course mth 164 Mr smith can you send me a a way to identify the query files by more than the location on the assignment page. I have found it hard to remember and think that I have completed all up to assignment 4 but am not sure.Thanks Tony Complete Assignment 2
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22:19:50 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 --> ok self critique assessment: 3
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22:23:24 `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 --> 0 1/4 1 1/4 0 -1/4 -1 -1/4 0 confidence assessment: 2
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22:24:08 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 --> estimated the y valus not gave the exact values self critique assessment: 3
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22:30:02 `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 --> 0 1/2 sqrt3/2 1 sqrt3/2 1/2 0 -1/2 sqrt3/2 -1 -sqrt3/2 -1/2 0 confidence assessment: 2
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22:30:19 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 --> ok self critique assessment: 3
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22:41:14 `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 --> this graph shows that the multiples of pi/6 and area plotted on the unit circle with a radius of 1and it completes the circle confidence assessment: 3
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22:41:33 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 --> ok self critique assessment: 2
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22:45:44 `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 --> because the unit circle makes a circle circular set of values. confidence assessment: 3
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22:48:59 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 --> ok self critique assessment: 3
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22:59:41 `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 --> 1/2 -1/2 sqrt2/2 -sqrt3/2 1/2 -sqrt3/2 confidence assessment: 3
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22:59:52 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 --> ok self critique assessment: 3
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23:01:26 `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 --> yes because the x and y coordinates correspond with eachother at the point on the unit circle confidence assessment: 2
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23:01:55 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 --> ok self critique assessment: 3
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23:05:11 `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 --> pi/6=1/2 pi/4=sqrt2/2 pi/3=sqrt3/2 confidence assessment: 2
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23:05:46 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 --> didnt double values to find sines self critique assessment: 3
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23:06:31 `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 --> this simply increases the min and max of the sin function graph confidence assessment: 2
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23:06:55 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 --> ok self critique assessment: 3
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23:08:27 `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 --> 0*3=0 pi/6*3=pi/3 pi/3*3=pi/6 pi/2*3=-3 confidence assessment: 1
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23:08:44 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 --> ok self critique assessment: 3
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23:13:11 `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 t sin pi/6 1 1/2 pi/6 3 pi/3 confidence assessment: 1
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23:13:59 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 --> now understand that the denominator is devided by the multiplier self critique assessment: 3
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23:14:46 `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 --> the x values change but the circle still completes itself confidence assessment: 2
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23:15:00 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 --> ok self critique assessment: 3
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23:16:35 `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 --> x=1/y*3 confidence assessment: 1
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23:17:29 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 --> i guess that multplying by the inverse states in the previous result would be the same self critique assessment: 3
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23:22:22 `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 --> theta would have to equal 0,pi/6,2pi/3,pi/3 confidence assessment: 2
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23:22:33 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 --> ok self critique assessment: 3
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23:23:43 `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 --> already completed? confidence assessment: 0
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23:23:54 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 --> ok self critique assessment: 3
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23:24:11 `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 does complete the circle confidence assessment: 2
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23:24:17 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 --> ok self critique assessment: 3
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Deconstruct the given solution and explain in detail what you do and do not understand about every part.. &#