Assignment 2b

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course mth 164

9/19 4pm

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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Your solution:

I am a little confused on what you are asking for, do you mean pi/6, pi/3, pi/2, 2pi/3, 5pi/6, pi, 7pi/6, 4pi/3, 3pi/2, 5pi/3, 11pi/6, 2pi. Are these the multiples of pi/6 ? I don’t understand what magnitudes are, where does 0, ½, sqrt(3)/2, and 1 come from are they the sine and cosine of pi/6.

confidence rating #$&*: 0

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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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The magnitude of a number is its absolute value.

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You have listed the multiplies of pi/6.

On a unit circle the multiples of pi/6 give you x and y components which have values +- 1, +- sqrt(3) / 2, +- 1/2 and 0. It should be clear that when the point on the circle is above the x axis the sine has positive values, and when below the sine has negative values. It should also be clear that when the point is to the right of the y axis the cosine has positive values and the sine has negative values.

Assuming that you are able to see whether a given value is positive or negative, it is sufficient just to specify that the magnitudes of the values will always be 0, 1/2, sqrt(3) / 2 or 1.

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