007. Tangent Function
Goals for this Assignment include but are not limited to the following:
1. Understand and be able to explain how the vertical asymptotes of the graph of the tangent function occur.
2. Using exact values construct y vs. x graphs of y = A sin(theta) or y = A cos(theta), where theta is given as a function of x.
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Previous Assignments: Be sure you have completed Assignment 6 as instructed under the Assts link on the homepage and submitted the result of the Query and q_a_ from that Assignment.
Question: `q001. The tangent function is defined in terms of the unit circle (a circle of radius 1 centered at the origin):
For an angular position theta the tangent of theta is the ratio y / x of the y coordinate to the x coordinate at the corresponding point on the circle.
What are the values of the tangent(theta) for theta = 0, pi/6, pi/4, and pi/3?
Sketch a graph of tan(theta) vs. theta for 0 <= theta <= pi/3.
Are the slopes of the graph increasing or decreasing.
Your solution:
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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.
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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.
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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.
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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?
Your solution:
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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.
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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?
Your solution:
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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.
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Question: `q005. What happens to the tangent of theta as theta approaches pi/2?
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`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.
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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?
Your solution:
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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.
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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.
Your solution:
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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
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Question: `q008. What happens to the value of the tangent function as the angle approaches pi/2 through second-quadrant angles?
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`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) |.
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Question: `q009. What happens to the value of the tangent function as the angle approaches 3 pi/2 through third-quadrant angles?
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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) |.
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Question: `q010. Sketch a graph of the tangent function from theta = -pi/2 through theta = 3 pi / 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.
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Question: `q011. Sketch a graph of y = tan ( 2 x ), showing two complete cycles. What is the length of a cycle?
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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.
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Question: `q011. Sketch a graph of y = tan ( 2 x + pi/3 ), showing two complete cycles. What is the length of a cycle?
Your solution:
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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.
Complete Assignments 6, 7, including Class Notes, text problems and Web-based problems as specified on the Assts page.
When you have completed the entire assignment run the Query program. Submit SEND files from Query and q_a_. Note that Assignment 7 consists of a test covering Assignments 1-5.