Section 1:
DAY ONE: Examples 1  4 (27 minutes) 

Lecture 1 

08:03 

Example 1: Trigonometric Coordinate Grids  Label all tick marks in the following grids and state the coordinates of each point.
 What you see here is called a trigonometric coordinate grid. A trigonometric coordinate grid can be used to graph trigonometric functions.
 The xaxis is used to express angles, so we can label tick marks as exactvalue radians.
 The independent variable in trigonometric functions is often θ, so the xaxis is called the θaxis.
 The yaxis is measured in regular numbers.
 Before we can find the coordinates of the points, we need to express each tick mark on the xaxis as an exactvalue radian.
 The first thing we need to do is count the number of ticks to the first labeled tick. In this case, π is the 6th tick.
 π is being split into six equal intervals. Divide π by 6 to get π/6 radians, or 30°. This means the ticks are spaced π/6 radians apart.
 Now that we know the ticks are spaced π/6 radians apart, we can label each tick. You may wish to use the unit circle for help in labeling the radians.
 Finally, label the rest of the ticks.
 Now label the coordinates of each point.
 The first point is located at (5π/6, 3).
 The second point is located at (π/6, 4).
 The third point is located at (7π/6, 1).
 Now we'll move on to Part B.
 We need to label the tick marks on the xaxis.
 Count the number of ticks to the first labeled tick. In this case, π is the 4th tick.
 π is being split into four equal intervals. Divide π by 4 to get π/4 radians (45°). This means the ticks are spaced π/4 radians apart.
 Now that we know the ticks are spaced π/4 radians apart, we can label each tick. You may wish to use the unit circle for help in labeling the radians.
 Finally, label the rest of the ticks.
 On the yaxis, it takes 5 ticks to get to 20.
 20 ÷ 5 = 4, so the ticks are spaced 4 units apart.
 Now label the coordinates of each point.
 The first point is located at (3π/4, 12).
 The second point is located at (π/4, 16).
 The third point is located at (7π/4, 8).
 Now we'll move on to Part C.
 We need to label the tick marks on the xaxis.
 Count the number of ticks to the first labeled tick. In this case, 8π is the 4th tick.
 8π is being split into four equal intervals. Divide 8π by 4 to get 2π radians (360°). This means the ticks are spaced 2π radians apart.
 Now that we know the ticks are spaced 2π radians apart, we can label each tick.
 On the yaxis, it takes 3 ticks to get to 12.
 12 ÷ 3 = 4, so the ticks are spaced 4 units apart.
 Now label the coordinates of each point.
 The first point is located at (6π, 8).
 The second point is located at (2π, 8).
 The third point is located at (4π, 4).
 Now we'll move on to Part D.
 We need to label the tick marks on the xaxis.
 Count the number of ticks to the first labeled tick. In this case, 4π is the 8th tick.
 π is being split into eight equal intervals. Divide 4π by 8 to get π/2 radians (90°). This means the ticks are spaced π/2 radians apart.
 Now that we know the ticks are spaced π/2 radians apart, we can label each tick. You may wish to use the unit circle for help in labeling the radians.
 On the yaxis, it takes 4 ticks to get to 40.
 40 ÷ 4 = 10, so the ticks are spaced 10 units apart.
 Now label the coordinates of each point.
 The first point is located at (3π, 10).
 The second point is located at (3π/2, 30).
 The third point is located at (5π/2, 20).


Lecture 2 

07:51 

Example 2: Graphing y = sinθ  In this example, we'll draw the graph of y = sinθ.
 For Examples 2 and 3, we are going to construct a unit circle grid to help us graph sinθ and cosθ.
 The xaxis is measured in angles, so it can be called the θaxis. The yaxis will use regular numbers.
 Draw grid lines on the θaxis every 15°.
 The unit circle doesn't have 15° or 75° triangles, so remove the corresponding grid lines.
 Unit circle coordinates include: 0, ±1/2, ±root(2)/2, ±root(3)/2, and ±1. Draw grid lines on the yaxis at these values.
 We have now completed the construction of the unit circle grid.
 Now we'll use the unit circle to draw the graph of y = sinθ.
 The ycoordinate is the value of sinθ.
 Plot the values of sinθ from the first quadrant.
 Plot the values of sinθ from the second quadrant.
 Plot the values of sinθ from the third quadrant.
 Plot the values of sinθ from the fourth quadrant.
 Draw a smooth curve through the points. This is the graph of y = sinθ.
 If we go around the unit circle clockwise, we can extend the sine curve to include negative angles.
 In Part B, we'll state the amplitude.
 The amplitude is the distance between the midline of a sinusoidal graph and its maximum or minimum. Note that the term sinusoidal means "sine or cosine". It is often used when describing properties that apply to both sine and cosine graphs.
 By inspecting the graph, we can see that the distance between the midline and maximum is 1, so the amplitude is 1 unit.
 We can also find amplitude using the formula: amplitude = (maxmin)/2.
 The maximum value of the graph is 1, and the minimum value of the graph is 1.
 Plug the maximum and minimum into the amplitude formula.
 This gives us 2/2, which becomes 1. This is the same amplitude we found by inspecting the graph earlier.
 In Part C, we'll state the period.
 The period of a trigonometric function is the horizontal length of one complete cycle.
 By inspection, we can see that one cycle has a period of 2π.
 In Part D, state the horizontal displacement.
 The horizontal displacement (also called the phase shift), is simply the distance a trigonometric graph has been moved left or right.
 The sine graph is currently in its original position, so there is no phase shift.
 In Part E, state the vertical displacement.
 The vertical displacement is the distance a trigonometric graph has been moved up or down from its original position.
 The midline of the graph is along the θ axis, so there is no vertical displacement.
 We can find the vertical displacement using the equation: d = (min+max)/2.
 The maximum value of the graph is 1, and the minimum value of the graph is 1.
 Plug the maximum and minimum into the vertical displacement formula.
 This gives us 0/2, which becomes 0. There is no vertical displacement.
 In Part F, state the θintercepts. Write your answer using a general form expression.
 The graph has θintercepts every π units. They include 2π, π, 0, π, 2π, extending forever in both directions.
 We can write this in general form as: θ = ±nπ, nεW.
 This means that if we start at 0 radians, we can add or subtract π to get the adjacent θintercept.
 In Part G, state the yintercept.
 The yintercept is 0.
 In Part H, state the domain and range.
 The graph goes left and right forever, and there are no restrictions on the variable. The domain is θ ε R.
 The graph has a minimum value at 1, and a maximum value at +1. The range is 1 ≤ y ≤ 1.


Lecture 3 

05:25 

Example 3: Graphing y = cosθ  In this example, we'll draw the graph of y = cosθ.
 Use the unit circle to draw the graph of y = cosθ.
 The xcoordinate is the value of cosθ.
 Plot the values of cosθ from the first quadrant.
 Plot the values of cosθ from the second quadrant.
 Plot the values of cosθ from the third quadrant.
 Plot the values of cosθ from the fourth quadrant.
 Draw a smooth curve through the points. This is the graph of y = cosθ.
 If we go around the unit circle clockwise, we can extend the cosine curve to include negative angles.
 In the graph of y = sinθ, the primary building block of the graph is the shape below, which looks like a sideways and rounded Z. In the graph of y = cosθ, the primary building block of the graph is the shape below, which looks like a rounded V.
 In Part B, we'll state the amplitude.
 The amplitude is the distance from the midline to the top or bottom of the graph.
 By inspection, we can see that the distance from the midline to the maximum or minimum is 1 unit. This is the amplitude.
 In Part C, we'll state the period.
 The period is the horizontal length of one complete cycle.
 By inspection, we can see that one cycle has a period of 2π.
 In Part D, state the horizontal displacement.
 The phase shift is the left or right displacement of the graph.
 The cosine graph is currently in its original position, so there is no phase shift.
 In Part E, state the vertical displacement.
 The vertical displacement is the distance the graph has been moved up or down.
 The midline of the graph is along the θ axis, so there is no vertical displacement.
 In Part F, state the θintercepts. Write your answer using a general form expression.
 The θintercepts occur at 3π/2, π/2, π/2, and 3π/2, extending forever in both directions.
 The general form is θ = π/2 ± n(π), nεW.
 This means that if we start at π/2 radians, we can add or subtract π to get the adjacent θintercept.
 In Part G, state the yintercept.
 The yintercept is 1.
 In Part H, state the domain and range.
 The graph goes left and right forever, and there are no restrictions on the variable. The domain is θ ε R.
 The graph has a minimum value at 1, and a maximum value at +1. The range is 1 ≤ y ≤ 1.


Lecture 4 

05:39 

Example 4: Graphing y = tanθ  In this example, we'll draw the graph of y = tanθ.
 The unit circle grid that we used to graph sinθ and cosθ does not work well for tanθ.
 Using the identity tanθ = sinθ/cosθ, we can get the values of tanθ from the unit circle.
 The exactvalues of tanθ include: 0, ±root(3)/3, ±1, ±root(3), and undefined.
 Reconstruct the unit circle grid, but change the yaxis gridlines so they match the tan values.
 Plot the values of tanθ from the first quadrant. Note that we have an asymptote when the value of tanθ is undefined.
 Plot the values of tanθ from the second quadrant.
 Plot the values of tanθ from the third quadrant. We have another asymptote, since tanθ is undefined.
 Plot the values of tanθ from the fourth quadrant.
 Draw a curve through the points. This is the graph of y = tanθ.
 We can extend the graph to negative angles if we go around the unit circle clockwise.
 Now we'll move on to part B. Is it correct to say a tangent graph has an amplitude?
 It is incorrect to say the graph of y = tanθ has an amplitude, since the graph has no top or bottom.
 In Part C, state the period.
 A tangent graph can be thought of as multiple copies of the curve shown.
 The period is the length of one of these shapes. Inspecting the graph, we can see that the period of the tan graph is π.
 In Part D, state the horizontal displacement.
 The phase shift is the distance a trigonometric graph has been moved left or right.
 The tan graph is in its original position, so there is no phase shift.
 In Part E, state the vertical displacement.
 The vertical displacement is the distance a trigonometric graph has been moved up or down from its original position.
 The midline of the graph is along the θ axis, so there is no vertical displacement.
 In Part F, state the θintercepts. Write your answer using a general form expression.
 The graph has θintercepts every π units. They include 2π, π, 0, π, 2π, extending forever in both directions.
 We can write this in general form as: θ = ±nπ, nεW.
 This means that if we start at 0 radians, we can add or subtract π to get the adjacent θintercept.
 In Part G, state the yintercept.
 The yintercept is 0.
 In Part H, state the domain and range.
 The domain is noncontinuous because of the asymptotes. We can write this as θ ε R, θ ≠ π/2±nπ, and n belongs to the set of whole numbers.
 The range is y ε R.

Section 2:
DAY TWO: Examples 5  8 (27 minutes) 

Lecture 5 

03:04 

Example 5: The "a" Parameter. (drawing graphs)  In this example we will explore the aparameter. Graph each function over the domain 0 ≤ θ < 2π. The base graph is provided as a convenience. In Part A, we'll draw the graph of y = 3sinθ.
 The number 3 in front of sinθ indicates that the amplitude of the graph is 3 units.
 It can also be thought of as a vertical stretch applied to the graph of y = sinθ.
 Let's bring up the graph of y = sinθ.
 Stretch the graph vertically by a scale factor of 3 to get y = 3sinθ.
 In Part B, draw the graph of y = 2cosθ.
 Let's bring up the graph of y = cosθ.
 Stretch the graph vertically by a scale factor of 2. This is the intermediate graph, y = 2cosθ
 The negative tells us to reflect the graph across the xaxis. This is the final graph, y = 2cosθ.
 In Part C, draw the graph of y = 1/2sinθ.
 Let's bring up the graph of y = sinθ.
 Stretch the graph vertically by a scale factor of 1/2. This is the intermediate graph, y = (1/2)sinθ.
 The negative tells us to reflect the graph across the xaxis. This is the final graph, y = (1/2)sinθ.
 In Part D, draw the graph of y = 5/2cosθ.
 Let's bring up the graph of y = cosθ.
 Stretch the graph vertically by a scale factor of 5/2 to get the graph of y = (2/5)cosθ.


Lecture 6 

09:19 

Example 6: The "a" Parameter. (deriving equations)  In this example, we will continue exploring the aparameter. Determine the trigonometric function corresponding to each graph. In Part A, write a sine function for the graph shown.
 On the yaxis, it takes 4 ticks to get to 8. 8/4 = 2, so the ticks are spaced 2 units apart.
 In Method 1, we'll inspect the graph to get the sine function.
 By inspection, we can see that the graph has an amplitude of 6 units.
 The function is y = 6sinθ.
 In Method 2, we'll use formulas to get the sine function.
 The general form of a sinusoidal graph is y = asinb(θ  c) + d, or y = acosb(θ  c) + d. We can use these formulas to construct a trigonometric function when we know the values of a, b, c, and d.
 The maximum value of the graph is 6, and the minimum value is 6.
 Use the amplitude formula, a = (maxmin)/2.
 Plug the maximum and minimum values into the formula.
 This gives us 12/2, which becomes 6.
 We have not covered b, c or d yet, so just use b = 1, c = 0, and d = 0.
 Plug the parameters into the sine function.
 Simplify to get the function y = 6sinθ. This is the same function we obtained using method 1.
 In Part B, write a sine function for the graph shown.
 On the yaxis, it takes 7 ticks to get to 28. 28/7 = 4, so the ticks are spaced 4 units apart.
 In Method 1, we'll inspect the graph to get the sine function.
 The amplitude of the graph is 12.
 The graph is upsidedown relative to y = sinθ, so insert a negative sign.
 The function is y = 12sinθ.
 In Method 2, we'll use formulas to get the sine function.
 The minimum value of the graph is 12, and the maximum value is 12.
 Use the amplitude formula, a = (maxmin)/2.
 Plug the maximum and minimum values into the formula.
 This gives us 24/2, which becomes 12.
 The graph is upsidedown relative to sinθ, so remember to insert a negative when we write the function.
 We have not covered b, c or d yet, so just use b = 1, c = 0, and d = 0.
 Plug the parameters into the sine function.
 Simplify to get the function y = 12sinθ. This is the same function we obtained using method 1.
 In Part C, write a cosine function for the graph shown.
 On the yaxis, it takes 5 ticks to get to 1. 1/5 = 0.2, so the ticks are spaced 0.2 units apart.
 In Method 1, we'll inspect the graph to get the cosine function.
 The amplitude is 0.4 units, or 2/5 as a fraction.
 The function is y = 2/5cosθ.
 In Method 2, we'll use formulas to get the cosine function.
 The maximum value of the graph is 0.4, and the minimum value is 0.4.
 Use the amplitude formula, a = (maxmin)/2.
 Plug the maximum and minimum values into the formula.
 This gives us 0.8/2, which becomes 0.4.
 We have not covered b, c or d yet, so just use b = 1, c = 0, and d = 0.
 Plug the parameters into the cosine function.
 Simplify to get the function is y = 2/5cosθ. This is the same function we obtained using method 1.
 In Part D, write a cosine function for the graph shown.
 On the yaxis, each tick is 1 unit.
 In Method 1, we'll inspect the graph to get the cosine function.
 The amplitude is 0.25, or 1/4 as a fraction.
 The graph is upsidedown relative to y = cosθ.
 The function is y =1/4cosθ.
 In Method 2, we'll use formulas to get the cosine function.
 The minimum value of the graph is 0.25, and the maximum value is 0.25.
 Use the amplitude formula, a = (maxmin)/2.
 Plug the maximum and minimum values into the formula.
 This gives us 0.5/2, which becomes 0.25.
 The graph is upsidedown relative to cosθ, so remember to insert a negative when we write the function.
 Plug the parameters into the cosine function.
 Simplify to get the function y =1/4cosθ. This is the same function we obtained using method 1.


Lecture 7 

03:15 

Example 7: The "d" Parameter. (drawing graphs)  In this example we will explore the dparameter. Graph each function over the domain 0 ≤ θ < 2π. The base graph is provided as a convenience. In Part A, we'll draw the graph of y = sinθ  2.
 The number in this position is the vertical displacement of the graph.
 This is the same as a vertical translation.
 Let's bring up the graph of y = sinθ.
 Translate the graph down 2 units. This is the graph of y = sinθ  2.
 In Part B, we'll draw the graph of y = cosθ + 4.
 Let's bring up the graph of y = cosθ.
 Translate the graph up 4 units. This is the graph of y = cosθ + 4.
 In Part C, we'll draw the graph of y = 1/2sinθ + 2.
 There are multiple transformations we need to apply to y = sinθ.
 There is a vertical stretch by a scale factor of 1/2. This is the intermediate graph, y = 1/2sinθ.
 There is a reflection about the xaxis. This is the second intermediate graph, y = 1/2sinθ.
 There is a translation up 2 units. This is the final graph, y = 1/2sinθ + 2. This is the final graph.
 In Part D, we'll draw the graph of y = 1/2cosθ – 1/2.
 There are multiple transformations we need to apply to y = cosθ.
 There is a vertical stretch by a scale factor of 1/2. This is the intermediate graph, y = 1/2cosθ.
 The graph is translated down by half a unit. This is the final graph, y = 1/2cosθ – 1/2.


Lecture 8 

11:16 

Example 8: The "d" Parameter. (deriving equations)  In this example, we will continue exploring the dparameter. Determine the trigonometric function corresponding to each graph. In Part A, write a sine function for the graph shown.
 In method 1, we'll find the sine function by inspecting the graph.
 The amplitude of the graph is 1.
 The midline of the graph is at y = 1, so the graph was translated down one unit.
 The function is y = sinθ – 1.
 In method 2, we'll use formulas to get the sine function.
 The maximum value of the graph is 0, and the minimum value is 2.
 Use the amplitude formula, a = (maxmin)/2.
 Plug the maximum and minimum values into the formula.
 This gives us 2/2, which becomes 1.
 Now use the vertical displacement formula, d = (min+max)/2.
 Plug the minimum and maximum values into the formula.
 This gives us 2/2, which becomes 1.
 We know a = 1 and d = 1. We`ll cover b & c later, so for now just use b = 1 and c = 0.
 Plug the parameters into the sine function.
 The function is y = sinθ – 1. This is the same function we obtained using method 1.
 In Part B, write a cosine function for the graph shown.
 On the yaxis, it takes 7 ticks to get to 35. 35/7 = 5, so the ticks are spaced 5 units apart.
 In method 1, we'll find the cosine function by inspecting the graph.
 The graph has an amplitude of 20 units.
 The midline of the graph is y = 15, so the graph was translated up 15 units.
 The function is y = 20cosθ + 15.
 In method 2, we'll use formulas to get the cosine function.
 The maximum value of the graph is 35, and the minimum value is 5.
 Use the amplitude formula, a = (maxmin)/2.
 Plug the maximum and minimum values into the formula.
 This gives us 40/2, which becomes 20.
 Now use the vertical displacement formula, d = (min+max)/2.
 Plug the minimum and maximum values into the formula.
 This gives us 30/2, which becomes 15.
 We have not covered b & c yet, so use b = 1 and c = 0.
 Plug the parameters into the cosine function.
 The function is y = 20cosθ + 15. This is the same function we obtained using method 1.
 In Part C, write a cosine function for the graph shown.
 On the yaxis, it takes 8 ticks to get to 32. 32/8 = 4, so the ticks are spaced 4 units apart.
 In method 1, we'll find the cosine function by inspecting the graph.
 The graph has an amplitude of 8 units.
 The midline of the graph is y = 16, so the graph was translated up 16 units.
 The graph is upsidedown compared to the graph of y = cosθ, so use a negative.
 The function is y = 8cosθ + 16.
 In method 2, we'll use formulas to get the cosine function.
 The minimum value of the graph is 8, and the maximum value is 24.
 Use the amplitude formula, a = (maxmin)/2.
 Plug the maximum and minimum values into the formula.
 This gives us 16/2, which becomes 8.
 The graph is upsidedown relative to cosθ, so remember to insert a negative when we write the function.
 Now use the vertical displacement formula, d = (min+max)/2.
 Plug the minimum and maximum values into the formula.
 This gives us 32/2, which becomes 16.
 We have not covered b & c yet, so use b = 1 and c = 0.
 Plug the parameters into the cosine function. Remember the negative since the cosine curve is upsidedown.
 The function is y = 8cosθ + 16. This is the same function we obtained using method 1.
 In Part D, write a sine function for the graph shown.
 On the yaxis, it takes 8 ticks to get to 4. 4/8 = 0.5, so the ticks are spaced 0.5 units apart.
 In method 1, we'll find the sine function by inspecting the graph.
 The amplitude of the graph is 0.5. (Or 1/2 as a fraction.)
 The midline of the graph is y = 2.5, so the graph was translated up 2.5 units. (Or 5/2 as a fraction.)
 The graph is upsidedown compared to the graph of y = sinθ, so use a negative.
 The function is y = 1/2sinθ + 5/2.
 In method 2, we'll use formulas to get the sine function.
 The minimum value of the graph is 2, and the maximum value is 3.
 Use the amplitude formula, a = (maxmin)/2.
 Plug the maximum and minimum values into the formula.
 This gives us 1/2.
 The graph is upsidedown relative to sinθ, so remember to insert a negative when we write the function.
 Now use the vertical displacement formula, d = (min+max)/2.
 Plug the minimum and maximum values into the formula.
 This gives us 5/2.
 We have not covered b & c yet, so use b = 1 and c = 0.
 Plug the parameters into the sine function. Remember the negative since the sine curve is upsidedown.
 The function is y = 1/2sinθ + 5/2. This is the same function we obtained using method 1.

Section 3:
DAY THREE: Examples 9  12 (31 minutes) 

Lecture 9 

06:30 

Example 9: The "b" Parameter. (drawing graphs)  In this example we will explore the bparameter. Graph each function over the stated domain. The base graph is provided as a convenience. In part A we'll draw the graph y = cos2θ.
 The 2 in this position is the bparameter.
 The bparameter is also known as the angular wave number, or angular frequency.
 The bvalue can be thought of in three unique ways:
 As a transformation, the reciprocal of b is the horizontal stretch factor.
 As the angular wave number, b is the number of wavelengths per 2π units of distance.
 b is NOT the period! Use the formula: P = (2π)/b to get the period.
 Let's bring up the graph of y = cosθ.
 As a transformation, the reciprocal of the bparameter is the horizontal stretch factor.
 The reciprocal of 2 is 1/2, so multiply all the xvalues of the graph by onehalf.
 If we think of b as the angular wave number, There are two complete wavelengths within the span of 2π.
 b is not the period, but we can use the formula P = (2π)/b to get the period.
 plug in 2 for b to get a period of π units. The wavelength of the graph is π.
 In Part B we'll draw the graph of y = sin3θ.
 Let's bring up the graph of y = sinθ.
 As a transformation, the reciprocal of the bparameter is the horizontal stretch factor.
 The reciprocal of 3 is 1/3, so multiply all the xvalues of the graph by onethird.
 If we think of b as the angular wave number, there are three complete wavelengths within the span of 2π.
 b is not the period, but we can use the formula P = (2π)/b to get the period.
 plug in 3 for b to get a period of 2π/3 units. The wavelength of the graph is 2π/3.
 It's a good idea to label the tick marks on the xaxis so we can verify that our period is correct.
 It takes 3 ticks to get to π. Divide to get the tick spacing, π/3. Label the ticks.
 Now that we have labeled the ticks, we can verify that the period is 2π/3.
 In Part C, we'll draw the graph of y = cos(1/3)θ.
 Let's bring up the graph of y = cosθ.
 As a transformation, the reciprocal of the bparameter is the horizontal stretch factor.
 The reciprocal of 1/3 is 3, so multiply all the xvalues of the graph by 3.
 If we think of b as the angular wave number, there is now onethird of a complete wavelength within the span of 2π.
 b is not the period, but we can use the formula P = (2π)/b to get the period.
 plug in 1/3 for b to get a period of 6π units. The wavelength of the graph is 6π.
 In Part D, we'll draw the graph of y = sin(1/5)θ.
 Let's bring up the graph of y = sinθ.
 As a transformation, the reciprocal of the bparameter is the horizontal stretch factor.
 The reciprocal of 1/5 is 5, so multiply all the xvalues of the graph by 5.
 If we think of b as the angular wave number, there is now onefifth of a complete wavelength within the span of 2π.
 b is not the period, but we can use the formula P = (2π)/b to get the period.
 plug in 1/5 for b to get a period of 10π units. The wavelength of the graph is 10π.


Lecture 10 

07:29 

Example 10: The "b" Parameter. (mixed parameters)  In this example we will continue exploring the bparameter. Graph each function over the stated domain. The base graph is provided as a convenience. In part A we'll draw the graph y = sin3θ.
 Let's bring up the graph of y = sinθ.
 Before we begin, let's label the tick marks on the xaxis.
 It takes three ticks to get to π. Divide π by 3 to get the tick interval, π/3.
 Label the ticks.
 The bvalue is 3. The horizontal stretch factor is the reciprocal of the bvalue, 1/3.
 Multiply all the xvalues on the graph by onethird.
 We can use the formula P = (2π)/b to get the period. Plug in 3 for b to get the period, 2π/3.
 The negative in front of the expression tells us to reflect the graph across the xaxis.
 We are asked to draw the graph over the domain 2π ≤ θ < 2π, so extend the graph leftwards to 2π. This is the completed graph.
 In Part B we'll draw the graph of y = 4cos2θ + 6.
 Let's bring up the graph of y = cosθ.
 Before we begin, let's label the tick marks on the xaxis.
 It takes four ticks to get to π. Divide π by 4 to get the tick interval, π/4.
 Label the ticks.
 The bvalue is 2. The horizontal stretch factor is the reciprocal of the bvalue, 1/2.
 Multiply all the xvalues on the graph by onehalf.
 We can use the formula P = (2π)/b to get the period. Plug in 2 for b to get the period, π.
 There is a vertical stretch by a scale factor of 4. Multiply all the yvalues of the graph by 4.
 There is a vertical translation 6 units up.
 We are asked to draw the graph over the domain 2π ≤ θ < 2π, so extend the graph leftwards to 2π. This is the completed graph.
 In Part C we'll draw the graph of y = 2cos(1/2)θ  1.
 Let's bring up the graph of y = cosθ.
 Before we begin, let's label the tick marks on the xaxis.
 It takes six ticks to get to π. Divide π by 6 to get the tick interval, π/6.
 Label the ticks.
 The bvalue is 1/2. The horizontal stretch factor is the reciprocal of the bvalue, 2.
 Multiply all the xvalues on the graph by 2.
 We can use the formula P = (2π)/b to get the period. Plug in 1/2 for b to get the period, 4π.
 If we think of b as the angular wave number, There is half a wavelength within the span of 2π.
 There is a vertical stretch by a scale factor of 2. Multiply all the yvalues of the graph by 2.
 There is a vertical translation 1 unit down.
 We are asked to draw the graph over the domain 2π ≤ θ < 2π, so extend the graph leftwards to 2π. This is the completed graph.
 In Part D we'll draw the graph of y = sin(4/3)θ.
 Let's bring up the graph of y = sinθ.
 Before we begin, let's label the tick marks on the xaxis.
 It takes four ticks to get to 2π. Divide 2π by 4 to get the tick interval, π/2.
 Label the ticks.
 The bvalue is 4/3. The horizontal stretch factor is the reciprocal of the bvalue, 3/4.
 Multiply all the xvalues on the graph by threequarters.
 We are asked to draw the graph over the domain 0 ≤ θ < 6π, so extend the graph rightwards to 6π.
 If we think of b as the angular wave number, there are 4/3 wavelengths (1.33 wavelengths) within the span of 2π.
 We can use the formula P = (2π)/b to get the period. Plug in 4/3 for b to get the period, 3π/2.


Lecture 11 

09:57 

Example 11: The "b" Parameter. (deriving equations)  In this example we will continue exploring the bparameter. Determine the trigonometric function corresponding to each graph. In Part A, write a cosine function for the graph shown.
 Before we begin, let's label the tick marks on the xaxis.
 It takes four ticks to get to π. Divide π by 4 to get the tick interval, π/4.
 Label the ticks.
 Let's create a box to store the graph data.
 The amplitude of the graph is 1 unit, so a = 1.
 There is no vertical displacement, so d = 0.
 The period of the graph is π/4.
 The period is not the bvalue!
 We can use the formula b = (2π)/P to get the angular wave number.
 Plug in π/4 for the period.
 Multiply the numerator by the reciprocal of the denominator.
 The π's cancel, giving us b = 8.
 An alternative way to get the bvalue is to count the number of wavelengths within a span of 2π.
 We have not covered the cvalue yet, so just use zero.
 Plug the a, b, c, and d parameters into the cosine formula.
 The function is y = cos8θ.
 In Part B, write a cosine function for the graph shown.
 Before we begin, let's label the tick marks on the xaxis.
 It takes six ticks to get to 6π. Divide 6π by 6 to get the tick interval, π.
 Label the ticks.
 Let's create a box to store the graph data.
 The amplitude of the graph is 4 units, so a = 4.
 There is no vertical displacement, so d = 0.
 The period of the graph is 12π.
 The period is not the bvalue!
 We can use the formula b = (2π)/P to get the angular wave number.
 Plug in 12π for the period.
 The π's cancel giving us b = 1/6.
 An alternative way to get the bvalue is to count the number of wavelengths within a span of 2π.
 We have not covered the cvalue yet, so just use zero.
 Plug the a, b, c, and d parameters into the cosine formula.
 The function is y = 4cos(1/6)θ.
 In Part C, write a sine function for the graph shown.
 Before we begin, let's label the tick marks on the xaxis.
 It takes three ticks to get to 3π. Divide 3π by 3 to get the tick interval, π.
 Label the ticks.
 Let's create a box to store the graph data.
 The amplitude of the graph is 1 unit, so a = 1.
 The vertical displacement is 1 unit down, so d = 1.
 The period of the graph is 3π.
 The period is not the bvalue!
 We can use the formula b = (2π)/P to get the angular wave number.
 Plug in 3π for the period.
 The π's cancel giving us b = 2/3.
 An alternative way to get the bvalue is to count the number of wavelengths within a span of 2π.
 We have not covered the cvalue yet, so just use zero.
 Plug the a, b, c, and d parameters into the sine formula.
 The function is y = sin(2/3)θ  1.
 In Part D, write a sine function for the graph shown.
 Before we begin, let's label the tick marks on the xaxis.
 It takes six ticks to get to π. Divide π by 6 to get the tick interval, π/6.
 Label the ticks.
 Let's create a box to store the graph data.
 The curve shown here is half of a sine curve.
 The amplitude of the graph is half a unit, so a = 1/2.
 The vertical displacement is half a unit up, so d = 1/2.
 The period is 4π.
 The period is not the bvalue!
 We can use the formula b = (2π)/P to get the angular wave number.
 Plug in 4π for the period.
 Reduce the fraction and cancel the π's to get b = 1/2.
 An alternative way to get the bvalue is to count the number of wavelengths within a span of 2π.
 We have not covered the cvalue yet, so just use zero.
 Plug the a, b, c, and d parameters into the sine formula.
 The function is y =1/2sin(1/2)θ + 1/2.


Lecture 12 

06:49 

Example 12: The "c" Parameter. (drawing graphs)  In this example we will explore the cparameter. Graph each function over the stated domain. The base graph is provided as a convenience. In part A we'll draw the graph y = sin(θ – π/2).
 The cparameter is the phase shift. It specifies the horizontal displacement of the graph.
 In horizontal translations, the graph moves in the opposite direction of the sign.
 For example, if we had y = sin(θ – π), the graph moves to the right π units.
 If we rewrite the function as y = sin(θ – (+π)), we can observe that c = +π.
 If we had y = sin(θ + π), the graph moves to the left π units.
 If we rewrite the function as y = sin(θ – (π)), we can observe that c = π.
 Let's bring up the graph of y = sinθ.
 Before we begin, let's label the tick marks on the xaxis.
 It takes four ticks to get to 2π. Divide 2π by 4 to get the tick interval, π/2.
 Label the ticks.
 Rewrite the function to get the cparameter. c = +π/2, so we have a horizontal translation of π/2 units right.
 Translate the graph π/2 units right.
 We are asked to draw the graph over the domain 4π ≤ θ < 4π. Extend the graph to fill the domain. This is the completed graph.
 In Part B, draw the graph of y = cos(θ + π).
 Let's bring up the graph of y = cosθ.
 Before we begin, let's label the tick marks on the xaxis.
 It takes four ticks to get to 2π. Divide 2π by 4 to get the tick interval, π/2.
 Label the ticks.
 Rewrite the function as y = cos(θ – (π)) to get the cparameter. c is –π, so we have a horizontal translation of π units left.
 Translate the graph π units left.
 We are asked to draw the graph over the domain 4π ≤ θ < 4π. Extend the graph to fill the domain. This is the completed graph.
 In Part C, draw the graph of y = cos(θ – π/6).
 Let's bring up the graph of y = cosθ.
 Before we begin, let's label the tick marks on the xaxis.
 It takes six ticks to get to 2π. Divide 2π by 6 to get the tick interval, π/6.
 Label the ticks.
 Rewrite the function to get the cparameter. c = +π/6, so we have a horizontal translation of π/6 units right.
 Translate the graph π/6 units right.
 We are asked to draw the graph over the domain 2π ≤ θ < 2π. Extend the graph to fill the domain. This is the completed graph.
 In Part D, draw the graph of y = 3sin(θ + 2π/3).
 Let's bring up the graph of y = sinθ.
 Before we begin, let's label the tick marks on the xaxis.
 It takes three ticks to get to π. Divide π by 3 to get the tick interval, π/3.
 Label the ticks.
 There is a vertical stretch by a scale factor of 3. Multiply all the yvalues of the graph by 3.
 Rewrite the function to get the cparameter. c is –2π/3, so we have a horizontal translation of 2π/3 units left.
 Translate the graph 2π/3 units left.
 We are asked to draw the graph over the domain 2π ≤ θ < 2π. Extend the graph to fill the domain. This is the completed graph.

Section 4:
DAY FOUR: Examples 13  16 (40 minutes) 

Lecture 13 

12:17 

Example 13: The "c" Parameter. (mixed parameters)  In this example we will continue exploring the cparameter. Graph each function over the stated domain. The base graph is provided as a convenience. In part A we'll draw the graph y = sin(2θ + π).
 We should NOT try to graph y = sin(2θ + π) in its current form. Factor the θcoefficient out of the brackets before continuing.
 We have a problem, however. How do we factor 2 out of (2θ + π)?
 When we factor 2 out of the brackets, we are dividing each term within the brackets by the quantity being pulled out. You've probably been doing this in your head for years, but never actually wrote it out.
 Consider the binomial 3x + 6.
 We can factor out a 3 to get 3(x + 2).
 Now we'll write the intermediate step. Divide each term within the brackets by the quantity being pulled out.
 Now we'll factor 2θ + π.
 Now we'll write the intermediate step. Divide each term within the brackets by the quantity being pulled out.
 The factored expression is 2(θ + π/2).
 Rewrite the original function using the factored expression.
 Let's bring up the graph of y = sinθ.
 Let's label the tick marks on the xaxis.
 It takes two ticks to get to π. Divide π by 2 to get the tick interval, π/2.
 Label the ticks.
 The bvalue is 2, so there is a horizontal stretch by a factor of 1/2.
 Multiply all the xvalues of the graph by 1/2.
 Rewrite the function so it matches y = asinb(θ  c) + d.
 The cparameter is –π/2. There is a horizontal translation π/2 units to the left.
 Translate the graph π/2 units left.
 We are asked to draw the graph over the domain π/2 ≤ θ < 2π, so extend the graph to fill this region. This is the completed graph.
 In Part B, draw the graph of y = cos(1/2θ + π).
 We should NOT try to graph this function in its current form. Factor the θcoefficient out of the brackets before continuing.
 Rewrite the expression in the brackets.
 Factor out 1/2. Divide each term within the brackets by the quantity being pulled out.
 For each division, multiply the numerator by the reciprocal of the denominator.
 The factored form is 1/2(θ + 2π).
 Rewrite the original function using the factored expression.
 Let's bring up the graph of y = cosθ.
 Let's label the tick marks on the xaxis.
 It takes four ticks to get to 2π. Divide 2π by 4 to get the tick interval, π/2.
 Label the ticks.
 The bvalue is 1/2, so there is a horizontal stretch by a factor of 2.
 Multiply all the xvalues of the graph by 2.
 Rewrite the function so it matches y = acosb(θ  c) + d.
 The cparameter is 2π. There is a horizontal translation 2π units to the left.
 Translate the graph 2π units left.
 We are asked to draw the graph over the domain 2π ≤ θ < 6π, so extend the graph to fill this region. This is the completed graph.
 In Part C, draw the graph of y = 1/2sin(2θ  3π) + 1.
 We should NOT try to graph this function in its current form. Factor the θcoefficient out of the brackets before continuing.
 Rewrite the expression in the brackets.
 Factor out 2. Divide each term within the brackets by the quantity being pulled out.
 The factored form is 2(θ  3π/2).
 Rewrite the original function using the factored expression.
 Let's bring up the graph of y = sinθ.
 Let's label the tick marks on the xaxis.
 It takes four ticks to get to 2π. Divide 2π by 4 to get the tick interval, π/2.
 Label the ticks.
 There is a vertical stretch by a factor of 1/2. There is also a reflection across the xaxis.
 Multiply the yvalues by 1/2 and reflect across the xaxis.
 The bvalue is 2, so there is a horizontal stretch by a factor of 1/2.
 Transform the graph.
 Multiply the xvalues by 1/2.
 Rewrite the function so it matches y = asinb(θ  c) + d.
 The cparameter is +3π/2. There is a horizontal translation 3π/2 units to the right.
 Translate the graph 3π/2 units right.
 There is a vertical translation 1 unit up.
 Translate the graph 1 unit up.
 We are asked to draw the graph over the domain π ≤ θ < 4π, so extend the graph to fill this region. This is the completed graph.
 In Part D, draw the graph of y = 2cos(4θ + 4π/3)  2.
 We should NOT try to graph this function in its current form. Factor the θcoefficient out of the brackets before continuing.
 Rewrite the expression in brackets, 4θ + 4π/3.
 We can factor out the 4 directly. There is no need to write out the intermediate step.
 Rewrite the original function using the factored expression.
 Let's bring up the graph of y = cosθ.
 Let's label the tick marks on the xaxis.
 It takes six ticks to get to π. Divide π by 6 to get the tick interval, π/6.
 Label the ticks.
 There is a vertical stretch by a factor of 2. There is also a reflection across the xaxis.
 Multiply the yvalues by 2 and reflect across the xaxis.
 The bvalue is 4, so there is a horizontal stretch by a factor of 1/4.
 Multiply the xvalues by 1/4.
 Rewrite the function so it matches y = acosb(θ  c) + d.
 The cparameter is π/3. There is a horizontal translation π/3 units to the left.
 Translate the graph π/3 units left.
 There is a vertical translation 2 units down.
 Translate the graph 2 units down.
 We are asked to draw the graph over the domain 2π ≤ θ < 2π, so extend the graph to fill this region. This is the completed graph.


Lecture 14 

08:27 

Example 14: The "c" Parameter. (deriving equations)  In this example we will continue exploring the cparameter. Determine the trigonometric function corresponding to each graph. In Part A, write a cosine function for the graph shown.
 Before we begin, label the tick marks.
 We are asked to find a cosine function, so draw a rectangle around the cosine shape.
 Let's create a box to store the graph data.
 The amplitude is 1 unit, so a = 1.
 There is no vertical displacement, so d = 0.
 The period is 2π.
 The period is not the bvalue! We can use the formula b = (2π)/P to get the angular wave number.
 Plug in 2π for the period to get a bvalue of 1.
 The cosine shape has been shifted π/2 units left, so c = π/2.
 Plug the a, b, c, and d parameters into the cosine formula.
 The function is y = cos(θ + π/2).
 In Part B, write a sine function for the graph shown.
 Before we begin, label the tick marks.
 We are asked to find a sine function, so draw a rectangle around the sine shape.
 Let's create a box to store the graph data.
 The amplitude is 6 units, so a = 6.
 There is no vertical displacement, so d = 0.
 The period is 2π.
 The period is not the bvalue! We can use the formula b = (2π)/P to get the angular wave number.
 Plug in 2π for P to get b = 1.
 The sine shape has been shifted 4π/3 units right, so c = 4π/3.
 Plug the a, b, c, and d parameters into the sine formula.
 The function is y = 6sin(θ  4π/3).
 In Part C, write a sine function for the graph shown.
 Before we begin, label the tick marks.
 We are asked to find a sine function, so draw a rectangle around the sine shape.
 There are two options for the sine shape! Which one should be used?
 Both shapes will give us a correct function, but its good form to use the shape that begins closest to θ = 0.
 Let's create a box to store the graph data.
 The amplitude is 1 unit, so a = 1.
 There is no vertical displacement, so d = 0.
 The period is π.
 The period is not the bvalue! We can use the formula b = (2π)/P to get the angular wave number.
 Plug in π for P to get b = 2.
 The sine shape has been shifted π/4 units left, so c = π/4.
 Plug the a, b, c, and d parameters into the sine formula.
 The function is y = sin2(θ + π/4).
 In Part D, write a cosine function for the graph shown.
 Before we begin, label the tick marks.
 We are asked to find a cosine function, so draw a rectangle around the cosine shape.
 Let's create a box to store the graph data.
 Label the maximum and minimum points on the graph. The maximum occurs when y = 3, and the minimum occurs when y = 2.
 Use the amplitude formula, a = (maxmin)/2.
 Plug the maximum and minimum values into the formula.
 This gives us an amplitude of 5/2.
 Now use the vertical displacement formula, d = (min+max)/2.
 Plug the minimum and maximum values into the formula.
 This gives us a vertical displacement of 1/2.
 The period is 8π.
 The period is not the bvalue! We can use the formula b = (2π)/P to get the angular wave number.
 Plug in 8π for P to get b = 1/4.
 The cosine shape has been shifted π units left, so c = π.
 Plug the a, b, c, and d parameters into the cosine formula.
 The function is y = 5/2cos1/4(θ + π) + 1/2.


Lecture 15 

09:45 

Example 15: Trigonometric Graphs (a, b, c, and d).  Graph each function over the stated domain. The base graph is provided as a convenience. In part A we'll draw the graph y = 2cos(1/2θ – π/8) + 3.
 Before we begin, label the tick marks.
 It takes eight ticks to get to 2π. Divide 2π by 8 to get the tick interval, π/4.
 Label the ticks.
 We need to factor out the θcoefficient before we can graph the function. This gives us y = 2cos(1/2(θ – π/4)) + 3.
 There is a vertical stretch by a factor of 2.
 Multiply the yvalues of the graph by 2.
 The bvalue is 1/2, so there is a horizontal stretch by a factor of 2.
 Multiply the xvalues of the graph by 2.
 Rewrite the function so it matches y = acosb(θ  c) + d.
 The cparameter is +π/4. There is a horizontal translation π/4 units to the right.
 Translate the graph π/4 units right.
 There is a vertical translation 3 units up.
 Translate the graph 3 units up.
 We are asked to draw the graph over the domain 0 ≤ θ < 6π, so extend the graph to fill this region. This is the completed graph.
 In part B we'll draw the graph y = 1/2sin(2θ – π/2).
 We should factor 2θ – π/2 before graphing the function.
 Factor out 2. Divide each term within the brackets by the quantity being pulled out.
 For the fraction division in the second term, multiply the numerator by the reciprocal of the denominator.
 Simplify to get 2(θ – π/4).
 Rewrite the original function using the factored expression.
 There is a vertical stretch by a factor of 1/2.
 Multiply the yvalues of the graph by 1/2.
 The bvalue is 2, so there is a horizontal stretch by a factor of 1/2.
 Multiply the xvalues by 1/2.
 Rewrite the function so it matches y = asinb(θ  c) + d.
 The cparameter is +π/4. There is a horizontal translation π/4 units to the right.
 Translate the graph π/4 units right.
 We are asked to draw the graph over the domain 0 ≤ θ < 2π, so extend the graph to fill this region. This is the completed graph.
 In part C we'll draw the graph y = 2sin(4θ – π)  3.
 Let's bring up the graph of y = sinθ.
 Label the ticks.
 We need to factor out the θcoefficient before we can graph the function. This gives us y = 2sin(4(θ – π/4))  3.
 There is a vertical stretch by a factor of 2. There is also a reflection across the xaxis.
 Multiply the yvalues by 2 and reflect across the xaxis.
 The bvalue is 4, so there is a horizontal stretch by a factor of 1/4.
 Multiply the xvalues by ¼.
 Rewrite the function so it matches y = asinb(θ  c) + d.
 The cparameter is +π/4. There is a horizontal translation π/4 units to the right.
 Translate the graph π/4 units right.
 There is a vertical translation 3 units down.
 Translate the graph 3 units down.
 We are asked to draw the graph over the domain 0 ≤ θ < 2π, so extend the graph to fill this region. This is the completed graph.
 In part D we'll draw the graph y = 5cos(2θ – π/3) + 1.
 Label the ticks.
 We should factor 2θ – π/3 before graphing the function.
 Factor out 2. Divide each term within the brackets by the quantity being pulled out.
 For the fraction division in the second term, multiply the numerator by the reciprocal of the denominator.
 Simplify to get 2(θ – π/6).
 Rewrite the original function using the factored expression.
 There is a vertical stretch by a factor of 5. There is also a reflection across the xaxis.
 Multiply the yvalues by 5 and reflect across the xaxis.
 The bvalue is 2, so there is a horizontal stretch by a factor of 1/2.
 Multiply the xvalues by 1/2.
 Rewrite the function so it matches y = acosb(θ  c) + d.
 The cparameter is +π/6. There is a horizontal translation π/6 units to the right.
 Translate the graph π/6 units right.
 There is a vertical translation 1 unit up.
 Translate the graph 1 unit up.
 We are asked to draw the graph over the domain 0 ≤ θ < 2π, so extend the graph to fill this region. This is the completed graph.


Lecture 16 

09:05 

Example 16: Trigonometric Functions (a, b, c, and d).  Write a trigonometric function for each graph.
 Before we begin, let's label the tick marks on the xaxis.
 It takes twelve ticks to get to 2π. Divide 2π by 12 to get the tick interval, π/6.
 Label the ticks.
 Let's create a box to store the graph data.
 The amplitude is 1 unit, so a = 1.
 The vertical displacement is 1 unit up, so d = 1.
 The period is tricky to spot in this graph. We know one period exists between 5π/6 and 17π/6.
 The difference between them is 2π, since 17π/6 – 5π/6 = 12π/6, which reduces to 2π.
 The period is not the bvalue! We can use the formula b = (2π)/P to get the angular wave number.
 Plug in 2π for P to get b = 1.
 Let's draw a rectangle around the sine shape.
 The sine curve has a phase shift of 4π/3 units to the right.
 Now let's draw a rectangle around the cosine shape.
 The cosine curve has a phase shift of 11π/6 units to the right.
 Now write out both the sine and cosine equations.
 The sine function is y = sin(θ  4π/3) + 1.
 The cosine function is y = cos(θ  11π/6) + 1.
 The a, P, b, and d values are the same for both the sine and cosine functions.
 The only parameter that is different is the cparameter.
 Note that the positions of the sine and cosine graphs differ by onequarter of a period, or π/2.
 Suppose our graph was extended backwards to 2π.
 We could write a second sine function, y = sin(θ + 2π/3) + 1.
 We could also write a second cosine function, y = cos(θ + π/6) + 1.
 All four of these functions represent the same graph!
 If we had to choose just one function to represent the graph, which one should we choose?
 While all four functions correctly represent the graph, good form dictates that we use the smallest possible phase shift. Under this criterion,the "best" function is: y = cos(θ + π/6) + 1.
 Now we'll move on to Part B.
 Label the tick marks.
 Let's create a box to store the graph data.
 Label the maximum and minimum points on the graph. The maximum occurs when y = 4, and the minimum occurs when y = 12.
 Use the amplitude formula, a = (maxmin)/2.
 Plug the maximum and minimum values into the formula.
 The double negatives become a positive.
 This gives us an amplitude of 4.
 Now use the vertical displacement formula, d = (min+max)/2.
 Plug the minimum and maximum values into the formula.
 The positive and negative combine to form a negative.
 This gives us a vertical displacement of 8.
 The period is π.
 The period is not the bvalue! We can use the formula b = (2π)/P to get the angular wave number.
 Plug in π for P to get b = 2.
 Let's draw a rectangle around the sine shape.
 The sine curve has a phase shift of π/4 units to the right.
 Now let's draw a rectangle around the cosine shape.
 The cosine curve has a phase shift of π/2 units to the right.
 The sine function is y = 4sin(2(θ  π/4))  8.
 The cosine function is y = 4cos(2(θ  π/2))  8.
 The a, b, and d values are the same for both the sine and cosine functions. The only parameter that will be different is the cparameter.
 Note that the positions of the sine and cosine graphs differ by onequarter of a period, or π/4.
 We could also get functions using negative phase shifts.
 Our second sine function is y = 4sin(2(θ + 3π/4))  8.
 Our second cosine function is y = 4cos(2(θ + π/2))  8.
 Yet another option would be to use the upsidedown cosine shape starting when θ = 0.
 This function is y = 4cos2θ  8.
 All of these functions correctly describe the graph.
 If we had to choose just one function to represent the graph, which one should we choose?
 Choose the function with the smallest phase shift. Under this criterion, the "best" function is: y = 4cos2θ – 8 since the phase shift is zero.

Section 5:
DAY FIVE: Examples 17  20 (18 minutes) 

Lecture 17 

04:47 

Example 17: Graphing y = secθ  In this example, we will explore the graph of y = secθ. In Part A, draw y = secθ.
 Bring up the unit circle for secθ.
 We need to create a grid so we can draw the graph of y = secθ.
 We have a line representing 2root(3)/3.
 We have another line representing root(2).
 By symmetry, we also have lines at 2root(3)/3 and –root(2).
 Draw the points from the first quadrant.
 Draw the points from the second quadrant.
 Draw the points from the third quadrant.
 Draw the points from the fourth quadrant.
 Draw the graph and asymptotes.
 If we go around the unit circle clockwise, we can draw the left side of the graph.
 In Part B, state the period of y = secθ.
 Bring up the graph of y = secθ.
 The period is 2π.
 In Part C, state the domain and range.
 Bring up the graph of y = secθ.
 Domain: The graph exists for all real numbers, with the exception of the asymptotes, where secθ is undefined.
 The domain is θεR, θ≠π/2 + nπ, nεI.
 The graph does not exist between yvalues of 1 and 1.
 The graph exists when y ≤ 1 or y ≥ 1. This is the range.
 In Part D, write the general equation of the asymptotes.
 Bring up the graph and the domain.
 The restriction of the domain for y = secθ becomes the equation of the asymptotes.
 The asymptotes occur at θ=π/2 + nπ, nεI.
 Now we'll move on to Part E. Given the graph of f(θ) = cosθ, draw y = 1/f(θ).
 Drawing the reciprocal graph of y = cosθ is a quick way to create the graph of y = secθ.
 In Step 1, draw asymptotes at the xintercepts.
 In Step 2, points where y = 1 or +1 are invariant. This means that they occupy the same position in both graphs.
 Finally, in Step 3, fill in the secant shape.


Lecture 18 

04:36 

Example 18: Graphing y = cscθ  In this example, we will explore the graph of y = cscθ. In Part A, draw y = cscθ.
 Bring up the unit circle for cscθ.
 We need to create a grid so we can draw the graph of y = cscθ.
 As in the previous example, we have lines representing 2root(3)/3, root(2), 2root(3)/3 and –root(2).
 Draw the points from the first quadrant.
 Draw the points from the second quadrant.
 Draw the points from the third quadrant.
 Draw the points from the fourth quadrant.
 Draw the graph and asymptotes.
 If we go around the unit circle clockwise, we can draw the left side of the graph.
 In Part B, state the period of y = cscθ.
 Bring up the graph of y = cscθ.
 The period is 2π.
 In Part C, state the domain and range.
 Bring up the graph of y = cscθ.
 Domain: The graph exists for all real numbers, with the exception of the asymptotes, where cscθ is undefined.
 The domain is θεR, θ≠ nπ, nεI.
 The graph does not exist between yvalues of 1 and 1.
 The graph exists when y ≤ 1 or y ≥ 1. This is the range.
 In Part D, write the general equation of the asymptotes.
 Bring up the graph and the domain.
 The restriction of the domain for y = cscθ becomes the equation of the asymptotes.
 The asymptotes occur at θ= nπ, nεI.
 Now we'll move on to Part E. Given the graph of f(θ) = sinθ, draw y = 1/f(θ).
 Drawing the reciprocal graph of y = sinθ is a quick way to create the graph of y = cscθ.
 In Step 1, draw asymptotes at the xintercepts.
 In Step 2, points where y = 1 or +1 are invariant. This means that they occupy the same position in both graphs.
 Finally, in Step 3, fill in the cosecant shape.


Lecture 19 

04:22 

Example 19: Graphing y = cotθ  In this example, we will explore the graph of y = cotθ. In Part A, draw y = cotθ.
 Bring up the unit circle for cotθ.
 We need to create a grid so we can draw the graph of y = cotθ.
 We have lines representing root(3), root(3)/3, root(3)/3 and –root(3).
 Draw the points from the first quadrant.
 Draw the points from the second quadrant.
 Draw the points from the third quadrant.
 Draw the points from the fourth quadrant.
 Draw the graph and asymptotes.
 If we go around the unit circle clockwise, we can draw the left side of the graph.
 In Part B, state the period of y = cotθ.
 Bring up the graph of y = cotθ.
 The period is π.
 In Part C, state the domain and range.
 Bring up the graph of y = cotθ.
 Domain: The graph exists for all real numbers, with the exception of the asymptotes, where cotθ is undefined.
 The domain is θεR, θ≠ nπ, nεI.
 The graph exists for all real numbers. Therange is yεR.
 In Part D, write the general equation of the asymptotes.
 Bring up the graph and the domain.
 The restriction of the domain for y = cotθ becomes the equation of the asymptotes.
 The asymptotes occur at θ = nπ, nεI.
 Now we'll move on to Part E. Given the graph of f(θ) = tanθ, draw y = 1/f(θ).
 Drawing the reciprocal graph of y = tanθ is a quick way to create the graph of y = cotθ.
 In Step 1, draw asymptotes at the xintercepts.
 In Step 2, points where y = 1 or +1 are invariant. This means that they occupy the same position in both graphs.
 Finally, in Step 3, fill in the cotangent shape.


Lecture 20 

03:44 

Example 19: Transformations of Reciprocal Trignometric Graphs  Graph each function over the domain 0 ≤ θ ≤ 2π. The base graph is provided as a convenience. State the new domain and range. In Part A, draw the graph of y = 1/2secθ.
 There is a vertical stretch by a scale factor of 1/2.
 Multiply all the yvalues by 1/2.
 The domain of the transformed graph is θεR, θ≠π/2 ± nπ, nεW.
 The range is y ≤ 1/2 or y ≥ 1/2.
 In Part B, draw the graph of y = sec2θ.
 There is a horizontal stretch by a scale factor of 1/2.
 Multiply all the xvalues by 1/2.
 The domain of the transformed graph is θεR, θ≠π/4 ± nπ/2, nεW.
 The range is y ≤ 1 or y ≥ 1.
 In Part C, draw the graph of y = csc(θ  π/4).
 There is a phase shift of π/4 units right.
 Translate the graph π/4 units right.
 The domain of the transformed graph is θεR, θ≠π/4 ± nπ, nεW.
 The range is y ≤ 1 or y ≥ 1.
 In Part D, draw the graph of y = cot(1/2θ).
 There is a horizontal stretch by a scale factor of 2.
 Multiply all the xvalues by 2.
 The domain of the transformed graph is θεR, θ ≠ ±n(2π), nεW.
 The range is y ε R.

Section 6:
Other Lesson Materials 

Lecture 21 

24 pages 

Lecture 22 

01:05 

Trigonometric Functions I: Introduction  Welcome to Trigonometric Functions I. This topic will take five days to complete.
 In Day 1 we will learn how to graph the primary trigonometric ratios using the unit circle. We will also define the properties of these graphs.
 In Day 2 we will study the amplitude and vertical displacement of sine and cosine graphs.
 In Day 3 we will study the angular wave number, period, and phase shifts of sine and cosine graphs.
 In Day 4 we will graph trigonometric functions that use all 4 parameters: a, b, c, and d.
 In Day 5 we will explore the reciprocal trigonometric ratios.


Lecture 23 

01:01 

Trigonometric Functions I: Summary  You have now completed Trigonometric Functions I. Over the past five days you have learned:
 how to find the coordinates of a trigonometric grid.
 how to graph trigonometric functions using the unit circle.
 the definitions of amplitude (a), vertical displacement (d), angular wave number (b), period (P), and phase shift (c).
 the general form of trigonometric functions.
 how to use parameters to graph a trigonometric function.
 how to derive a trigonometric function from its graph.
 how to work with different combinations of parameters.
 And finally, how to graph the reciprocal trigonometric ratios.


Quiz 1 
Trigonometric Functions One

10 questions 
Full curriculum

Simply Great
A really great course with understandable material and a good instructor. Worth learning from!
good one
Very Nice...