assignment 3

Excellent work. See my notes.

Let me know the specifics if anything is not clear.

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18:48:10 Previous Assignments: Be sure you have completed Assignment 1 as instructed under the Assts link on the homepage at 164.106.222.236 and submitted the result of the Query and q_a_ from that Assignment.

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RESPONSE --> Sine Questions

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18:54:28 `q001. Note that this assignment has 15 activities. Figure 93 shows the angular positions which are multiples of pi/4 superimposed on a grid indicating the scale relative to the x y coordinate system. Estimate the y coordinate of each of the points whose angular positions correspond to 0, pi/4, pi/2, 3 pi/4, pi, 5 pi/4, 3 pi/2, 7 pi/4, and 2 pi.

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RESPONSE --> sin 0 = 0 sin pi/4 = 0.71 using calculator sin pi/2 = 1 sin 3pi/4 = 0.71 sin pi = 0 sin 5pi/4 = -0.71 sin 3pi/2 = -1 sin 7pi/4 = -0.71 sin 2pi = 0

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18:55:51 The angular positions of the points coinciding with the positive and negative x axes all have y coordinate 0; these angles include 0, pi and 2 pi. At angular position pi/4 the point of circle appears to be close to (.7,.7), perhaps a little beyond at (.71,.71) or even at (.72,.72). Any of these estimates would be reasonable. Note for reference that, to two decimal places the coordinate so are in fact (.71,.71). To 3 decimal places the coordinates are (.707, .707), and the completely accurate coordinates are (`sqrt(2)/2, `sqrt(2) / 2). The y coordinate of the pi/4 point is therefore .71. The y coordinate of the 3 pi/4 point is the same, while the y coordinates of the 5 pi/4 and 7 pi/4 points are -.71.

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RESPONSE --> okay

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19:00:30 `q002. Figure 31 shows the angular positions which are multiples of pi/6 superimposed on a grid indicating the scale relative to the x y coordinate system. Estimate the y coordinate of each of the points whose angular positions correspond to 0, pi/6, pi/3, pi/2, 2 pi/3, 5 pi/6, pi, 7 pi/6, 4 pi/3, 3 pi/2, 5 pi/3, 11 pi/6 and 2 pi.

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RESPONSE --> sin 0 = 0 sin pi/6 = 0.5 sin pi/3 = .87 sin pi/2 = 1 sin 2pi/3 = .87 sin 5pi/6 = 0.5 sin pi = 0 sin 7pi/6 = -.05 sin 4pi/3 = -.87 sin 3pi/2 = -1 sin 5pi/3 = -.87 sin 11pi/6 = -0.5 sin 2pi = 0

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19:01:59 The angular positions of the points coinciding with the positive and negative x axes all have y coordinate 0; these angles include 0, pi and 2 pi. At angular position pi/6 the point on the circle appears to be close to (.9,.5); the x coordinate is actually a bit less than .9, perhaps .87, so perhaps the coordinates of the point are (.87, .5). Any estimate close to these would be reasonable. Note for reference that the estimate (.87, .50) is indeed accurate to 2 decimal places. The completely accurate coordinates are (`sqrt(3)/2, 1/2). The y coordinate of the pi/6 point is therefore .5. The coordinates of the pi/3 point are (.5, .87), just the reverse of those of the pi/6 point; so the y coordinate of the pi/3 point is approximately .87. The 2 pi/3 point will also have y coordinate approximately .87, while the 4 pi/3 and 5 pi/3 points will have y coordinates approximately -.87. The 5 pi/6 point will have y coordinate .5, while the 7 pi/6 and 11 pi/6 points will have y coordinate -.5.

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RESPONSE --> okay

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19:17:39 `q003. Make a table of y coordinate vs. angular position for points which lie on the unit circle at angular positions theta which are multiples of pi/4 with 0 <= theta <= 2 pi (i.e., 0, pi/4, pi/2, 3 pi/4, pi, 5 pi/4, 3 pi/2, 7 pi/4, 2 pi). You may use 2-significant-figure approximations for this exercise. Sketch a graph of the y coordinate vs. angular position. Give your table and describe the graph.

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RESPONSE --> a wave of sort. --^v^v-- (0,0) (pi/4,0.70 to 0.73) (pi/2,1) (3pi/4,0.70 to 0.73) (pi, 0) (5pi/4,-0.70 to -0.73) (3pi/2,-1) (7pi/4,-0.70 to -0.73) (2pi, 1) I'm am uncertain, but if the copy picture button is to be used, I wasn't aware. Thus, I have constructed a text graphic.

The instruction was 'Give your table and describe the graph.' Emphasis here on the word 'describe'.

In any case it's clear that you understand.

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19:20:17 The table is theta y coordinate 0 0.0 pi/4 0.71 pi/2 1.0 3 pi/4 0.71 pi 0.0 5 pi/4 -0.71 3 pi/2 -1.0 7 pi/4 -0.71 2 pi 0.0. We note that the slope from (0,0) to (pi/4,.71) is greater than that from (pi/4,.71) to (pi/2,1), so is apparent that between (0,0) and (pi/2,1) the graph is increasing at a decreasing rate. We also observe that the maximum point occurs at (pi/2,1) and the minimum at (3 pi/2,-1). The graph starts at (0,0) where it has a positive slope and increases at a decrasing rate until it reaches the point (pi/2,1), at which the graph becomes for an instant horizontal and after which the graph begins decreasing at an increasing rate until it passes through the theta-axis at (pi, 0). It continues decreasing but now at a decreasing rate until reaching the point (3 pi/2,1), for the graph becomes for an instant horizontal. The graph then begins increasing at an increasing rate until it again reaches the theta-axis at (2 pi, 0).

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RESPONSE --> oh, table... I was a little to concerned with the line graph. still, the coordinates should match approximately.

Your format had the information, but table form with columns is generally easier on the reader. No problem here, though.

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19:22:39 `q004. In terms of the motion of the point on the unit circle, why is it that the graph between theta = 0 and pi/2 increases? Why is that that the graph increases at a decreasing rate?

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RESPONSE --> taking a circle with the origin at (0,0), the y increases less as X approaches the origin. At the origin, Y is at it's max and min.

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19:23:29 As we move along the unit circle from the theta = 0 to the theta = pi/2 position it is clear that the y coordinate increases from 0 to 1. At the beginning of this motion the arc of the circle take us mostly in the y direction, so that the y coordinate changes quickly. However by the time we get near the theta = pi/2 position at the 'top' of the circle the arc is carrying us mostly in the x direction, with very little change in y. If we continue this reasoning we see why as we move through the second quadrant from theta = pi/2 to theta = pi the y coordinate decreases slowly at first then more and more rapidly, reflecting the way the graph decreases at an increasing rate. Then as we move through the third quadrant from theta = pi to theta = 3 pi/2 the y coordinate continues decreasing, but at a decreasing rate until we reach the minimum y = -1 at theta = 3 pi/2 before begin beginning to increase an increasing rate as we move through the fourth quadrant. If you did not get this answer, and if you did not draw a sketch of the circle and trace the motion around the circle, then you should do so now and do your best to understand the explanation in terms of your picture. You should also document in the notes whether you have understood this explanation.

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RESPONSE --> check

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19:34:56 `q005. The table and graph of the preceding problems describe the sine function between = 0 and theta = pi/2. The sine function can be defined as follows: The sine of the angle theta is the y coordinate of the point lying at angular position theta on a unit circle centered at the origin. We write y = sin(theta) to indicate the value of this function at angular position theta. Make note also of the definition of the cosine function: The cosine of the angle theta is the x coordinate of the point lying at angular position theta on a unit circle centered at the origin. We write x = cos(theta) to indicate the value of this function at angular position theta. We can also the line tangent function to be tan(theta) = y / x. Since for the unit circle sin(theta) and cos(theta) are respectively y and x, it should be clear that tan(theta) = sin(theta) / cos(theta). Give the following values: sin(pi/6), sin(11 pi/6), sin(3 pi/4), sin(4 pi/3), cos(pi/3), cos(7 pi/6).

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RESPONSE --> sin pi/6 = 0.5 sin 11pi/6 = -0.5 sin 3pi/4 = 0.71 sin 4pi/3 = -.85 cos pi/3 = 0.5 cos 7pi/6 = -0.87

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19:35:42 sin(pi/6) is the y coordinate on the unit circle of the point at the pi/6 position. We have seen that this coordinate is .5. sin(11 pi/6) is the y coordinate on the unit circle of the point at the 11 pi/6 position, which lies in the fourth quadrant and in angular displacement of pi/6 below the positive x-axis. We have seen that this coordinate is -.5. sin(3 pi/4) is the y coordinate on the unit circle of the point at the 3 pi/4 position. We have seen that this coordinate is .71. sin(4 pi/3) is the y coordinate on the unit circle of the point at the 4 pi/3 position, which lies in the third quadrant at angle pi/3 beyond the negative x axis. We have seen that this coordinate is -.87. cos(pi/3) is the x coordinate on the unit circle of the point at the pi/3 position, which lies in the first quadrant at angle pi/3 above the negative x axis. We have seen that this coordinate is .5. cos(7 pi/6) is the y coordinate on the unit circle of the point at the 7 pi/6 position, which lies in the third quadrant at angle pi/6 beyond the negative x axis. We have seen that this coordinate is -.87.

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RESPONSE --> check

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19:52:14 `q006. Suppose that the angle theta is equal to 2 x and that y = sine (theta). Given values of theta correspond to x = pi/6, pi/3, ..., pi, give the corresponding values of y = sin(theta). Sketch a graph of y vs. x. Not y vs. theta but y vs. x. Do you think your graph is accurate?

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RESPONSE --> y = sin(2x) x = pi/6 to pi when X = pi/6 Y = 0.87 when X = pi/3 y = 0.87 when X = pi/2 y = 0 when X = 2pi/3 y = -.87 when X = 5pi/6 y = -.87 when X = pi y = 0 This does make sense pi/2 x 2 = pi and 2pi x 2 = 4pi it would seem that 1 is no longer an option. additionaly, the wave length is shorter.

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19:55:28 The angles are in increments of pi/6, so we have angles pi/6, pi/3, pi/2, 2 pi/3, 5 pi/6 and pi. If x = pi/6, then 2x = 2 * pi/6 = pi/3. If x = pi/3, then 2x = 2 * pi/3 = 2 pi/3. If x = pi/2, then 2x = 2 * pi/2 = pi. If x = 2 pi/3, then 2x = 2 * 2 pi/3 = 4 pi/3. If x = 5 pi/6, then 2x = 2 * 5 pi/6 = 5 pi/3. If x = pi, then 2x = 2 * pi/6 = 2 pi. The values of sin(2x) are therefore sin(pi/3) = .87 sin(2 pi/3) = .87 sin(pi) = 0 sin(4 pi/3) = -.87 sin(5 pi/3) = -.87 sin(2 pi) = 0. We can summarize this in a table as follows: x 2x sin(2x) 0 0 0.0 pi/6 pi/3 0.87 pi/3 2 pi/3 0.87 pi/2 pi 0 2 pi/3 4 pi/3 -0.87 5 pi/6 5 pi/3 -0.87 0 2 pi 0.0. Figures 93 and 77 depict the graphs of y = sin(theta) vs. theta and y = sin(2x) vs. x. Note also that the graph of y = sin(2x) continues through another complete cycle as x goes from 0 to 2 pi; the incremental x coordinates pi/4 and 3 pi / 4 are labeled for the first complete cycle.

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RESPONSE --> so 1 is an option, but the wave length did shorten. it would seem that at pi/4 is 1 and it would be because 2(pi)/4 = pi/2

It's just that with the x values chosen, we didn't hit the 1's.

Lesson: Gotta be careful what x values you choose. As we'll see below.

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20:09:53 `q007. Now suppose that x = pi/12, 2 pi/12, 3 pi/12, 4 pi/12, etc.. Give the reduced form of each of these x values. Given x = pi/12, pi/6, pi/4, pi/3, 5 pi/6, ... what are the corresponding values of y = sin(2x)?

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RESPONSE --> ( X ) (2X) (sin (2X)) pi/12 pi/6 0.5 pi/6 pi/3 0.87 pi/4 pi/2 1 pi/3 2pi/3 0.87 5pi/12 5pi/6 0.5 6pi/12 pi 0 7pi/12 7pi/6 -0.5 and the negative matching values.

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20:10:26 pi / 12 doesn't reduce. 2 pi/12 reduces to pi/6. 3 pi/12 reduces to pi/4. 4 pi/12 reduces to pi/3. 5 pi/12 doesn't reduce. 6 pi/12 reduces to pi/2. 7 pi/12 doesn't reduce 8 pi/12 reduces to 2 pi/3 9 pi/12 reduces to 3 pi/4 10 pi/12 reduces to 5 pi/6 11 pi/12 doesn't reduce 12 pi/12 reduces to pi Doubling these values and taking the sines we obtain the following table: x 2x sin(2x) 0 0 0.0 pi / 12 pi/6 0.5 pi/6 pi/3 0.87 pi/4 pi/2 1.0 pi/3 2 pi/3 0.87 5 pi/12 5 pi/6 0.5 pi/2 pi 0.0 7 pi/12 7 pi/6 -0.5 2 pi/3 4 pi/3 -0.87 3 pi/4 3 pi/2 -1.0 5 pi/6 5 pi/3 -0.87 11 pi/12 11 pi/6 -0.5 pi/2 pi -0.0

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RESPONSE --> check

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20:24:13 `q008. Given the table of values obtained in the preceding problem, sketch a graph of y vs. x. Describe your graph. By how much does x change as the function sin(2x) goes through its complete cycle, and how does this compare with a graph of y = sin(x)?

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RESPONSE --> pi/12 = pi/6 = 0.5 3pi/12 = pi/2 = 1 6pi/12 = pi = 0 9pi/12 = 3pi/2 = 1 for every 6pi/12 it is 0 or 1 wave length seems shorter

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20:26:04 Your graph should pass through the origin (0,0) with a positive slope. It will have a peak at x = pi/4, pass back through the x-axis at x = pi/2, reach a minimum at x = 3 pi/4 and return to the x-axis at x = pi. More detail: The graph of y vs. x passes through the origin (0,0) with a positive slope and increases at a decreasing rate until it reaches a maximum value at the x = pi/4 point (pi/4,1). The graph is horizontal for an instant and then begins decreasing at an increasing rate, again reaching the x-axis at the x = pi/2 point (pi/2,0). The graph continues decreasing, but now at a decreasing rate until a reaches its minimum value at the x = 3 pi/4 point (3 pi/4,0). The graph is horizontal for an instant then begins increasing an increasing rate, finally reaching the point (pi, 0). The graph goes through its complete cycle as x goes from 0 to pi. A graph of y = sin(x), by contrast, would go through a complete cycle as x changes from 0 to 2 pi. We see that placing the 2 in front of x has caused the graph to go through its cycle twice as fast. Note that the values at multiples of the function at 0, pi/4, pi/2, 3 pi/4 and 2 pi are clearly seen on the graph. Note in Figure 3 how the increments of pi/12 are labeled between 0 and pi/4. You should complete the labeling of the remaining points on your sketch.

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RESPONSE --> check

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20:32:28 `q009. Now consider the function y = sin(theta) = sin(3x). What values must x take so that theta = 3x can take the values 0, pi/6, pi/3, pi/2, ... ?

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RESPONSE --> x=pi/3 is equal to pi (0) x=pi/18 is equal to pi/6 (.5) x=pi/9 is equal to pi/3 (.87) x=pi/6 is equal to pi/2 (1) exc...

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20:33:00 If theta = 3x then x = theta / 3. So if theta = 3x takes values 0, pi/6, pi/3, pi/2, 2 pi/3, 5 pi/6, pi, 7 pi/6, 2 pi/3, 3 pi/2, 5 pi/3, 11 pi/6 and 2 pi then x takes values 0, 1/3 * pi/6, 1/3 * pi/3, 1/3 * pi/2, 1/3 * 2 pi/3, 1/3 * 5 pi/6, 1/3 * pi, 1/3 * 7 pi/6, 1/3 * 2 pi/3, 1/3 * 3 pi/2, 1/3 * 5 pi/3, 1/3 * 11 pi/6, 1/3 * 2 pi, or 0, pi/18, pi/9, pi/6, 2 pi/9, 5 pi/18, pi/3, 7 pi/18, 4 pi/9, pi/2, 5 pi/9, 11 pi/18, and 2 pi/3.

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RESPONSE --> check

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11:34:15 `q010. Make a table consisting of 3 columns, one for x, one for theta and one for sin(theta). Fill in the column for theta with the values 0, pi/6, pi/3, ... which are multiples of pi/6, for 0 <= theta <= 2 pi. Fill in the column under sin(theta) with the corresponding values of the sine function. Now change the heading of the theta column to 'theta = sin(3x)' and the heading of the sin(theta) column to 'sin(theta) = sin(3x)'. Fill in the x column with those values of x which correspond to the second-column values of theta = 3x. The give the first, fifth and seventh rows of your table.

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RESPONSE --> ( X ) Theta = sin(3x) sin(Theta) = sin(3x) 0 0 0 pi/18 pi/6 0.5 pi/9 pi/3 .87 pi/6 pi/2 1 2pi/9 2pi/3 0.87 5pi/18 5pi/6 0.5 pi/3 pi 0 7pi/18 7pi/6 -0.5 4pi/9 4pi/3 -0.87 pi/2 3pi/2 -1 5pi/9 5pi/3 -0.87 11pi/18 11pi/6 -0.5 2pi/3 2pi 0

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11:34:35 The table originally reads as follows: x theta sin(theta) 0 0 0.0 0 pi/6 0.5 0 pi/3 0.87 0 pi/2 1.0 0 2 pi/3 0.87 0 5 pi/6 0.5 0 pi 0.0 0 7 pi/6 -0.5 0 4 pi/3 -0.87 0 3 pi/2 -1.0 0 5 pi/3 -0.87 0 11 pi/6 -0.5 0 2 pi -0.0 After inserting the values for x and changing column headings the table is x theta = 3x sin(3x) 0 0 0.0 pi/18 pi/6 0.5 pi/9 pi/3 0.87 pi/6 pi/2 1.0 2 pi/9 2 pi/3 0.87 5 pi/18 5 pi/6 0.5 pi/3 pi 0.0 7 pi/18 7 pi/6 -0.5 4 pi/9 4 pi/3 -0.87 pi/2 3 pi/2 -1.0 5 pi/9 5 pi/3 -0.87 11 pi/18 11 pi/6 -0.5 2 pi/3 2 pi -0.0

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RESPONSE --> check

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11:41:21 `q011. Sketch the graph corresponding to your table for sin(3x) vs. x. Does the sine function go through a complete cycle? By how much does x change as the sine function goes through its complete cycle?

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RESPONSE --> yes, 2pi/3

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11:41:49 Your graph should pass through the origin (0,0) with a positive slope. It will have a peak at x = pi/6, pass back through the x-axis at x = pi/3, reach a minimum at x = pi/2 and return to the x-axis at x = 2 pi/3. More detail: The graph of y vs. x passes through the origin (0,0) with a positive slope and increases at a decreasing rate until it reaches a maximum value at the x = pi/6 point (pi/6,1). The graph is horizontal for an instant and then begins decreasing at an increasing rate, again reaching the x-axis at the x = pi/3 point (pi/3,0). The graph continues decreasing, but now at a decreasing rate until a reaches its minimum value at the x = pi/2 point (pi/2,0). The graph is horizontal for an instant then begins increasing an increasing rate, finally reaching the point (2 pi/3, 0). The graph goes through its complete cycle as x goes from 0 to 2 pi/3. A graph of y = sin(x), by contrast, would go through a complete cycle as x changes from 0 to 2 pi. We see that placing the 3 in front of x has caused the graph to go through its cycle three times as fast.

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RESPONSE --> check

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11:44:58 `q012. For the function y = sin(3x), what inequality in the variable x corresponds to the inequality 0 <= theta <= 2 pi?

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RESPONSE --> 0 to 2pi/3 peak at pi/6

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11:45:50 If theta = 3x then the inequality 0 <= theta <= 2 pi becomes 0 <=3x <= 2 pi. If we multiply through by 1/3 we have 1/3 * 0 <= 1/3 * 3x <= 1/3 * 2 pi, or 0 <= x <= 2 pi/3. In the preceding problem our graph when through a complete cycle between x = 0 and x = 2 pi/3. This precisely correspond to the inequality we just obtained.

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RESPONSE --> check

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12:50:41 `q013. For y = sin(theta) = sin(2x - 2 pi/3), what values must x take so that theta = 2x - pi/3 will take the values 0, pi/6, pi/3, pi/2, ... ?

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RESPONSE --> (2x - 2 pi/3) I am not sure if something happened with the question. x = pi/3 = 0 x = 3pi/4 = pi/6 x = pi/2 = pi/3 x = 7pi/12 = pi/2 x = 2pi/3 = pi/3

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12:51:59 If theta = 2x - pi/3 then 2 x = theta + 2 pi/3 and x = theta/2 + pi/6. So if theta = 2x - pi/3 takes values 0, pi/6, pi/3, pi/2, 2 pi/3, 5 pi/6, pi, 7 pi/6, 2 pi/3, 3 pi/2, 5 pi/3, 11 pi/6 and 2 pi then x = theta/2 + pi/6 takes values x values: 0 + pi/6, pi/12 + pi/6, pi/3 + pi/6, pi/4 + pi/6,pi/3 + pi/6, 5 pi/12 + pi/6, pi/2 + pi/6, 7 pi/12 + pi/6,2 pi/6 + pi/6,3 pi/4 + pi/6,5 pi/6 + pi/6,11 pi/12 + pi/6,pi + pi/6, which are added in the usual manner and reduce to added and reduced x values: pi/6, pi/3, 5 pi/12, pi/2, 7 pi/12, 2 pi/3, 3 pi/4, 5 pi/6, 11 pi/12, pi, 13 pi/12, 7 pi/6.

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RESPONSE --> even though I was working with a different equation (I'm not sure how it happened) I still went through the same solving process.

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12:57:12 `q014. Make a table consisting of 3 columns, one for x, one for theta and one for sin(theta). Fill in the column for theta with the values 0, pi/6, pi/3, ... which are multiples of pi/6, for 0 <= theta <= 2 pi. Fill in the column under sin(theta) with the corresponding values of the sine function. Now change the heading of the theta column to 'theta = sin(2x - pi/3)' and the heading of the sin(theta) column to 'sin(theta) = sin(2x - pi/3)'. Fill in the x column with those values of x which correspond to the second-column values of theta = 2x - pi/3. Give the first, fifth and seventh rows of your table.

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RESPONSE --> x =0 y=-.87 x = 2pi/3 0 x = y=-.87

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12:59:39 Our first table is the same as before, as it will always be: x theta sin(theta) 0 0.0 pi/6 0.5 pi/3 0.87 pi/2 1.0 2 pi/3 0.87 5 pi/6 0.5 pi 0.0 7 pi/6 -0.5 4 pi/3 -0.87 3 pi/2 -1.0 5 pi/3 -0.87 11 pi/6 -0.5 2 pi -0.0 Our second table is as follows: x theta = 2x - pi/3 sin(2x-pi/3) pi/6 0 0.0 3 pi/12 pi/6 0.5 pi/3 pi/3 0.87 5 pi/12 pi/2 1.0 pi/2 2 pi/3 0.87 7 pi/12 5 pi/6 0.5 2 pi/3 pi 0.0 3 pi/4 7 pi/6 -0.5 5 pi/6 4 pi/3 -0.87 11 pi/12 3 pi/2 -1.0 pi 5 pi/3 -0.87 13 pi/12 11 pi/6 -0.5 7 pi/6 2 pi -0.0 The second table indicates that the function y = sin(2x - pi/3) goes through a complete cycle for x values running from pi/3 to 5 pi/3, with y running from 0 to 1 to 0 to -1 to 0.

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RESPONSE --> oh, I was setting x=0 and increasing by pi/6

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13:02:30 `q015. Sketch the graph corresponding to your table for sin(2x - pi/3) vs. x. Does the sine function go through a complete cycle? By how much does x change as the sine function goes through its complete cycle? For the function y = sin(2x - pi/3), what inequality in the variable x corresponds to the inequality 0 <= theta <= 2 pi?

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RESPONSE --> yes it goes through a complete cycle twice as fast, but it is shifted right by pi/3

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13:03:37 If theta = 2x - pi/3 then the inequality 0 <= theta <= 2 pi becomes 0 <=2x - pi/3 <= 2 pi. If we add pi/3 to both sides we get pi/3 <= 2x <= 2 pi + pi/3. If we then multiply through by 1/2 we have 1/2 * pi/3 <= 1/2 * 2x <= 1/2 * 2 pi + 1/2 * pi/3, or pi/6 <= x <= 7 pi/6. In the preceding problem our graph when through a complete cycle between x = pi/6 and x = 7 pi/6. This precisely corresponds to the inequality we just obtained. A graph of y = sin(2x - pi/3) vs. x is shown in Figure 43. This graph goes through its cycle in an x 'distance' of pi, between pi/6 and 7 pi/6. In this it is similar to the graph of y = sin(2x), which also requires an x 'distance' of pi. It differs in that the graph is 'shifted' pi/6 units to the right of that graph.

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RESPONSE --> 2pi/6 = pi/3 check

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13:04:00 Complete Assignment 2 Includes Class Notes #3 (Class Notes are accessed under the Lectures button at the top of the page and are included on the CDs starting with CD #1). Text Section 5.2 and Section 5.3, 'Blue' Problems (i.e., problems whose numbers are highlighted in blue) and odd multiples of 3 in text and the Web version of Ch 5 Problems Section 5.2 and 5.3 (use the link in the Assts page to access the problems). When you have completed the entire assignment run the Query program. Submit SEND files from Query and q_a_.

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RESPONSE --> check

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