are located.

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Given Solution:

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Question:** The answer can be pictured in terms of 2 ants, one going counterclockwise and the other clockwise.

The cosine is the x coordinate of the reference point. Since we start at the positive x axis, it doesn't matter whether we go clockwise or counterclockwise through the given angular distance, we end up with the same x coordinate.

The sine function being the y coordinate, clockwise motion takes us first to negative values of the sine while counterclockwise motion takes us first to positive values of the sine. Thus the sine is odd. **

STUDENT COMMENT: Ok, by looking at a complete unit circle I can see what you are talking about. Sometimes I have trouble visualizing this.

INSTRUCTOR RESPONSE: Sketch it every time you need it; you will probably soon begin to visualizing it without the need of the sketch, but if not it's always fine to revert to sketching.<

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00:16:36

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Question: **** Can you very quickly sketch on a reference circle the angles which are multiples of `pi/6 and immediately list the sine and cosine of each?

Can you do the same for multiples of `pi/4?

(It's OK to answer honestly.

You should be prepared to have to do this on a test, and remember that this task is central to understanding the trigonometric functions; if you've reached this point without that skill you have already wasted a lot of time by not knowing something you need to know to do what you're trying to do).

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00:17:51

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Your solution:

I can sketch a reference circle with the angles of pi/6 and pi/4. The sin and cos is a little more difficult, but I can do it.

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Given Solution:

Multiples of pi/6 give you magnitudes 0, 1/2, sqrt(3)/2 = .87 approx., and 1. It is clear from a decent sketch which gives you which, and when the result is positive and when negative.

Multiples of pi/4 given you magnitudes 0, sqrt(2)/2 = .71 and 1 approx., and again a good sketch makes it clear which is which. **

STUDENT QUESTION: I am quickly realizing how important this is. While I can slowly work my way through the calculations, I think I might be

better off memorizing the coordinates. Is that the goal? Should I try to memorize my pi coordinates and the sine and cosine

of each?

INSTRUCTOR RESPONSE: If you work them out every time you need them, you'll soon be able to work them out quickly, and in the process you will not only memorize them but will understand how to quickly confirm them out any time your memory fails.

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**** Query Add comments on any surprises or insights you experienced

as a result of this assignment.

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00:23:45

i was suprised by the fact that this assignment was shorter than the other

yet somehow it seemed more complex.

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Miscellaneous comments, questions, etc.:

117. If f(theta) = sec(theta) and f(a) = -4, find the exact value of

A) f(-a) = f(-(-4)) = 4 b) f(a) + f(a+2pi) +(a+4pi) = -8.8584

****( I checked the answer in the back of the book and I know it is -12. I read in the book that you can ignore 2pi and multiples of 2pi because they represent complete revolutions, but I’m not sure I understand why.

Think of the unit circle. If you start at a point and go completely around the circle you end up where you started.

A complete trip around the circle corresponds to a change of 2 pi radians in angular position.

n complete trips around the circle would correspond to a change of n * 2 pi = 2 pi n radians in angular position.

If f(theta) is a function whose value is determined by the 'theta' position, i.e., angular position, on the unit circle, then if theta changes by 2 pi radians, or by 2 pi n radians (where n is an integer), your position on the circle won't change. The function f(theta) won't know the difference. So we can safely say that

f(theta) = f(theta + 2 pi n).

So if f(a) = -4, it follows that f(a+2pi) = -4 and f(a+4pi) = -4, and

f(a) + f(a+2pi) +(a+4pi) = -4 + -4 + -4 = -12.

123. Show that the period of f(theta) =sin(theta) is 2 pi

**** don’t really understand this - If f(0) = sin(0) +p then p= 0

If f(pi/2) = sin(pi/2) + p then p= -.0274

Not sure if this is right and if so I don’t understand what it means

sin(theta) is the y coordinate of the unit-circle point for position theta. So as we've seen in the qa exercises that preceded this text assignment,

sin(pi/6) = 1/2

sin(pi/4) = sqrt(2)/2

sin(pi/3) = sqrt(3)/2

sin(pi/2) = 1

sin(2 pi / 3) = sqrt(3) / 2

etc..

Adding 2 pi to theta doesn't change where you are on the unit circle. So it doesn't change the value of sin(theta).

Thus sin(theta) has period 2 pi.

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@& In relation to the first question:

If you go around the circle once, you hit every point of the circle once. Your x and y coordinates go through a complete cycle, ending up where they started.

If you go around the circle again, you repeat the same points. Your x and y coordinates repeat in the same sequence as before. The values of the cosine and sine functions therefore repeat in the same sequence as before.

If you keep going around the circle, the same points keep repeating, and the same values of the cosine and sine functions continue to repeat in the same sequence.

So we say that the sine and cosine functions are periodic. The period is the time required to go around the circle. if we go around the circle, say, three times per second then the sine and cosine functions will go through their complete set of sequential values three times per second; the period will be 1/3 of a second.

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@& Let me know if you have additional questions.*@