Query_02 

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course Mth 164

1/28 11:13

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)......!!!!!!!!...................................

00:05:16

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Your solution:

F(-a) = 1/4

F(a) + f(a+2pi) + f(a - 2pi) = ¼ + ¼ + ¼ = ¾

The answer to this ¾ because f(a+2pi) = f(a) and also f(a - 2pi) = f(a)

confidence rating #$&*:2

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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. **

@& The first thing you need to understand is that the cosine is an even function and the sine is an odd function, as can be seen from their graphs. 
You can also see this from the unit circle. If you start from the positive x axis and move through some arc clockwise, while someone else moves through the same arc counterclockwise, you will both end up at the same x coordinate and opposite y coordinates (if this isn't clear ask yourself what would happen if you both moved through an arc corresponding to angle pi/6; you would end up in the first quadrant and your counterpart would end up in the fourth; your x coordinates would be the same and your y coordinates would be equal and opposite).
Since the x coordinate is the cosine of the angle, this shows that the cosine of your angle (corresponding to the x coordinate) would be equal to the cosine of the negative angle, while tje sine of your angle (correpsonding to the y coordinate) would be equal but opposite in sigh to the sine of the negative angle.
This is seen in the fact that the graph of the cosine function to the left of the y axis is a perfect reflection of the graph to the right of the y axis, while the graph of the sine function to the left of the y axis is a reflection through the origin (or a negative reflection) of the graph to the right of the y axis.*@


STUDENT QUESTION:

I’m not so sure I did this correctly. I can see that I answered 0.75 in decimal form and it was correct but I think I’m missing the overall method and reasoning.

INSTRUCTOR RESPONSE:


@& Another way of saying this is that cos(-theta) = - cos(theta), whille sin(-theta) = - sin(theta).
The current function being the cosine, we see that f(-a) = f(a).
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@& Both the sine and the cosine repeat for every cycle of 2 pi (if we go through 2 pi radians we've gone through the entire unit circle, and we're back where we started). So f(a + 2 pi) = f(a). It's also easy to see that the same thing happens if we go backwards through 2 pi radians, so f(a - 2 pi ) = f(a).
So:
Since the cosine is even, f(-a) = f(a).
Since the cosine has period 2 pi, f(a) = f(a+2pi) = f(a-2pi) = 1/4, and f(a) + f(a+2pi) + f(a - 2 pi ) = 1/4 + 1/4 + 1/4 = 3/4.
*@

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00:05:18

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Self-critique (if necessary):

Cosine is for positive values and my X values and Sine is for my negative values and Y values.

Also I did read ahead on this question because the question kind of confused me. How do we really know that f(a) is equal to (a+2`pi) and f(a - 2 `pi). Just still a little confused on this even after reading the solution.

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Self-critique Rating:3

@&

On the unit circle, the a position is the same as the a + 2 pi position, since adding 2 pi to any position on the circle takes you all the way around the circle to where you started. So starting at the a position and going through 2 pi radians, you end up back at the a position.

Similarly with a - 2 pi, You go a full circle in the negative direction, but you still end up where you started, at the a position.

Be sure you understand this. Submit a Question Form if you don't.

*@

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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:

The circular model demonstrate the periodic nature by showing you that if you are traveling in a circle you will always pass over the same points kind of like a car tire. No new points on made on the tire when it completes the circle once. The tire just keeps over lapping the same points but with higher values.

@&

Right. And to reinforce my comments on the preceding problem, once on every revolution you will hit the a radian position.

*@

confidence rating #$&*:3

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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):

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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 circular model demonstrates the even or odd nature of sine and cosine by moving clockwise or counterclockwise. When take the cosine of a degree or radian you will always get and even number whether you move clockwise or counterclockwise. It does however matter when taking the sine of a degree or radian because if you counterclockwise and take the sine you will get a negative number until you reach 180 degrees or pi. If you move clockwise when you take the sines of numbers you will get positive numbers.

confidence rating #$&*:3

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Given Solution:

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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Self-critique (if necessary):

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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:

Yes but I don’t know how fast.

To calculate the radians draw the unit circle and the degrees. One you have your degrees now start getting your radians.

Examples

Need the radian for 30 degree

Take 30/360 and reduce the fraction to 1/12

Now take 1/12 and multiply it by 2pi to get 2pi/12

Now reduce 2pi/12 to get your answer of pi/6

confidence rating #$&*:3

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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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Self-critique (if necessary):

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Self-critique Rating:

**** 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= 0If 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) = 1/2sin(pi/4) = sqrt(2)/2sin(pi/3) = sqrt(3)/2sin(pi/2) = 1sin(2 pi / 3) = sqrt(3) / 2etc..

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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Self-critique (if necessary):

This assignment may have been shorter than others but I think this is an important part of the class and you should spend some of your own time remembering and know how to make the unit circle correctly.

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&#Your work looks good. See my notes. Let me know if you have any questions. &#