Section 1:
DAY ONE: Examples 1  4 (31 minutes) 

Lecture 1 

06:41 

Example 1: Trigonometric Functions of Angles  We'll begin this lesson by graphing trigonometric functions of angles. In Part A, we'll graph f(θ) = cos(2(θ – π/4)).
 Start with the graph of y = cosθ.
 This function is called a trigonometric function of an angle. The independent variable is θ, and θ represents a radian angular measure.
 The reciprocal of the bvalue is the horizontal stretch factor.
 The reciprocal of 2 is 1/2, so multiply the xvalues of the graph by 1/2.
 Rewrite the function so it matches y = acosb(θ  c) + d.
 c = +π/4, so there is a phase shift of π/4 right.
 Draw the graph between 0 and 3π.
 Note: The graph of f(θ) = cos(2(θ – π/4)) is equivalent to the graph of f(θ) = sin2θ. A sine and cosine graph differ by onequarter of a period.
 Now we'll graph this function using technology.
 With the calculator in radian mode, use the following window settings: x: [0, 3π, π/4], and y: [2, 2, 1].
 Finally, graph the function.
 In Part B, we'll draw the graph of f(θ) = cos(2(θ – 45°)).
 Start with the graph of y = cosθ.
 This function is also called a trigonometric function of an angle. The independent variable is θ, and θ represents a degree angular measure.
 The reciprocal of the bvalue is the horizontal stretch factor.
 The reciprocal of 2 is 1/2, so multiply the xvalues of the graph by 1/2.
 NOTE: The formulae for the bvalue and period need to be adjusted for degrees. (Replace 2π with 360°) This gives us b = 360°/P and P = 360°/b.
 If we plug the bvalue of 2 into the period formula, we get a period of 180°.
 Rewrite the function so it matches y = acosb(θ  c) + d.
 c = +45°, so there is a phase shift of 45° right.
 Draw the graph between 0° and 540°.
 Now we'll graph this function using technology.
 With the calculator in degree mode, use the following window settings: x: [0, 540, 45], and y: [2, 2, 1].
 Finally, graph the function.


Lecture 2 

08:07 

Example 2: Trigonometric Functions of Real Numbers  In this example we'll graph trigonometric functions of real numbers. In Part A, we'll graph h(t) = cos(π/30(t – 15)).
 This function is called a trigonometric function of a real number. The independent variable is t, and t represents a physical quantity (time), not an angle.
 Let's set up a coordinate grid.
 Notice that time (t) is on the xaxis, not θ.
 Use the formula P = 2π/b to find the period of the graph.
 Plug in π/30 for b.
 Multiply the numerator by the reciprocal of the denominator.
 The period is 60 seconds.
 Create a box to store the graph data.
 Rewrite the function so it matches y = acosb(θ  c) + d.
 c = +15, so there is a phase shift of 15 seconds to the right.
 Draw the cosine graph with a period of 60 s and a phase shift of 15 s right.
 Now we'll graph this function using technology.
 With the calculator in radian mode, use the following window settings: x: [0, 60, 15], and y: [2, 2, 1].
 Finally, draw the graph.
 In Part B, we'll draw the graph of f(x) = cos(π/8(x – 4)).
 This function is also called a trigonometric function of a real number. The independent variable is x, and x represents a real number, not an angle.
 Let's set up a coordinate grid.
 Use the formula P = 2π/b to find the period of the graph.
 Plug in π/8 for b.
 Multiply the numerator by the reciprocal of the denominator.
 The period is 16 units.
 Create a box to store the graph data.
 Rewrite the function so it matches y = acosb(θ  c) + d.
 c = +4, so there is a phase shift of 4 units right.
 Draw a cosine graph with a period of 16 units and a phase shift of 4 units right.
 Now we'll graph this function using technology.
 With the calculator in radian mode, use the following window settings: x: [0, 24, 2], and y: [2, 2, 1].
 Finally, draw the graph.
 In Part C, what are three differences between trigonometric functions of angles and trigonometric functions of real numbers?
 Let's rewrite the functions from Examples 1 and 2.
 The first difference is that angular functions use θ as the independent variable, but real number functions use x or a physical quantity (time, distance, etc...) as the independent variable.
 The bvalue in angular functions does not contain π.
 The bvalue in real number functions does contain π.
 The cvalue in angular functions must be an angle, and is usually measured in radians or degrees.
 The cvalue in real number functions is not an angle, and is often a physical quantity like seconds.
 All of the examples in Lesson 3 (Trigonometric Functions I) used trigonometric functions of angles.
 The main reason why we care about the difference between angular functions and real number functions is that modeling realworld scenarios requires real number functions.
 For the rest of this lesson (Trigonometric Functions II), nearly all of work will involve trigonometric functions of real numbers.


Lecture 3 

08:50 

Example 3: Finding the View Window for a Trigonometric Graph  Determine the view window for each function and sketch each graph. In Part A, we'll graph f(x) = 12sin(π/3(x 2)) – 14.
 Before we can draw this graph, we need to inspect the parameters to determine an appropriate view window.
 Let's begin with the xaxis settings.
 The period can be found with the formula P = 2π/b.
 Plug in π/3 for b.
 Multiply the numerator by the reciprocal of the denominator.
 The π's cancel, giving us a period of 6 units.
 The cvalue is +2, so there is a phase shift 2 units right.
 The goal is to see at least one complete sine cycle in our graph.
 Start at x = 0. If the period is 6 units, and the graph is shifted 2 units right, we need 8 units on the xaxis.
 The xaxis window settings are: x: [0, 8, 1].
 There are not that many ticks, so use a scaling of 1 unit.
 Now we'll determine the yaxis settings.
 The amplitude of the graph is 12.
 The vertical displacement is 14.
 The goal is to see both the minimum and maximum when we draw the graph.
 The minimum occurs at d – a. 14 – 12 = 26.
 The maximum occurs at d + a. 14 + 12 = 2.
 The yaxis window settings are: y: [26, 0, 2].
 For ymax, use zero instead of 2. This will let us include the xaxis in our sketch.
 Use a scaling of 2 units to prevent tick crowding.
 Now that we have appropriate window settings, we can draw the axes for the graph.
 Sketch the graph.
 Now we'll draw the graph using technology. Adjust the window settings as shown.
 Next, input the function.
 Finally, draw the graph.
 In Part B we'll draw the graph of f(x) = 25cos(π/250(x+225)) + 50.
 Before we can draw this graph, we need to inspect the parameters to determine an appropriate view window.
 Let's begin with the xaxis settings.
 The period can be found with the formula P = 2π/b.
 Plug in π/250 for the bvalue.
 Multiply the numerator by the reciprocal of the denominator.
 Cancel the π's to get a period of 500 units.
 The cvalue is 225, so there is a phase shift 225 units to the left.
 The goal is to see at least one complete cosine cycle in our graph.
 The phase shift is 225 units left, so xmin = 225. The period is 500 units, so xmax = 225 + 500 = 275.
 The xaxis window settings are: x: [225, 275, 25].
 Use a scaling of 25 units to prevent tick crowding.
 Now we'll find the yaxis settings.
 The amplitude is 25 units. There is also a reflection in the xaxis.
 The vertical displacement is 50 units.
 The goal is to see both the minimum and maximum when we draw the graph.
 The minimum occurs at d  a. 50 – 25 = 25.
 The maximum occurs at d + a. 50 + 25 = 75.
 The yaxis window settings are: y: [0, 75, 25].
 Set ymin = 0 (instead of 25)so we can see the yaxis.
 Use a scaling of 25 units to prevent tick crowding.
 Now that we have the appropriate window settings, we can draw the axes for the graph.
 Sketch the graph.
 Now we'll draw the graph using technology. Adjust the window settings as shown.
 Next, input the function.
 Finally, draw the graph.


Lecture 4 

07:01 

Example 4: Finding View Windows (more practice)  Determine the view window for each function and sketch each graph. In Part A, we'll draw the graph of f(x) = 13.5cos(2π/96(x – 24)) + 6.5.
 Let's begin with the xaxis settings.
 The period can be found with the formula P = 2π/b.
 Plug in 2π/96 for the bvalue.
 Multiply the numerator by the reciprocal of the denominator.
 The period is 96 units.
 The cvalue is +24, so there is a phase shift of 24 units right.
 The goal is to see at least one complete cosine cycle in our graph.
 Start at x = 0. If the period is 96 units, and the graph is shifted 24 units right, we need 120 units on the xaxis.
 The xaxis window settings are: x: [0, 120, 10].
 Now we'll find the yaxis settings.
 The amplitude is 13.5.
 The vertical displacement is 6.5.
 The goal is to see both the minimum and maximum when we draw the graph.
 The minimum occurs at da. 6.5 – 13.5 = 7.
 The maximum occurs at d+a. 6.5 + 13.5 = 20.
 The yaxis window settings are: y: [8, 20, 2]. Use ymin = 8 (instead of 7) since we are using an even spacing for the ticks.
 Now that we have the appropriate window settings, we can draw the axes for the graph.
 Sketch the graph.
 Now we'll draw the graph using technology. Adjust the window settings as shown.
 Input the function.
 Finally, draw the graph.
 In Part B, we'll draw the graph of f(x)= 2.5sin0.25π(x + 3) + 16.
 Let's begin with the xaxis settings.
 The period can be found with the formula P = 2π/b.
 Plug in 0.25π for the bvalue.
 The π's cancel out, giving us a period of 8 units.
 The cvalue is 3, so there is a phase shift 3 units left.
 The goal is to see at least one complete sine cycle in our graph.
 The graph is shifted 3 units left, so xmin = 3. The period is 8 units, so xmax = 3 + 8 = 5.
 The xaxis window settings are: x: [3, 5, 1].
 Now we'll find the yaxis settings.
 The amplitude is 2.5.
 The vertical displacement is 16.
 The goal is to see both the minimum and maximum when we draw the graph.
 The minimum occurs at d  a. 16  2.5 = 13.5.
 The maximum occurs at d + a. 16 + 2.5 = 18.5.
 The yaxis window settings are:y: [0, 20, 2]. Use ymin = 0 so we can see the xaxis. Use ymax = 20 since we are using an even spacing for the ticks.
 Now that we have the appropriate window settings, we can draw the axes for the graph.
 Sketch the graph.
 Now we'll draw the graph using technology. Adjust the window settings as shown.
 Next, input the function.
 Finally, draw the graph.

Section 2:
DAY TWO: Examples 5  7 (23 minutes) 

Lecture 5 

08:19 

Example 5: Finding a Trigonometric Function from its Graph  Determine the trigonometric function corresponding to each graph. In part A, write a cosine function for the graph shown.
 We are asked to find a cosine function. Before we begin, draw a rectangle around the first cosine shape.
 The minimum value of the graph is 8, and the maximum value of the graph is 10.
 Use the amplitude formula, a = (max – min)/2, to calculate the amplitude.
 Plug in 10 for the maximum value, and 8 for the minimum value.
 This gives us an amplitude of 9.
 Use the vertical displacement formula, d = (min + max)/2 to find the dvalue.
 Plug in 8 for the minimum value, and 10 for the maximum value.
 This gives us a vertical displacement of 1 unit up.
 By inspection, we can see that the period of the graph is 3 units.
 Use b = 2π/P to get the bvalue.
 Plug in 3 for the period. This gives us b = 2π/3.
 The graph has a phase shift of 1 unit right, so the cvalue is +1.
 Plug the parameters into the cosine formula to get y = 9cos(2π/3(x1))+1.
 Now we'll move on to part B.
 We are asked to find a sine function. Before we begin, draw a rectangle around the sine shape closest to the yaxis.
 Use the amplitude formula, a = (max – min)/2, to calculate the amplitude.
 Plug in 1 for the maximum value, and 5 for the minimum value.
 This gives us an amplitude of 3.
 Use the vertical displacement formula, d = (min + max)/2 to find the dvalue.
 Plug in 5 for the minimum value, and 1 for the maximum value.
 This gives us a vertical displacement of 2 units down.
 By inspection, we can see that the period of the graph is 8 units.
 Use b = 2π/P to get the bvalue.
 Plug in 8 for the period.
 The bvalue is π/4.
 The graph has a phase shift of 2 units left, so the cvalue is 2.
 Plug the parameters into the sine formula to get y = 3sin(π/4(x+2))2.
 Now we'll move on to part C.
 We are asked to find a cosine function. Before we begin, draw a rectangle around the first cosine shape.
 Use the amplitude formula, a = (max – min)/2, to calculate the amplitude.
 Plug in 9 for the maximum value, and 3 for the minimum value.
 This gives us an amplitude of 6.
 Use the vertical displacement formula, d = (min + max)/2 to find the dvalue.
 Plug in 3 for the minimum value, and 9 for the maximum value.
 This gives us a vertical displacement of 3 units up.
 By inspection, the period of the graph is 16 units.
 Use b = 2π/P to get the bvalue.
 Plug in 16 for the period.
 The π's cancel, giving us b = π/8.
 The graph has a phase shift of 8 units right, so the cvalue is +8.
 Plug the parameters into the cosine formula to get y = 6cos(π/8(x8))+3.
 Now we'll move on to part D.
 We are asked to find a sine function. Before we begin, draw a rectangle around the first sine shape.
 By inspection, we can see that the midline exists at d = 50.
 Also by inspection, we can see that the amplitude is 200 units.
 The distance between the given points is 1125 units. This corresponds to 1.25 cycles on the graph.
 Divide 1125 by 1.25 to get the length of 1 cycle, 900 units.
 Use b = 2π/P to get the bvalue.
 Plug in 900 for the period.
 The bvalue is π/450.
 There is a phase shift of 300 units right, so c= +300.
 Plug the parameters into the sine formula to get y = 200sin(π/450(x300))50.


Lecture 6 

07:22 

Example 6: Trigonometric Functions (challenging problems)  If the transformation g(θ)  3 = f(2θ) is applied to the graph of f(θ) = sinθ, find the new range.
 We'll begin by drawing the graph of y = sinθ to use as a reference.
 Rewrite the transformation as g(θ) = f(2θ) + 3.
 The bvalue is 2, so there is a horizontal stretch by a factor of 1/2.
 There is also a vertical displacement 3 units up.
 The new range is between 2 and 4, inclusive. Remember that the bracket notation [2, 4] means the same thing as 2 ≤ y ≤ 4.
 In Part B, find the range of f(θ) = ksin(θ – π/4) – 3.
 We'll begin by drawing the graph of y = sinθ to use as a reference.
 The amplitude of the transformed graph is k.
 We don't have a numerical value for the amplitude, so it could be anything!
 For the purposes of illustration, we'll assign a numerical value of 2.5 for k.
 There is a vertical displacement of 3 units down.
 There is a phase shift π/4 units right.
 Note: The phase shift has no impact on the range of the graph.
 The minimum value of the graph occurs at 3  k.
 The maximum value of the graph occurs at 3 + k.
 The range is: 3k ≤ y ≤ 3+k.
 Now we'll move on to Part C. If the range of y = 3cosθ + d is [4, k], determine the values of d and k.
 We'll begin by drawing the graph of y = cosθ to use as a reference.
 The amplitude of the graph is 3 units.
 The minimum value of the graph is 4, so move the graph down until the lowest point is 4.
 The graph needs to be moved down 1 unit to have a minimum value at 4.
 The vertical displacement is 1 unit down, so d = 1.
 The maximum value is 2, so this is the value of k.
 The equation is y = 3cosθ  1, and the range is [4, 2]. So, the final answer is d = 1, k = 2.
 In Part D, state the range of f(θ)  2 = msin(2θ) + n.
 We'll begin by drawing the graph of y = sinθ to use as a reference.
 The amplitude of the graph is m units.
 For the purposes of illustration, we'll assume m = 1.5.
 There is a horizontal stretch by a factor of 1/2. Note that the range is unaffected by the horizontal stretch.
 The vertical displacement of the graph is n units, so the minimum becomes m + n, and the maximum becomes m + n.
 For the purposes of illustration, we'll assume n = 1.
 Rewrite the transformation as: f(θ) = msin(2θ) + n + 2.
 There is an additional vertical displacement 2 units up.
 When the graph is moved 2 units up, all points on the graph must have 2 added to them. The minimum becomes m + n + 2, and the maximum becomes m + n + 2.
 The range of the graph is: m+n+2 ≤ y ≤ m+n+2.
 Now we'll move on to Part E. The graphs of f(θ) and g(θ) intersect at the points (π/8, root(2)/2) and (5π/8, root(2)/2). If the amplitude of each graph is quadrupled, determine the new points of intersection.
 Let's draw some graphs to visualize the information from the question.
 Quadruple the amplitude of the graph by multiplying all of the yvalues by 4.
 root(2)/2 × 4 = 2root(2). –root(2)/2×4 = 2root(2).
 These are the ycoordinates of the transformed points.


Lecture 7 

06:30 

Example 7: Trigonometric Functions (more challenging problems)  If the point (π/2, 2) lies on the graph of f(θ) = acos(θπ/4)4, find the value of a.
 We know the point (π/2, 2) exists on the graph. Plug in π/2 for θ and 2 for f(θ).
 This gives us 2 = acos(π/2 – π/4) – 4.
 Add 4 to both sides and subtract the radian fractions to get 2 = acos(π/4).
 cos(π/4) evaluates to root(2)/2.
 cross multiply to get 4 =root(2)a.
 Divide both sides of the equation by root(2) to get a = 4/root(2).
 Rationalize the denominator. Multiply the numerator and denominator by root(2).
 Multiply the fractions.
 Simplify. The value of a is 2root(2).
 In Part B, find the yintercept of f(θ) = 3cos(kθ + π/2) – b.
 To find the yintercept, set θ = 0.
 Simplify.
 cos(π/2) = 0.
 Simplify further to get –b.
 The yintercept occurs at the point (0, b).
 Now we'll move on to Part C. The graphs of f(θ) and g(θ) intersect at the point (m, n). Find the value of f(m) + g(m).
 Bring up the graph showing f(θ)and g(θ).
 f(m) means "What is the yvalue of function f when x = m?".
 When x = m...
 The ycoordinate is n. This is written as f(m) = n.
 g(m) means "What is the yvalue of function g when x = m?".
 When x = m...
 The ycoordinate is n. This is written as g(m) = n.
 Now evaluate f(m) + g(m).
 Both f(m) and g(m) are equivalent to n.
 Add to get the answer, 2n.
 Now we'll move on to Part D. The graph of f(θ) = kcosθ is transformed to the graph of g(θ) = bcosθ by a vertical stretch about the xaxis. If the point (5π/6, 3root(3)k/10) exists on the graph of g(θ), state the vertical stretch factor.
 Start with g(θ) = bcosθ.
 Plug in 3root(3)k/10 for g, and plug in 5π/6 for θ.
 cos(5π/6) = root(3)/2.
 Isolate b by dividing both sides by –root(3)/2.
 Multiply the numerator by the reciprocal of the denominator.
 Simplify.
 Reduce the fraction to get 3/5k.
 We can find the vertical stretch factor by dividing the amplitudes of each graph.
 The vertical stretch factor is the amplitude of the transformed graph divided by the amplitude of the original graph.
 The amplitude of the transformed graph is b, and the amplitude of the original graph is k.
 The value of b is 3/5k.
 The k's cancel, giving us the vertical stretch factor, 3/5. This is the answer.

Section 3:
DAY THREE: Examples 8  12 (26 minutes) 

Lecture 8 

03:34 

Example 8: Trigonometry Applications (pendulum)  The graph shows the height of a pendulum bob as a function of time. One cycle of a pendulum consists of two swings  a right swing and a left swing. In Part A, write a function that describes the height of the pendulum bob as a function of time.
 In the animation, we can see that the height of the pendulum bob oscillates between a minimum of 4 cm and a maximum of 8 cm.
 The midline corresponds to a vertical displacement 6 m up.
 The amplitude is 2 m.
 The period is 2 seconds.
 Use the formula b = 2π/P to get the bvalue.
 Plug in 2 for the period.
 The bvalue is equivalent to π.
 The graph has no phase shift.
 Plug the parameters into the general cosine formula y = acosb(θ  c) + d.
 The independent variable is time, and the dependent variable is height. The function is h(t) = 2cos(π*t) + 6.
 Now we'll move on to Part B. If the period of the pendulum is halved, how will this change the parameters in the function you wrote in part (a)?
 Bring up the graph we drew in Part A.
 If the period of the pendulum is halved, this has the same effect as a horizontal stretch by a factor of ½.
 The new bvalue is 2π.
 The bparameter is doubled when the period is halved. The a, c, and d parameters remain the same.
 Now we'll move on to Part C. If the pendulum is lowered so its lowest point is 2 cm above the ground, how will this change the parameters in the function you wrote in part (a)?
 Bring up the graph we drew in Part A.
 Move the graph down until the lowest point is 2 m above the ground.
 The dparameter decreases by 2 units, giving us d = 4. All other parameters remain unchanged.


Lecture 9 

03:24 

Example 9: Trigonometry Applications (wind turbine)  A wind turbine has blades that are 30 m long. An observer notes that one blade makes 12 complete rotations (clockwise) every minute. The highest point of the blade during the rotation is 105 m. Using Point A as the starting point of the graph, draw the height of the blade over two rotations.
 Before we can draw the graph, we need to find the duration of one complete rotation.
 In 60 seconds, there are 12 rotations. This gives us: 60 seconds/12 rotations, which is equal to 5 seconds/rotation. This means the period is 5 seconds.
 We want to graph two rotations, so draw tick marks on the taxis at 5 s and 10 s.
 The maximum height is 105 m. The blade is 30 m long, so we can get the minimum height by subtracting two blade lengths. the minimum height is 105 m  2×(30 m) = 45 m.
 In the animation, we can see that the height oscillates between a minimum value of 45 m and maximum value of 105 m. The period of one complete rotation is 5 seconds.
 In Part B, write a function that corresponds to the graph.
 Bring up the graph from Part A.
 The vertical displacement is 75 m.
 The amplitude is 30 m.
 The period is 5 seconds.
 The bvalue is 2π/5.
 There is no phase shift, so the cvalue is zero.
 Plug the parameters into the general cosine formula y = acosb(θ  c) + d.
 The independent variable is time, and the dependent variable is height.
 The function is h(t) = 30cos(2π/5 t) + 75.
 In Part C, do we get a different graph if the wind turbine rotates counterclockwise?
 If the wind turbine rotates counterclockwise, we still get the same graph.


Lecture 10 

02:24 

Example 10: Trigonometry Applications (helicopter)  A person is watching a helicopter ascend from a distance 150 m away from the takeoff point. In Part A, write a function, h(θ), that expresses the height as a function of the angle of elevation. Assume the height of the person is negligible.
 Begin with tanθ = opp/adj.
 The opposite side is the height, h, and the adjacent side is 150 m.
 Crossmultiply to get h = 150tanθ.
 This can be written using function notation as h(θ) = 150tanθ.
 In Part B, draw the graph, using an appropriate domain.
 Bring up our work from Part A.
 We can restrict the domain to 0° ≤ θ < 90°, since the helicopter will always be in front of the observer.
 In Part C, explain how the shape of the graph relates to the motion of the helicopter.
 Bring up the graph we drew in Part B.
 The observer must tilt their head upwards to see the helicopter rising. The angle of elevation increases quickly at first, but slows down as the helicopter reaches greater heights.
 The helicopter will never be behind the observer if it goes straight up, so the maximum angle of elevation is 90° (exclusive). The angle never actually reaches 90°.


Lecture 11 

08:12 

Example 11: Trigonometry Applications (mass oscillating on a spring)  A mass is attached to a spring 4 m above the ground and allowed to oscillate from its equilibrium position. The lowest position of the mass is 2.8 m above the ground, and it takes 1 s for one complete oscillation. In Part A, draw the graph for two full oscillations of the mass.
 Each complete oscillation of the mass takes 1 second. We want to graph two cycles, so draw tick marks at 1 s and 2 s.
 The mass begins at a height of 4 m. The minimum height is 2.8 m, and by symmetry, the maximum height is 5.2 m.
 The animation shows the mass oscillating between its minimum height of 2.8 m, and its maximum height of 5.2 m. The period is 1 second.
 In Part B, write a sine function that gives the height of the mass above the ground as a function of time.
 Bring up our work from Part A.
 The vertical displacement is 4 m.
 The amplitude is 1.2 m.
 The period is 1 s.
 The sine shape we selected is upsidedown, so remember to introduce a negative when we create the function.
 The bvalue is 2π.
 The cvalue is zero.
 Plug the parameters into the general sine formula: y = asinb(θ  c) + d.
 The independent variable is time, and the dependent variable is height.
 The function is h(t) = 1.2sin(2π*t) + 4.
 In Part C, calculate the height of the mass after 1.2 seconds.
 Bring up the graph and its function.
 When t = 1.2 s...
 The height is found by evaluating h(1.2).
 In Method 1 we'll evaluate the function.
 Plug in 1.2 s for the time.
 Evaluate to get the height 2.86 m. Make sure the calculator is in radian mode.
 We can also use technology to answer the question.
 Set the view window to: x: [0, 2, 1] and y: [0, 6, 1].
 Graph the function and then use 2nd  TRACE – VALUE to evaluate the function when x = 1.2.
 This gives us a height of 2.86 m, which is the same result we obtained in Method 1.
 Now we'll move on to Part D. In one oscillation, how many seconds is the mass lower than 3.2 m?
 Bring up the graph and its function.
 Draw the line y = 3.2.
 We are trying to find the length of time that the graph dips under this line (within the first cycle).
 We can find the points of intersection by solving the equation: 3.2 = 1.2sin(2π*t) + 4.
 Replace h(t) with the height 3.2 m...
 and solve for the times that make the function evaluate to 3.2.
 Doing this calculation algebraically is beyond the scope of this course, but we can use the graphing calculator to find the times quite easily.
 Set the view window to: x: [0, 2, 1] and y: [0, 6, 1] and graph each line.
 Use 2nd  TRACE – INTERSECT to find the points of intersection. The times are 0.12 s and 0.38 s.
 Subtract the two times to get the duration that the mass is lower than 3.2 m. 0.38 s  0.12 s = 0.26 s, which is the answer.


Lecture 12 

08:42 

Example 12: Trigonometry Applications (Ferris Wheel)  A Ferris wheel with a radius of 15 m rotates once every 100 seconds. Riders board the Ferris wheel using a platform 1 m above the ground. In Part A, draw the graph for two full rotations of the Ferris wheel.
 It takes 100 seconds for one complete rotation, so draw tick marks on the taxis at 25 s intervals up to 200 s.
 Riders board the Ferris wheel at a height of 1 m. The radius of the wheel is 15 m, so the midline is at 16 m (1 + 15), and the maximum height is at 31 m (1 + 15 + 15).
 The animation shows the rider oscillating between the minimum height of 1 m and the maximum height of 31 m. Two rotations of the Ferris wheel are shown, and each rotation has a period of 100 s.
 In Part B, write a cosine function that gives the height of the rider as a function of time.
 Bring up the graph from Part A.
 The vertical displacement is 16 m.
 The amplitude is 15 m.
 The period is 100 s.
 The cosine shape we selected is upsidedown, so remember to introduce a negative when we create the function.
 The bvalue is π/50.
 The cvalue is zero.
 Plug the parameters into the general cosine formula: y = acosb(θ  c) + d.
 The independent variable is time, and the dependent variable is height. The function is h(t) = 15cos(π/50t) + 16.
 In Part C, calculate the height of the rider after 1.6 rotations of the Ferris wheel.
 Bring up the graph and its function.
 We need to find the time required for 1.6 rotations.
 Multiply 1.6 rotations by the conversion multiplier 100 s/ 1 rotation. This gives us 160 s.
 When t = 160 s.
 The height is found by evaluating h(160).
 The first way to find the height is to evaluate the function.
 Plug in 160 s for the time.
 This evaluates to 28.14 m.
 The second way is to use technology.
 Set the view window to: x: [0, 200, 25] and y: [0, 32, 2].
 Graph the function and then use 2nd  TRACE – VALUE to evaluate the function when x = 160.
 This gives us y = 28.14, which is the same result we obtained in Method 1.
 In one rotation, how many seconds is the rider higher than 26 m?
 Bring up the graph and its function.
 Draw the line y = 26.
 We are trying to find the length of time that the graph rises over this line (within the first cycle).
 We can find the points of intersection by solving the equation: 26 = 15cos(π/50t) + 16.
 Replace h(t) with the height 26 m...
 and solve for the times that make the function evaluate to 26.
 Now we'll use technology to find the points of intersection.
 Set the view window to: x: [0, 200, 25] and y: [0, 32, 2] and graph each line.
 Use 2nd  TRACE – INTERSECT (two times) to find the points of intersection. The times are 36.61 s and 63.39 s.
 Subtract the two times to get the duration that the mass is higher than 26 m. 63.39 s  36.61 s = 26.78 s, which is the answer.

Section 4:
DAY FOUR: Examples 13  17 (37 minutes) 

Lecture 13 

13:46 

Example 13: Trigonometry Applications (daylight hours)  The following table shows the number of daylight hours in Grande Prairie. In Part A, Convert each date and time to a number that can be used for graphing.
 When we model a realworld situation with mathematics, it is often necessary to convert measurements, like dates and times, to numeric values that can be used for graphing and analysis.
 We can convert dates to day numbers using the following rules:
 January 1 is day zero.
 Dates in the previous year are negative.
 If no year is specified, assume a nonleap year.
 Let's look at the days surrounding January 1.
 December 21 is 11 days back (into the previous year), so we'll use a day number of 11.
 A consequence of assigning January 1 as day zero is that all days in the future will be offset by one number.
 To account for this offset, remember to subtract 1 when calculating day numbers after January 1.
 The calculation shows how we can get the day number for March 21.
 There are 31 days in January, 28 days in February, and 21 days in March (to get to Mar. 21).
 Subtract 1 to account for the Jan. 1 = day zero offset.
 The day number is 79.
 The day number for June 21 is 171.
 The day number for September 21 is 263.
 And the day number for December 21 is 354. We have now converted all of the dates to day numbers.
 Time that is expressed using hours and minutes can't be used for graphing purposes. It must be converted to decimal time.
 6 hours and 46 minutes is equivalent to 6.77 hours.
 Leave the hour value unchanged.
 Divide the minutes by 60 to get the decimal hour equivalent. 46/60 = 0.77.
 This gives us 6.77 as the decimal hour.
 12 hours and 17 minutes is equivalent to 12.28 hours.
 17 hours and 49 minutes is equivalent to 17.82 hours.
 12 hours and 17 minutes is equivalent to 12.28 hours.
 And finally, 6 hours and 46 minutes is equivalent to 6.77 hours. We have now converted all of the times to decimal hours.
 In Part B, draw the graph for one complete cycle.
 Bring up the table we completed in Part A.
 The day number, n, is the independent variable. It goes on the xaxis. The number of daylight hours, d(n), depends on the day number, so it goes on the yaxis.
 One year is 365 days, so the xaxis should have tick marks that will let us view one complete year.
 In this example, we'll use x: [50, 400, 50]. Note: We should start at 50 instead of 0 so we can include the point with a day number of 11.
 There are 24 hours in a day, so the yaxis can be set to y: [0, 24, 4].
 Use the table to plot the points.
 Now draw the graph.
 In Part C, write a cosine function that relates the number of daylight hours, d, to the day number, n.
 Bring up the table and the graph.
 The minimum value of the graph is 6.77 hours, and the maximum value is 17.82 hours.
 Use the amplitude formula, a = (maxmin)/2.
 Plug in 17.82 for the maximum, and 6.77 for the minimum.
 This gives us an amplitude of a = 5.525.
 Now use the vertical displacement formula, d = (min+max)/2.
 This gives us d = 6.77 + 17.82/2.
 The vertical displacement is d = 12.295.
 The period is 365 days.
 The bvalue is 2π/365.
 There is a phase shift 11 days back, so c = 11.
 Plug the parameters into the general cosine formula: y = acosb(θ  c) + d.
 The independent variable is the day number (n), and the dependent variable is the daylight hours (d).
 The function is d(n) = 5.525cos(2π/365(n+11))+12.295. Remember the negative since the cosine shape is upsidedown.
 In Part D, how many daylight hours are there on May 2?
 Bring up the graph and its function.
 Find the day number for May 2: (31 + 28 + 31 + 30 + 2)  1 = 121.
 Remember to subtract 1 to account for the 1day offset (Jan. 1 is day zero).
 On day number 121, the number of daylight hours can be found by evaluating d(121).
 In Method 1, we'll evaluate d(121) directly.
 Use 121 as the day number in our function.
 This gives us 15.86 hours.
 In Method 2, we'll use technology.
 Set the view window to x: [50, 400, 50] , y: [0, 24, 4] and draw the graph.
 Use 2nd  TRACE  VALUE (x = 121) to get 15.86 daylight hours. This is the same result we obtained in Method 1.
 Now we'll move on to Part E. In one year, approximately how many days have more than 17 daylight hours?
 Bring up the graph and its function.
 We are trying to find the number of days the graph is above the line y = 17.
 To find the points of intersection, we need to solve the equation shown. Plug in 17 for the number of daylight hours.
 Use technology to find the intersection points, 139.45 and 203.55.
 Subtract. 203.55  139.45 ≈ 64 days. Each year, Grande Prairie has approximately 64 days with more than 17 hours of daylight.


Lecture 14 

11:48 

Example 14: Trigonometry Applications (tidal height)  The highest tides in the world occur between New Brunswick and Nova Scotia, in the Bay of Fundy. Each day, there are two low tides and two high tides. The chart below contains tidal height data that was collected over a 24hour period. In Part A, convert each time to a decimal hour. Note that actual tides at the Bay of Fundy are 6 hours and 13 minutes apart due to daily changes in the position of the moon. In this example, we will use 6 hours for simplicity.
 The times are given based on a 12hour clock. This won't work for analysis, so we need to convert each time to the decimal hours elapsed since midnight.
 2:12 AM is equal to 2.20 hours.
 Leave the hour value unchanged.
 Divide the minutes by 60 to get the decimal hour equivalent.12/60 = 0.20.
 The decimal hour is 2.20.
 8:12 AM = 8.20 hours.
 2:12 PM should be converted to 24hour time by adding 12 hours. This gives us 14:12, which is equivalent to 14.20 decimal hours.
 8:12 PM should be converted to 24hour time by adding 12 hours. This gives us 20:12, which is equivalent to 20.20 decimal hours.
 In Part B, graph the height of the tide for half a day. In other words, from low tide to low tide.
 Bring up the table we completed in Part A.
 Time is the independent variable, so it goes on the xaxis. Height is the dependent variable, so it goes on the yaxis.
 The length of one day is 24 hours. Set the view window x: [0, 24, 4].
 The height of the water varies from 3.48 m to 13.32 m. Set the view window y: [0, 16, 2].
 Plot the points.
 We can draw a complete cycle using just three points, so the fourth point is not required.
 In Part C, write a cosine function that relates the height of the water to the elapsed time.
 Bring up the table and the graph.
 The minimum is 3.48, and the maximum is 13.32.
 The amplitude is a = (maxmin)/2.
 Plug in 13.32 for the maximum, and 3.48 for the minimum.
 The amplitude is 4.92.
 The vertical displacement is d = (min+max)/2.
 Plug in 3.48 for the minimum and 13.32 for the maximum.
 This gives us d = 8.40.
 The period of the graph is 12 hours. We can find this by subtracting the leftmost time (2.20) from the rightmost time (14.20).
 The bvalue is π/6.
 There is a phase shift of 2.20 hours right.
 Plug the parameters into the general cosine formula: y = acosb(θ  c) + d.
 The independent variable is the time (t), and the dependent variable is the water height (h).
 The function is h(t) = 4.92cos(π/6(t2.20)) + 8.40. The cosine graph is upsidedown, so use a negative.
 In Part D, what is the height of the water at 6:09 AM?
 Bring up the graph and its function.
 Before we can answer this question, convert 6:09 AM to a decimal hour.
 6:09 AM = 6.15 hours.
 Leave the hour value unchanged.
 Divide the minutes by 60 to get the decimal hour equivalent.9/60 = 0.15.
 The decimal hour is 6.15.
 We can find the height of the water at 6.15 hours by evaluating h(6.15).
 Method 1: Evaluate h(6.15) directly.
 Plug in 6.15 for the time.
 Evaluate to get 10.75 m. This is the height of the water.
 In Method 2, use technology to find the height of the water.
 Set the view window to x: [0, 24, 4] , y: [0, 16, 4] and draw the graph.
 Use 2nd  TRACE  VALUE (x = 6.15) to get a tidal height of 10.75 m. This is the same result we obtained in Method 1.
 In Part E, For what percentage of the day is the height of the water greater than 11 m?
 For this question, draw the second cycle of the graph.
 Draw the line y = 11. We need to determine the total time the graph is above this line.
 To find the points of intersection, we need to solve the equation shown. The height of the water is 11 m, so replace h(t) with 11.
 Use technology to find the intersection points. They occur at the times: 6.26 h, 10.14 h, 18.26 h, and 22.14 h.
 Subtract the first two times to get 3.88 hours.
 Subtract the next two times to get 3.88 hours.
 The height of the water is greater than 11 m for 7.76 h. This can be found by adding the two times (3.88 h + 3.88 h).
 Divide 7.76 hours by 24 hours to get 0.3233.
 The water is higher than 11 m for approximately 32.3% of the day.


Lecture 15 

03:08 

Example 15: Trigonometry Applications (ecosystem populations)  A wooded region has an ecosystem that supports both owls and mice. Owl and mice populations vary over time according to the equations shown, where O is the population of owls, M is the population of mice, and t is the time in years. In Part A, Graph the population of owls and mice over six years.
 We'll begin by graphing the owl population.
 The midline is 250.
 The amplitude is 50.
 The period is 6 years. Draw a sine graph in the box shown.
 There is a phase shift of 1.5 years forward. Move the graph.
 Now we'll graph the mouse population.
 The midline is 12000.
 The amplitude is 4000.
 The period is 6 years. Draw a sine graph in the box.
 There is no phase shift.
 Extend the graph so it fills its domain.
 Bring back the owl graph so we can compare the owl graph with the mouse graph.
 In Part B, describe how the graph shows the relationship between owl and mouse populations.
 Bring up the graphs we completed in Part A.
 Both the owl and mouse populations are increasing in this region.
 As the owl population continues to increase, the mouse population decreases.
 Both the owl and mouse populations are decreasing in this region.
 As the owl population continues to decrease, the mouse population increases.
 Finally, we return to the original state. Both the owl and mouse populations are increasing in this region.


Lecture 16 

04:30 

Example 16: Trigonometry Applications (clock)  The angle of elevation between the 6:00 position and the 12:00 position of a historical building's clock, as measured from an observer standing on a hill, is π/444. The observer also knows that he is standing 424 m away from the clock, and his eyes are at the same height as the base of the clock. The radius of the clock is the same as the length of the minute hand. If the height of the minute hand's tip is measured relative to the bottom of the clock, what is the height of the tip at 5:08, to the nearest tenth of a metre?
 Let's outline a strategy that can be used to solve the problem.
 First, use basic trigonometry to calculate the height of the clock using the given diagram.
 The radius of the clock is the same as the length of the minute hand. Divide the clock height by 2 to get the length of the minute hand.
 Model the movement of the minute hand with a trigonometric function.
 Finally, use the function from step 3 to calculate the height of the minute hand's tip at 5:08.
 Let's begin with the first step in our strategy. Redraw the triangle.
 Use tanθ = opp/adj.
 Replace the angle with π/444, and replace the adjacent side with 424.
 Cross multiply to get the opposite side.
 The opposite side is 3 m.
 Now we'll move on to the second step in our strategy. The radius of the clock is the same as the length of the minute hand. Divide the clock height by 2 to get the length of the minute hand.
 The minute hand is 1.5 m long.
 Next, model the movement of the minute hand with a trigonometric function.
 time goes on the xaxis, and height goes on the yaxis.
 Introduce ticks on the time axis.
 Introduce ticks on the height axis.
 The animation shows the height of the minute hand's tip over one hour.
 The vertical displacement is 1.5 m.
 The amplitude is 1.5 m.
 The period is 60 minutes, and the bvalue is π/30.
 There is no phase shift.
 The function is h(t) = 1.5cos(π/30t)+1.5.
 Finally, use the function from step 3 to calculate the height of the minute hand's tip at 5:08.
 At 5:08, the tvalue is 8 minutes, and the height is found by plugging 8 into the cosine function.
 Plug in 8 minutes for the time.
 The height of the minute hand's tip at 5:08 is 2.5 m.


Lecture 17 

03:37 

Example 17: Trigonometry Applications (Ferris Wheel)  Shane is on a Ferris wheel, and his height can be described by the equation h(t) = 9cos(π/30t) + 10. Tim, a baseball player, can throw a baseball with a speed of 20 m/s. If Tim throws a ball directly upwards, the height can be determined by the equation h(t) = 4.905t^{2} + 20t + 1. If Tim throws the baseball 15 seconds after the ride begins, when are Shane and the ball at the same height?
 Let's outline a strategy that can be used to solve the problem.
 First, graph the motion of the Ferris wheel and the baseball.
 Next, find a way to deal with the 15 second delay.
 Finally, use a graphing calculator to find the points where the two graphs intersect.
 Let's begin by graphing Shane's height on the Ferris wheel.
 The amplitude is 9 m, and the graph is upsidedown.
 The vertical displacement is 10 m.
 Introduce tick marks on the height axis.
 The period is 60s.
 There is no phase shift.
 Introduce tick marks on the time axis.
 We can now draw the cosine graph corresponding to the Ferris wheel.
 Now use a graphing calculator (or complete the square) to graph the height of the baseball.
 The motion of the ball can be graphed as shown.
 Now find a way to deal with the 15 second delay.
 If the ball is thrown 15 seconds after the Ferris wheel begins, translate the graph of the ball 15 units right.
 As a function, we can represent this horizontal translation with h(t – 15).
 Use a graphing calculator to find the points where the two graphs intersect.
 The first intersection point occurs at 15.6 seconds.
 The second intersection point occurs at 18.4 seconds.
 Shane is the same height as the baseball at 15.6 s and 18.4 s.

Section 5:
Other Lesson Materials 

Lecture 18 

22 pages 

Lecture 19 

00:54 

Trigonometric Functions II: Introduction  Welcome to Trigonometric Functions II. This topic will take four days to complete.
 In Day 1 we will learn the difference between trigonometric functions of an angle and trigonometric functions of a real number.
 In Day 2 we will use our knowledge of trigonometric functions to solve challenging problems.
 In Day 3 we will study trigonometric functions in realworld contexts.
 In Day 4 we will continue to study trigonometric functions in realworld contexts.


Lecture 20 

00:39 

Trigonometric Functions II: Summary  You have now completed Trigonometric Functions II. Over the past four days you have learned:
 the difference between trigonometric functions of an angle and trigonometric functions of a real number.
 how to determine an appropriate view window for a trigonometric function.
 how to solve abstract problems involving trigonometric functions.
 how to model realworld scenarios with trigonometric functions.


Quiz 1 
Trigonometric Functions Two

11 questions 
Full curriculum

Its great
This helped me so much with my math ia for IB thank youu