#$&* course MTH 164 Question: `q001. The tangent function is defined in terms of the unit circle (a circle of radius 1 centered at the origin):
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Given Solution: `aThe unit-circle points corresponding to the given angles are (1,0), (sqrt(3)/2, 1/2), (sqrt(2)/2, sqrt(2)/2) and (1/2, sqrt(3)/2). So the values of the tangent, each calculated as y / x, are as follows: tan(0) = 0/1 = 0, tan(pi/6) = 1/2 / (sqrt(3)/2) = 1/2 * 2/sqrt(3) = 1/sqrt(3) = sqrt(3)/3, tan(pi/4) = sqrt(2)/2 / (sqrt(2)/2) = 1 and tan(pi/3) = sqrt(3)/2 / (1/2) = sqrt(3). The corresponding graph is shown in Figure 73. The slopes of the graph between 0 and pi/6, between pi/6 and pi/4, and between pi/4 and pi/3 are, respectively: (sqrt(3)/3 - 0)/ (pi/6 - 0) = 6 sqrt(3) / pi = 1.10, (1 - sqrt(3)/3)/(pi/4-pi/6) = 1.62 and (sqrt(3) - 1)/(pi/3 - pi/4) = 2.80. The slopes are increasing, slowly at first, then more quickly.  &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*Ok ********************************************* Question: `q002. Figure 52 indicates the unit circle positions corresponding to the angles which are multiples of pi/18. The grid shows intervals of .5. Estimate the x coordinates of the first-quadrant points then make a table of tangent (theta) vs. theta for the data from 0 to 8 pi/18. Give your values and use them to extend the graph of tan(theta) through theta = 8 pi/18.  YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: we can factor some of the values down into things we already know like pi/6 and pi/3 to get some values. So using this we start at the pi/18 and go up in value of the tan like so: .18, .36, .58, .84, 1.19, 1.7, 2.7, 5.7. confidence rating #$&*3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe angles 3 pi/18 = pi/6 and 6 pi/18 = pi/3 are known to be approximately sqrt(3)/3 = .58 and sqrt(3) = 1.73, respectively. The values at the remaining points can be estimated more or less accurately. The actual values, to 2 significant figures, strarting with angle pi/18, are .18, .36, .58, .84, 1.19, 1.7, 2.7 and 5.7. The rapidly increasing slopes as theta exceeds pi/3 are apparent from these numbers. The extended graph shown in Figure 97 depicts only the results for angles through 7 pi/18; the figure would have to be twice as high to include the point (8 pi/18, 5.7), or in reduced for (4 pi/9, 5.6). The colored line segments just below the x axis indicate the multiples of pi/18.  &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*OK ********************************************* Question: `q003. Sketch a series of points on the unit circle approaching the pi/2 position. As we approach closer and closer to pi/2, what happens to the y coordinate? What happens to the x coordinate? Does the y coordinate approach a limiting value? Does the x coordinate approach a limiting value? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I looked at my unit circle and it is clear that when we get closer to pi/2 the y will get closer to 1 and the x will get closer to 0. Neither of these are limiting values. confidence rating #$&*3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aFigure 48 shows a series of points on the unit circle approaching the pi/2 position. It should be clear that the y coordinate approaches 1 and the x coordinate approaches 0.  &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*Ok ********************************************* Question: `q004. As the angle approaches pi/2 from the first quadrant the y coordinate approaches 1 and the x coordinate approaches 0, as we saw in the last problem. What happens to the ratio 1/x as x take values 0.1, 0.01, 0.001, 0.0001? What happens as x continues to approach 0? Is there a limit to how large 1/x can get? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: we see the ratio go up by tenth 10, 100, 1000, and 10,000. As x continues to approach 0 it will continue to get larger and can continue as long as we are dividing by a smaller number. confidence rating #$&*3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIf x = 0.1, 0.01, 0.001 and 0.0001 we see that 1/x = 10, 100, 1000 and 10,000. There is no limit to how large 1/x can get; we can make it as large as we wish by choosing x small enough. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):Ok ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q005. What happens to the tangent of theta as theta approaches pi/2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: tan(theta) = y/x so as x approaches 0 and y approaches 1 the number will continue to get larger and larger so it is going toward infinity. confidence rating #$&*3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSince the x coordinate approaches zero and the y coordinate approaches 1, it follows that tan(theta) = y / x gets larger and larger, without bound. We say that this quantity approaches infinity. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*Ok ********************************************* Question: `q006. As we have seen the value of tangent (theta) exceeds all bounds as theta approaches pi/2 from within the first quadrant. How do we extend the graph of tangent (theta) vs. theta to cover the domain 0 <= theta <= pi/2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: since tan is y/x at pi/2 we will have no solution because we are going to be dividing by 0 therefore we will see an asymptote at pi/2. confidence rating #$&*3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aPreceding solutions and graphs showed how the slopes of the graph increased as we moved from theta = 0 to theta = 8 pi / 18 (which reduces to 4 pi / 9, just a little less than pi/2). The slopes will continue to increase, and the ratio y / x will increase without bound so that the values of the function increase without bound as theta approaches pi / 2. The graph therefore forms a vertical asymptote at theta = pi/2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*Ok ********************************************* Question: `q007. What are the values of tan(theta) as theta changes from -8 pi/18 to 0 in increments of pi/18? Use your estimates of the coordinates of the first-quadrant points as a basis for your estimates of the values of the tangent function. Extend your graph of the function so that it is graphed for -8 pi / 18 <= theta <= 8 pi / 18, then show what happens as theta approaches -pi/2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We have already determined the values for 0 to pi/18 so those number would just be negative in this case so we would have: -.18, -.36, -.58, -.84, -1.19, -1.7, -2.7, -5.7 we also saw that at pi/2 a vertical asymptote was there so there will also be one at -pi/2 confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aFigure 71 shows the multiples of pi/18 from -8 pi/18 to 0. It is clear that for all values of theta the x coordinates are positive and for all values of theta except 0 the y coordinates negative, which make the values of tan(theta) = y / x negative. Otherwise the values are the same as those obtained for theta = 0 to 8 pi/18. The actual values, to 2 significant figures, starting with angle -pi/18, are -.18, -.36, -.58, -.84, -1.19, -1.7, -2.7 and -5.7. The graph is shown in Figure 8. Note how the graph now approaches the vertical line theta = -pi/2 as an asymptote. The values of y = tan(theta) for -pi/2 < theta < pi/2 combines the graph just obtained with the graph obtained in the preceding exercise, and is shown in Figure 97. Note that the graph as shown is terminated at y values exceeding | tan( 7 pi / 18) |. fig 8  fig 97 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*Ok ********************************************* Question: `q008. What happens to the value of the tangent function as the angle approaches pi/2 through second-quadrant angles? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: tan = y/x and since we are in the second quadrant the x is negative so then too will the tan. Also since x will be approaching zero the tan will be getting more negative. confidence rating #$&*3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIn the second quadrant, as we approach theta = pi/2 on the circle the x coordinates are negative as the approach 0 while the y value approaches +1. Thus tan(theta) = y / x will be negative, with the magnitude of the quantity approaching infinity. We say that tan(theta) approaches -infinity. The graph will be asymptotic to the negative portion of the line theta = pi/2, as indicated in Figure 69, which combines this portion of the graph with the existing graph of the full cycle of the tangent function. Note that the graph as shown is terminated at y values exceeding | tan( 7 pi / 18) |.  &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*Ok ********************************************* Question: `q009. What happens to the value of the tangent function as the angle approaches 3 pi/2 through third-quadrant angles? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The x values are negative and are getting closer to 0 and the y values are also negative getting closer to 1. Therefore tan will be a positive value going toward infinity. confidence rating #$&*3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIn the third quadrant, as we approach theta = 3 pi/2 on the circle the x coordinates are negative as the approach 0 while the y value approaches -1. Thus tan(theta) = y / x will be positive, with the magnitude of the quantity approaching infinity. We see that tan(theta) approaches +infinity. The graph will be asymptotic to the positive portion of the line theta = 3 pi/2, as in Figure 30. Note that the graph as shown is terminated at y values exceeding | tan( 7 pi / 18) |.  &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*Ok ********************************************* Question: `q010. Sketch a graph of the tangent function from theta = -pi/2 through theta = 3 pi / 2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: we will see on the graph that there are two asymptotes at -pi/2 and pi/2 and then we will be approaching infinity as stated in the previous questions in other places. confidence rating #$&*2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe graph is shown in Figure 30. Note how the tangent function goes through its complete cycle of values, from -infinity to +infinity, between theta = -pi/2 and theta = pi/2, then again between theta = pi/2 and theta = 3 pi/2. The horizontal 'distance' corresponding to a complete cycle is thus seen to be pi.  &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): couldve described more than just the asymptotes. ------------------------------------------------ Self-critique rating #$&*Ok ********************************************* Question: `q011. Sketch a graph of y = tan ( 2 x ), showing two complete cycles. What is the length of a cycle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: After drawing the graph we can look and see that one cycle will have a length of pi/2. confidence rating #$&*3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aWe begin by sketching a graph of y = tan(theta) for -pi/2 <= theta <= 3 pi/2, as shown in Figure 30. We then let theta = 2x and relabel the graph as indicated in Figure 79, with pi/2 <= theta <= 3 pi/2 replaced by -pi/2 <= 2x <= 3 pi/2 so that we have -pi/4 <= x <= 3 pi/4. This shows us that x goes through a complete cycle between -pi/4 and pi/4 and again between pi/4 and 3 pi/4. Each cycle has length pi/2. Note how the period of the function tan(2x) is 1/2 the period of the tan(x) function, which is pi.  &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did not give the numbers that added up to be the pi/2 that were the answer but it is clear to see on the graph. ------------------------------------------------ Self-critique rating #$&*Ok ********************************************* Question: `q011. Sketch a graph of y = tan ( 2 x + pi/3 ), showing two complete cycles. What is the length of a cycle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: After drawing the graph and looking at the values we can see that each cycle has a length of pi/2 as well. confidence rating #$&*3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aWe begin by sketching a graph of y = tan(theta) for -pi/2 <= theta <= 3 pi/2, as shown in Figure 30. We then let theta = 2x + pi/3 and relabel the graph as indicated in Figure 62, with pi/2 <= theta <= 3 pi/2 replaced by -pi/2 <= 2x + pi/3 <= 3 pi/2 so that we have -5 pi/12 <= x <= 7 pi/12. This graph shows us that x goes through a complete cycle between -5 pi/12 and pi/12 and again between pi/12 and 7 pi/12. Each cycle has length pi/2. Note how the period of the function tan(2x + pi/3) is 1/2 the period of the tan(x) function, which is pi. "