course MTH 164 2/7/10 11:59 a.m. If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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00:05:16 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: all these terms = 1/4 this would be 1/4 + 1/4 + 1/4 = 3/4 confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** the cosine is an even function, with f(-a) = f(a) so if f(a) = 1/4, f(-a) = 1/4. The idea of periodicity is that f(a+2`pi) = f(a), and the same for f(a-2`pi). Since f(a) = 1/4, all these terms are 1/4 and f(a) + f(a+2`pi) + f(a - 2 `pi) = 1/4 + 1/4 + 1/4 = 3/4. It is helpful to visualize the situation on a unit circle. If a is the angular position on the unit circle, then at angular position -a the x coordinate will be the same. So the cosine of a and of -a is the same. Angular positions a + 2 `pi and a - 2 `pi put you at the same location on the circle, since 2 pi corresponds to one complete revolution. So any trigonometric function will be the same at a, a + 2 pi and a - 2 pi. **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* 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 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Every time you go around the circle you still end up at the same points with the same (x,y) values. The angles may change but the coordinates stay the same confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ********************************************* Question:** the circular model demonstrates the periodic nature of the trigonometric functions because if you go all the way around the circle you end up at the same point, giving you the same values of the trigonometric functions, even though in going around an additional time the angle has changed by 2 `pi. This is the case no matter how many times you go around. Every time the angle changes by 2 `pi you find yourself at the same point with the same values. **
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00:13:54
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* 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 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The sine (y) in quadrant one or two will be positive, and negative in quadrant 3 and 4 so clockwise we end up negative first The cosine (x) will be positive in quadrant 1 negative in quadrant 2 and 3 positive again in quadrant 4 confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Question:** The answer can be pictured in terms of 2 ants, one going counterclockwise and the other clockwise. The cosine is the x coordinate of the reference point. Since we start at the positive x axis, it doesn't matter whether we go clockwise or counterclockwise through the given angular distance, we end up with the same x coordinate. The sine function being the y coordinate, clockwise motion takes us first to negative values of the sine while counterclockwise motion takes us first to positive values of the sine. Thus the sine is odd. **
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00:16:36
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* 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 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: pi/4 you have (sqrt2/2) and 1 which are 45deg. pi/6 you have (sqrt3/2, 1/2 and 1) which are 60 deg. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Multiples of pi/6 give you magnitudes 0, 1/2, sqrt(3)/2 = .87 approx., and 1. It is clear from a decent sketch which gives you which, and when the result is positive and when negative. Multiples of pi/4 given you magnitudes 0, sqrt(2)/2 = .71 and 1 approx., and again a good sketch makes it clear which is which. **
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00:17:54
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* **** 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.