Section 1:
DAY ONE: Examples 1  6 (38 minutes) 

Lecture 1 

08:08 

Example 1: Primary Trigonometric Equations: Solve with the Unit Circle  Find all angles in the domain 0 ≤ θ ≤ 2π that satisfy the given equation. Write the general solution. In Part A, we'll solve sinθ = root(3)/2 using the unit circle.
 Bring up the unit circle.
 sinθ = root(3)/2 can be paraphrased as "which angle(s) correspond to a ycoordinate of root(3)/2?"
 Draw the line y = root(3)/2 and highlight the points of intersection.
 The answer is θ = π/3 and 2π/3.
 The solution we just found is for 0 ≤ θ ≤ 2π.
 In the general solution, we need to account for all coterminal angles.
 If we start at π/3 and rotate counterclockwise, we get coterminal angles at 7π/3 and 13π/3.
 The first part of the general solution is θ = π/3 + n(2π), nεI.
 Now position the terminal arm at 2π/3.
 If we start at 2π/3 and rotate counterclockwise, we get coterminal angles at 8π/3 and 14π/3.
 The second part of the general solution is θ = 2π/3 + n(2π), nεI.
 We have now found the specific solution for the domain where 0 ≤ θ ≤ 2π, and the general solution for all real numbers.
 In Part B, we'll solve cosθ = 1/2 using the unit circle.
 Bring up the unit circle.
 cosθ = 1/2 can be paraphrased as "which angle(s) correspond to an xcoordinate of 1/2?"
 Draw the line x = 1/2 and highlight the points of intersection.
 The answer is θ = 2π/3 and 4π/3.
 The solution we just found is for 0 ≤ θ ≤ 2π.
 In the general solution, we need to account for all coterminal angles.
 If we start at 2π/3 and rotate counterclockwise, we get coterminal angles at 8π/3 and 14π/3.
 The first part of the general solution is θ = 2π/3 + n(2π), nεI.
 Now position the terminal arm at 4π/3.
 If we start at 4π/3 and rotate counterclockwise, we get coterminal angles at 10π/3 and 16π/3.
 The second part of the general solution is θ = 4π/3 + n(2π), nεI.
 We have now found the specific solution for the domain where 0 ≤ θ ≤ 2π, and the general solution for all real numbers.
 In Part C, we'll solve tanθ = 0 using the unit circle.
 Bring up the unit circle.
 Recall that tanθ = sinθ/cosθ.
 tanθ = 0 when the numerator, sinθ, is 0.
 Draw the line y = 0 and highlight the intersection points with the unit circle.
 The solution is θ = 0, π, and 2π for the domain 0 ≤ θ ≤ 2π.
 In the general solution, we need to account for all coterminal angles.
 In this example, the solutions are a halfrotation apart. This means we only need one equation for the general solution.
 The general solution is θ = nπ, nεI.
 We have now found the specific solution for the domain where 0 ≤ θ ≤ 2π, and the general solution for all real numbers.
 In Part D, we'll solve tan^{2}θ = 1 using the unit circle.
 Before we solve for θ, we need to square root both sides of the equation.
 This gives us tanθ = ±1.
 Recall that tanθ = sinθ/cosθ.
 tanθ = ±1 when both sinθ and cosθ have the same magnitude.
 The answer is θ = π/4, 3π/4, 5π/4, and 7π/4 for the domain where 0 ≤ θ ≤ 2π.
 In this example, the solutions are a quarterrotation apart. This means we only need one equation for the general solution.
 The general solution is θ = π/4 + n(π/2), nεI.
 We have now found the specific solution for the domain where 0 ≤ θ ≤ 2π, and the general solution for all real numbers.


Lecture 2 

08:29 

Example 2: Primary Trigonometric Equations: Solve with Points of Intersection  Find all angles in the domain 0 ≤ θ ≤ 2π that satisfy the given equation. Write the general solution. In Part A, solve sinθ = 1/2 by graphing and finding the points of intersection.
 Begin by drawing the graph of y = sinθ.
 Next, draw the line y = 1/2.
 Determine the angles of the intersection points. They occur at π/6 and 5π/6. This is the specific solution for the domain 0 ≤ θ ≤ 2π.
 Before we find the general solution, extend the graph so we can see two complete cycles.
 If we add one period (2π) to the first solution at π/6, we get a repeat solution at 13π/6.
 The first part of the general solution is θ = π/6 + n(2π), nεI.
 If we add one period (2π) to the second solution at 5π/6, we get a repeat solution at 17π/6.
 The second part of the general solution is θ = 5π/6 + n(2π), nεI.
 If we zoom out, we can see that the points of intersection repeat infinitely.
 In Part B, solve sinθ = 1 by graphing and finding the points of intersection.
 Begin by drawing the graph of y = sinθ.
 Next, draw the line y = 1.
 Determine the angle of the intersection point.
 The intersection point occurs at 3π/2.
 The only solution over the domain 0 ≤ θ ≤ 2π is 3π/2.
 Before we find the general solution, extend the graph so we can see two complete cycles.
 If we add one period (2π) to the solution at 3π/2, we get a repeat solution at 7π/2.
 The general solution is θ = 3π/2 + n(2π), nεI.
 If we zoom out, we can see that the points of intersection repeat infinitely.
 In Part C, solve cosθ = root(2)/2 by graphing and finding the points of intersection.
 Begin by drawing the graph of y = cosθ.
 Next, draw the line y = root(2)/2.
 Determine the angles of the intersection points. They occur at 3π/4 and 5π/4. This is the specific solution for the domain 0 ≤ θ ≤ 2π.
 Before we find the general solution, extend the graph so we can see two complete cycles.
 If we add one period (2π) to the solution at 3π/4, we get a repeat solution at 11π/4.
 The first part of the general solution is θ = 3π/4 + n(2π), nεI.
 If we add one period (2π) to the solution at 5π/4, we get a repeat solution at 13π/4.
 The second part of the general solution is θ = 5π/4 + n(2π), nεI.
 If we zoom out, we can see that the points of intersection repeat infinitely.
 In Part D, solve cosθ = 2 by graphing and finding the points of intersection.
 Begin by drawing the graph of y = cosθ.
 Next, draw the line y = 2.
 The graph and line do not intersect, so there is no solution.
 In Part E, solve tanθ = sqrt(3) by graphing and finding the points of intersection.
 Begin by drawing the graph of y = tanθ.
 Next, draw the line y = sqrt(3).
 Determine the angles of the intersection points. They occur at 2π/3 and 5π/3. This is the specific solution for the domain 0 ≤ θ ≤ 2π.
 If we add one period (π) to the solution at 2π/3, we get a repeat solution at 5π/3.
 The general solution is θ = 2π/3 + nπ, nεI.
 If we zoom out, we can see that the points of intersection repeat infinitely.
 In Part F, solve tanθ = undefined by graphing and finding the points of intersection.
 Begin by drawing the graph of y = tanθ.
 The graph of y = tanθ is undefined at the asymptotes. The asymptotes occur at π/2 and 3π/2. This is the specific solution for the domain 0 ≤ θ ≤ 2π.
 If we add one period (π) to the solution at π/2, we get a repeat solution at 3π/2.
 The general solution is θ = π/2 + nπ, nεI.


Lecture 3 

10:49 

Example 3: Primary Trigonometric Equations: Solve with a Calculator  Find all angles in the domain 0° ≤ θ ≤ 360° that satisfy the given equation. Write the general solution. In Part A, solve sinθ = 1/2 nongraphically, using a calculator in degree mode.
 Before we find the solution in a calculator, let's look at the unit circle and the graph.
 From the unit circle, the solution is π/6 and 5π/6.
 If we graph y = sinθ and y = 1/2 we get the same result.
 Remember: Solving a trigonometric equation means:
 Find the value(s) of θ that make the left side...
 evaluate to the right side.
 We can solve this equation by using a calculator's inverse sine operation. Make sure the calculator is in degree mode.
 The result is 30°.
 On the left side, the inverse sine operation undoes the sine operation, leaving us with just θ.
 On the right side, sin^{1}(1/2) = 30°. The result of an inverse sine operation is always an angle.
 The following animation illustrates the role of sine and inverse sine in solving trigonometric equations.
 Start with the angle 30°.
 The sine of 30° is 1/2.
 The inverse sine operation returns the original angle.
 A calculator's inverse sine operation returns only one angle. We'll call this result the first solution of the trigonometric equation.
 We can use reasoning to find the value of the second angle.
 Draw a coordinate grid and label the axis angles.
 sinθ is positive in quadrants I & II.
 We can draw two triangles with an opposite side of 1 and a hypotenuse of 2.
 Each triangle has a reference angle of 30°.
 The second angle is located at 150°, measured in standard position.
 The solution over the domain 0° ≤ θ ≤ 360° is θ = 30° and 150°.
 The general solution is θ = 30° + n(360°), nεI and θ = 150° + n(360°), nεI.
 The general solution says that when the terminal arm of either 30° or 150° is rotated by a multiple of 360°, in either direction, we arrive at a coterminal angle that is another solution of the equation.
 The general solution gives us all angles that will make sinθ evaluate to 1/2.
 Starting from 30°, we get: sin30° = 1/2.
 sin390° = 1/2.
 sin750° = 1/2 and so on.
 Starting from 150°, we get: sin150° = 1/2.
 sin510° = 1/2.
 sin870° = 1/2 and so on.
 In Part B, solve cosθ = root(3)/2 nongraphically, using a calculator in degree mode.
 Take the inverse cosine of each side.
 The result is θ = 150°.
 It's possible that more than one angle will satisfy the equation, so we'll call 150° our first solution.
 cosθ is negative in quadrants II & III.
 We can draw two triangles with an adjacent side of –root(3) and a hypotenuse of 2.
 Both of the reference angles are 30°.
 The second solution is θ = 210°.
 The solution over the domain 0° ≤ θ ≤ 360° is θ = 150° and 210°.
 The general solution is θ = 150° + n(360°), nεI and θ = 210° + n(360°), nεI.
 The general solution gives us all angles that will make cosθ evaluate to –root(3)/2.
 Starting from 150°, we get cos150° = root(3)/2.
 cos510° = root(3)/2.
 cos870° = root(3)/2 and so on.
 Starting from 210°, we get cos210° = root(3)/2.
 cos570° = root(3)/2.
 cos930° = root(3)/2 and so on.
 In Part C, solve tanθ = 1 nongraphically, using a calculator in degree mode.
 Take the inverse tan of each side.
 The result is θ = 45°.
 It's possible that more than one angle will satisfy the equation, so we'll call 45° our first solution.
 tanθ is positive in quadrants I & III.
 We can draw two triangles.
 The reference angle in each triangle is 45°.
 The second solution is θ = 225°.
 The solution over the domain 0° ≤ θ ≤ 360° is θ = 45° and 225°.
 In the case of tan = 1, all of the solutions are exactly 180° apart. Only one equation is needed to describe this.
 The general solution is θ = 45° + n(180°).
 The general solution gives us all angles that will make tanθ evaluate to 1.
 tan45° = 1.
 tan225° = 1.
 tan405° = 1and so on.


Lecture 4 

03:04 

Example 4: Primary Trigonometric Equations: Solve with θIntercepts  Find all angles in the domain 0 ≤ θ ≤ 2π that satisfy the given equation. In Part A, solve sinθ = 1 using θintercepts.
 One way we could solve this equation is to graph both sides and find the points of intersection. This would yield one solution at π/2.
 An alternative way to solve this is to set the equation equal to zero, then find the θintercepts.
 Rewrite sinθ = 1.
 Bring all terms to the left side of the equation so it equals zero. Draw the graph.
 The θintercept occurs at π/2. This is the same result we obtained before.
 Notice that both techniques (intersection & θintercepts) give the same answer.
 Rearranging an equation using algebra does not change the solution of the equation.
 In Part B, solve cosθ = 1/2 using θintercepts.
 One way we could solve this equation is to graph both sides and find the points of intersection.
 The points of intersection occur at π/3 and 5π/3.
 An alternative way to solve this is to set the equation equal to zero, then find the θintercepts.
 Rewrite cosθ = 1/2.
 Multiply both sides of the equation by 2. Notice that in this intermediate step, we still have the same solutions.
 Bring all terms to the left side of the equation so it equals zero. The θintercepts are π/3 and 5π/3.
 Notice that in this final step, we still have the same solutions.
 The solutions are π/3 and 5π/3. This is the same result we obtained before.


Lecture 5 

03:43 

Example 5: Primary Trigonometric Equations: Multiple Solution Strategies  Solve cosθ = 1/2, where 0 ≤ θ ≤ 2π nongraphically, using the cos^{1} feature of a calculator.
 Get θ by itself by taking the inverse cosine of both sides.
 The result is 120°.
 cosθ is negative in quadrants II & III.
 Use symmetry to draw the angle in quadrant III.
 The second angle is 240°.
 The solution over the domain 0° ≤ θ ≤ 360° is 120° and 240°.
 Or, using radians, 2π/3 and 4π/3.
 In Part B, solve cosθ = 1/2 nongraphically, using the unit circle.
 Bring up the unit circle.
 We are trying to find the angles with an xcoordinate of 1/2.
 The solution is 2π/3 and 4π/3. This is the same result we found in Part A.
 In Part C, solve cosθ = 1/2 graphically, using the points of intersection.
 Draw the graph of y = cosθ.
 Draw y = 1/2. Place dots at the points of intersection.
 The solution is 2π/3 and 4π/3. This is the same result we obtained in parts a & b.
 In Part D, solve cosθ = 1/2 graphically, using θintercepts.
 We need to rearrange the equation so it equals zero.
 Multiply both sides of the equation by 2.
 Bring all terms to the left side so the equation equals zero.
 Draw the graph and place dots at the θintercepts.
 The solution is 2π/3 and 4π/3. This is the same result we obtained in parts a, b & c.


Lecture 6 

03:23 

Example 6: Primary Trigonometric Equations: Multiple Solution Strategies  Solve sinθ = 0.3, θ ε R nongraphically, using the sin^{1} feature of a calculator.
 Take the inverse sine of each side.
 The result is 17.46°. This angle is found by rotating the terminal arm clockwise 17.46°.
 sinθ is negative in quadrants III & IV.
 Use symmetry to draw the angle in quadrant III.
 The solution over the domain 0° ≤ θ ≤ 360° is 197.46° and 342.54°.
 Solve sinθ = 0.3 nongraphically, using the unit circle.
 Bring up the unit circle.
 The line y = 0.3 does not intersect the unit circle at a 30°45°60° triangle. Therefore, the unit circle is not useful to us for this question.
 In Part C, solve sinθ = 0.3 graphically, using points of intersection.
 In a graphing calculator, graph the functions: Y_{1} = sinθ and Y_{2} = 0.3.
 The angles of the points of intersection are θ = 197.46° and θ = 342.54° (Use the intersect feature of the calculator.)
 In Part D, solve sinθ = 0.3 graphically, using θintercepts.
 Rearrange the equation so it equals zero. This gives us sinθ + 0.30 = 0.
 In a graphing calculator, graph the function: Y_{1} = sinθ.
 The angles at the θintercepts are 197.46° and 342.54° (Use the zero feature of the calculator).

Section 2:
DAY TWO: Examples 7  12 (26 minutes) 

Lecture 7 

04:14 

Example 7: Reciprocal Trigonometric Equations: Solve with the Unit Circle  Find all angles in the domain 0 ≤ θ ≤ 2π that satisfy the given equation. Write the general solution. In Part A, solve secθ = 2 using the unit circle.
 An equation with reciprocal trigonometric ratios can be solved by rewriting it so it involves only primary trigonometric ratios.
 Rewrite secθ as 1/cosθ.
 Multiply both sides by cosθ to get 1 = 2cosθ.
 Divide both sides by 2 to get cosθ = 1/2.
 Bring up the unit circle.
 cosθ = 1/2 at 2π/3 and 4π/3.
 The general solution is 2π/3 + n(2π), nεI. and 4π/3 + n(2π), nεI.
 In Part B, solve cscθ = undefined using the unit circle.
 An equation with reciprocal trigonometric ratios can be solved by rewriting it so it involves only primary trigonometric ratios.
 Rewrite the equation as sinθ = 0.
 sinθ is the reciprocal of cscθ.
 Zero is the reciprocal of undefined.
 Bring up the unit circle.
 The solution is 0, π, and 2π.
 The general solution is θ = nπ, nεI.
 In Part C, solve cotθ = 1 using the unit circle.
 An equation with reciprocal trigonometric ratios can be solved by rewriting it so it involves only primary trigonometric ratios.
 Rewrite the equation as tanθ = 1.
 tanθ is the reciprocal of cotθ.
 1 is the reciprocal of 1.
 Bring up the unit circle.
 tanθ = 1 when sinθ and cosθ have the same magnitude, but opposite signs.
 The solution is θ = 3π/4 and 7π/4.
 The general solution is θ = 3π/4 + nπ, nεI.


Lecture 8 

07:05 

Example 8: Reciprocal Trigonometric Equations: Solve with Points of Intersection  Find all angles in the domain 0 ≤ θ ≤ 2π that satisfy the given equation. Write the general solution. In Part A, solve cscθ = 1/2 by graphing and finding the points of intersection.
 If we draw the graph of y = cscθ and y = 1/2, we can see that there are no intersection points. Therefore, there is no solution.
 Alternatively, solve sinθ = 2.
 If we draw the graph of y = sinθ and y = 2, we can see that there are no intersection points. Therefore, there is no solution.
 In Part B, solve cscθ = undefined by graphing and finding the points of intersection.
 If we draw the graph of y = cscθ, we can see that there are vertical asymptotes at 0, π, and 2π.
 The solution for 0 ≤ θ ≤ 2π is 0, π, and 2π.
 The general solution is θ = nπ, nεI.
 Alternatively, solve sinθ = 0.
 If we draw the graph of y = sinθ, and y = 0, we get intersection points at 0, π, and 2π. This is the same result we obtained above.
 In Part C, solve secθ = 2 by graphing and finding the points of intersection.
 If we draw the graph of y = secθ and y = 2, we can see that there are two intersection points at π/3 and 5π/3.
 The solution for 0 ≤ θ ≤ 2π is π/3 and 5π/3.
 The general solution is θ = π/3 + n(2π), nεI, and θ = 5π/3 + n(2π), nεI.
 Alternatively, solve cosθ = 1/2.
 If we draw the graph of y = cosθ and y = 1/2, we can see that there are two intersection points at π/3 and 5π/3. This is the same result we found above.
 In Part D, solve secθ = 1 by graphing and finding the points of intersection.
 If we draw the graph of y = secθ and y = 1, we can see that there is one intersection point at π.
 The solution for 0 ≤ θ ≤ 2π is θ = π.
 The general solution is θ = π + n(2π), nεI.
 Alternatively, solve cosθ = 1.
 If we draw the graph of y = cosθ and y = 1, we can see that there is one intersection point at π. This is the same result we found above.
 In Part E, solve cotθ = root(3)/3 by graphing and finding the points of intersection.
 If we draw the graph of y = cotθ and y = root(3)/3, we can see that there are intersection points at π/3 and 4π/3.
 The solution for 0 ≤ θ ≤ 2π is θ = π/3 and 4π/3.
 The general solution is θ = π/3 + nπ, nεI.
 Alternatively, solve tanθ = 3/root(3).
 If we draw the graph of y = tanθ and y = 3/root(3), we can see that there are intersection points at π/3 and 4π/3. This is the same result we found above.
 In Part F, solve cotθ = 0 by graphing and finding the points of intersection.
 If we draw the graph of y = cotθ and y = 0, we can see that there are intersection points at π/2 and 3π/2. This is the same result we found above.
 The solution for 0 ≤ θ ≤ 2π is θ = π/2 and 3π/2.
 The general solution is θ = π/2 + nπ, nεI.
 Alternatively, solve tanθ = undefined.
 If we draw the graph of y = tanθ, the graph is undefined at π/2 and 3π/2. This is the same result we found above.


Lecture 9 

04:37 

Example 9: Reciprocal Trigonometric Equations: Solve with a Calculator  Find all angles in the domain 0° ≤ θ ≤ 360° that satisfy the given equation. Write the general solution. In Part A, solve secθ = 2 nongraphically, using a calculator in degree mode.
 It would be difficult to solve this in a calculator using secθ. The calculator doesn't even have a secθ button!
 Rewrite this as cosθ = 1/2.
 Isolate θ by taking the inverse cosine of both sides.
 The result is 120°.
 cosθ is negative in quadrants II and III.
 By symmetry, we have a second angle at 240°.
 The solution over the domain 0° ≤ θ ≤ 360° is 120° and 240°.
 The general solution is 120° + n(360°), nεI and θ = 240° + n(360°), nεI.
 In Part B, solve cscθ = 2root(3)/3 nongraphically, using a calculator in degree mode.
 Rewrite this as sinθ = 3/2root(3).
 Take the inverse sine of both sides.
 The result is 60°.
 sinθ is positive in quadrants I and II.
 By symmetry, we have a second angle at 120°.
 The solution over the domain 0° ≤ θ ≤ 360° is 60° and 120°.
 The general solution is 60° + n(360°), nεI and θ = 120° + n(360°), nεI.
 In Part C, solve cotθ = root(3)/3 nongraphically, using a calculator in degree mode.
 Rewrite this as tanθ = 3/root(3).
 Isolate θ by taking the inverse tan of each side.
 The result is 60°.
 tanθ is positive in quadrants I and III.
 By symmetry, we have a second angle at 240°.
 Draw 240°.
 The solution over the domain 0° ≤ θ ≤ 360° is 60° and 240°.
 The general solution is 60° + n(180°), nεI.


Lecture 10 

02:32 

Example 10: Reciprocal Trigonometric Equations: Solve with θIntercepts  Find all angles in the domain 0 ≤ θ ≤ 2π that satisfy the given equation. Write the general solution. In Part A, solve cscθ = 1/2 using θintercepts.
 One way we could solve this equation is to graph both sides and find the points of intersection.
 There are no intersection points, so there is no solution.
 An alternative way to solve this is to set the equation equal to zero, then find the θintercepts.
 Rewrite the equation as 2cscθ = 1.
 Bring all terms to the left side so the equation equals zero.
 The graph of y = 2cscθ + 1 has no θintercepts, so there is no solution.
 In Part B, solve secθ = 1 using θintercepts.
 One way we could solve this equation is to graph both sides and find the points of intersection.
 The solution over the domain 0 ≤ θ ≤ 2π is 0 and 2π.
 The general solution is θ = n(2π), nεI.
 An alternative way to solve this is to set the equation equal to zero, then find the θintercepts.
 Rewrite the original equation as secθ  1 = 0.
 The θintercepts are 0 and 2π. This is the same solution we found above.


Lecture 11 

03:18 

Example 11: Reciprocal Trigonometric Equations: Multiple Solution Strategies  Solve cscθ = 2, where 0 ≤ θ ≤ 2π, nongraphically, using the sin^{1} feature of a calculator.
 Rewrite the equation as sinθ = 1/2.
 Isolate θ by taking the inverse sine of each side of the equation.
 The result is 30°. This is found by rotating the terminal arm clockwise 30°.
 The angle 30° is the same as 330°.
 sinθ is negative in quadrants III & IV.
 By symmetry, the second angle is found at 210°.
 The solution over the domain 0 ≤ θ ≤ 2π is 7π/6 and 11π/6.
 In Part B, solve cscθ = 2 nongraphically, using the unit circle.
 Rewrite the equation as sinθ = 1/2.
 Bring up the unit circle.
 The solution is 7π/6 and 11π/6.This is the same result we found in Part A.
 In Part C, solve cscθ = 2, graphically, using the points of intersection.
 Graph the functions Y_{1} = cscθ, which goes into the calculator as 1/sinθ, and Y_{2} = 2.
 The points of intersection are 7π/6 and 11π/6.This is the same result we found in Parts A & B.
 In Part D, solve cscθ = 2, graphically, using θintercepts.
 Rewrite the equation as cscθ + 2 = 0.
 The θintercepts occur at the angles 7π/6 and 11π/6. This is the same result we found in Parts A, B & C.


Lecture 12 

03:17 

Example 12: Reciprocal Trigonometric Equations: Multiple Solution Strategies  Solve secθ = 2.3662 nongraphically, using the cos^{1} feature of a calculator.
 Rewrite the equation as cosθ = 0.42261854 continuing.
 Take the inverse cosine of each side to isolate θ.
 The result is 115°.
 cosθ is negative in quadrants II & III.
 By symmetry, the second angle is found at 245°.
 The solution over the domain 0° ≤ θ ≤ 360° is 115° and 245°.
 In Part B, solve secθ = 2.3662 nongraphically, using the unit circle.
 Rewrite the equation as cosθ = 0.4226 and draw x = 0.4226.
 The unit circle is not useful for this question, since x = 0.4226 does not correspond to a 30°45°60° triangle.
 In Part C, solve secθ = 2.3662 graphically, using the points of intersection.
 Graph the functions Y_{1} = secθ, which goes into the calculator as 1/cosθ, and Y_{2} = 2.3662.
 The points of intersection occur at 115° and 245°. This is the same result we found in Part A.
 In Part D, solve secθ = 2.3662 graphically, using θintercepts.
 Rewrite the equation as secθ + 2.3662 = 0.
 The θintercepts occur at 115° and 245°. This is the same result we found in Parts A & C.

Section 3:
DAY THREE: Examples 14  17 (41 minutes) 

Lecture 13 

07:33 

Example 13: FirstDegree Trigonometric Equations (with one trigonometric ratio)  Find all angles in the domain 0 ≤ θ ≤ 2π that satisfy the given equation. Write the general solution. In Part A, solve cosθ – 1 = 0 algebraically.
 Pick the simplest strategy to solve this equation. We can solve with the unit circle, solve with points of intersection, solve with a calculator, or solve with θintercepts.
 Before we choose a strategy, isolate cosθ. Add 1 to each side of the equation to get cosθ = 1.
 We know that cosθ = 1 is on the unit circle, so let's solve this equation using the unit circle.
 Bring up the unit circle.
 cosθ = 1 at 0 and 2π.
 The solution over the domain where 0 ≤ θ ≤ 2π is 0 and 2π.
 The general solution is θ = n(2π), where nεI.
 In Part B, solve 2sinθ – root(3) = 0 algebraically.
 Pick the simplest strategy to solve this equation. We can solve with the unit circle, solve with points of intersection, solve with a calculator, or solve with θintercepts.
 Before we choose a strategy, isolate sinθ. This gives us sinθ = root(3)/2.
 We know that sinθ = root(3)/2 is on the unit circle, so let's solve this equation using the unit circle.
 Bring up the unit circle.
 sinθ = root(3)/2 at π/3 and 2π/3.
 The solution over the domain where 0 ≤ θ ≤ 2π is π/3 and 2π/3.
 The general solution is θ = π/3 + n(2π), where nεI, and θ = 2π/3 + n(2π), where nεI.
 In Part C, solve 3tanθ – 5 = 0 algebraically.
 Pick the simplest strategy to solve this equation. We can solve with the unit circle, solve with points of intersection, solve with a calculator, or solve with θintercepts.
 Before we choose a strategy, isolate tanθ. This gives us tanθ = 5/3.
 We'll use a calculator to solve tanθ = 5/3.
 Take the inverse tan of each side to isolate θ.
 The angle is 59°.
 tanθ is positive in quadrants I and III.
 By symmetry, we can draw a second angle at 239°.
 The solution for 0° ≤ θ ≤ 360° is 59° and 239°.
 The general solution is θ = 59° + n(180°).
 Note: The domain was given in radians, so it's good form to express our answer in radians too.
 Use the conversion multiplier π/180° to get:
 θ = 1.03 and 4.17 rad, with a general solution of θ = 1.03 + nπ.
 In Part D, solve 4secθ + 3 = 3secθ + 1 algebraically.
 Pick the simplest strategy to solve this equation. We can solve with the unit circle, solve with points of intersection, solve with a calculator, or solve with θintercepts.
 Before choosing a strategy, simplify the equation.
 Bring all the terms with secθ to the left side of the equation, and take all the constants to the right side.
 Simplify to get secθ = 2.
 Take the reciprocal of each side to get cosθ = 1/2.
 We know that cosθ = 1/2 is on the unit circle, so let's solve this equation using the unit circle.
 Bring up the unit circle.
 cosθ = 1/2 at 2π/3 and 4π/3.
 The solution over the domain where 0 ≤ θ ≤ 2π is 2π/3 and 4π/3.
 The general solution is θ = 2π/3 + n(2π), where nεI, and θ = 4π/3 + n(2π), where nεI.


Lecture 14 

09:15 

Example 14: FirstDegree Trigonometric Equations (with two different trigonometric ratios)  Find all angles in the domain 0 ≤ θ ≤ 2π that satisfy the given equation. In Part A, we'll solve 2sinθcosθ = cosθ algebraically.
 Before we can solve this equation, we need to bring all of the terms to the left side and factor.
 This gives us 2sinθcosθ  cosθ = 0.
 Factor out cosθ to get cosθ(2sinθ  1) = 0.
 We can solve this equation by setting each factor equal to zero.
 Bring up the unit circle.
 cosθ = 0 when θ = π/2 and 3π/2.
 Moving on to the next factor, we can simplify to get sinθ = 1/2.
 sinθ = 1/2 when θ = π/6 and 5π/6.
 The solution over the domain where 0 ≤ θ ≤ 2π is π/6, π/2, 5π/6, and 3π/2.
 Let's see how this looks graphically.
 Graph the original function, Y_{1} = cosθ(2sinθ – 1), in a calculator.
 The θintercepts of the factors will give us back the roots of the original equation.
 Graph the first factor, Y_{2} = cosθ. This gives us back the roots π/2 and 3π/2.
 Now graph the second factor, Y_{3} = 2sinθ – 1. This gives us back the roots π/6 and 5π/6.
 Watch Out! Don't cancel identical terms across the equals. You'll lose solutions!
 If we cancel cosθ from the original equation, we get 2sinθ = 1.
 This will make us lose the solutions π/2 and 3π/2.
 In Part B, solve 7sinθ = 4sinθ algebraically.
 Before we can solve this equation, we need to bring all of the terms to the left side and simplify.
 This gives us 7sinθ  4sinθ = 0.
 Simplify to get 3sinθ = 0.
 Divide both sides by 3 to get sinθ = 0.
 Bring up the unit circle.
 The solution over the domain where 0 ≤ θ ≤ 2π is 0, π, and 2π.
 The general solution is θ = nπ, where nεI.
 If we bring up the graph, we can see that the θintercepts match the solution.
 In Part C, solve sinθtanθ = sinθ algebraically.
 Bring all of the terms to the left side.
 Factor out sinθ to get sinθ(tanθ – 1) = 0.
 We can solve this equation by setting each factor equal to zero.
 Bring up the unit circle.
 sinθ = 0 when θ = 0, π, and 2π.
 Moving on to the next factor, we can simplify to get tanθ = 1.
 tanθ = 1 when θ = π/4 and 5π/4.
 The solution over the domain where 0 ≤ θ ≤ 2π is 0, π/4, π, 5π/4, and 2π.
 Let's see how this looks graphically.
 Graph the original function, Y_{1} = sinθ(tanθ – 1), in a calculator.
 The θintercepts of the factors will give us back the roots of the original equation.
 Graph the first factor, Y_{2} = sinθ. This gives us back the roots 0, π, and 2π.
 Now graph the second factor, Y_{3} = tanθ  1. This gives us back the roots π/4 and 5π/4.
 In Part D, solve tanθ + cosθtanθ = 0 algebraically.
 Rewrite the equation.
 Factor out tanθ to get tanθ(1 + cosθ) = 0.
 We can solve this equation by setting each factor equal to zero.
 Bring up the unit circle.
 tanθ = 0 when 0, π, and 2π.
 Moving on to the next factor, we can simplify to get cosθ = 1.
 cosθ = 1 when θ = π.
 The solution over the domain where 0 ≤ θ ≤ 2π is 0, π, and 2π.
 Notice that even though π occurs twice, we only include it once in our solution.
 Let's see how this looks graphically.
 Graph the original function, Y_{1} = tanθ(1 + cosθ), in a calculator.
 The θintercepts of the factors will give us back the roots of the original equation.
 Graph the first factor, Y_{2} = tanθ. This gives us back the roots 0, π, and 2π.
 Now graph the second factor, Y_{3} = 1 + cosθ  1. This gives us back a duplicate root of π.


Lecture 15 

07:30 

Example 15: SecondDegree Trigonometric Equations (simple factoring)  Find all angles in the domain 0 ≤ θ ≤ 2π that satisfy the given equation. In Part A, we'll solve sin^{2}θ = 1 algebraically.
 Take the square root of both sides of the equation to isolate sinθ.
 We get sinθ = ±1.
 Bring up the unit circle.
 sinθ = ±1 when θ = π/2 and 3π/2.
 If we look at the graph, we can see that y = sinθ intersects the lines y = 1 and y = 1 at the angles in our solution.
 In Part B, solve 4cos^{2}θ  3 = 0 algebraically.
 Add 3 to both sides to get 4cos^{2}θ = 3.
 Divide both sides of the equation by 4.
 Square root both sides.
 This gives us cosθ = ±root(3)/2.
 Bring up the unit circle.
 cosθ = ±root(3)/2 when θ = π/6, 5π/6, 7π/6, and 11π/6.
 If we look at the graph, we can see that y = cosθ intersects the lines y = root(3)/2 and y = root(3)/2 at the angles in our solution .
 In Part C, solve 2cos^{2}θ = cosθ algebraically.
 Remember: Don't cancel across the equals sign or solutions will be lost!
 Bring all terms to the left side to get 2cos2θ – cosθ = 0.
 Factor out cosθ to get cosθ(2cosθ – 1) = 0.
 Set each factor equal to zero.
 Bring up the unit circle.
 cosθ = 0 when θ = π/2, 3π/2.
 Moving on to the next factor, simplify to get cosθ = 1/2.
 cosθ = 1/2 when θ = π/3 and 5π/3.
 The solution over the domain where 0 ≤ θ ≤ 2π is π/2, π/3, 3π/2, and 5π/3.
 Let's see how this looks graphically.
 Graph the original function, Y_{1} = cosθ(2cosθ  1), in a calculator.
 The θintercepts of the factors will give us back the roots of the original equation.
 Now graph Y_{2} = cosθ. This gives us the roots π/2 and 3π/2.
 And finally, graph Y_{3} = 2cosθ – 1. This gives us the roots π/3 and 5π/3.
 In Part D, solve tan^{2}θ  tan^{2}θ = 0.
 Factor out tan^{2}θ to get tan^{2}θ(tan^{2}θ – 1) = 0.
 Set each factor equal to zero.
 Square root both sides.
 This gives us tanθ = 0.
 Bring up the unit circle.
 tanθ = sinθ/cosθ, so tanθ = 0 when the numerator, sinθ, is zero.
 This occurs at 0, π, and 2π.
 Moving on to the next factor, isolate tan2θ by adding 1 to both sides.
 Square root both sides.
 This gives us tanθ = ±1.
 tanθ = sinθ/cosθ, so tanθ = ±1 when the numerator and denominator have the same magnitude.
 This occurs when θ = π/4, 3π/4, 5π/4, and 7π/4.
 The solution over the domain where 0 ≤ θ ≤ 2π is 0, π/4, 3π/4, π, 5π/4, and 7π/4, and 2π.
 If we look at the graph, y = tanθ intersects the lines y = 1, y = 0, and y = 1 at the angles in our solution set.


Lecture 16 

16:20 

Example 16: SecondDegree Trigonometric Equations (with trinomial factoring)  Find all angles in the domain 0 ≤ θ ≤ 2π that satisfy the given equation. In Part A, we'll solve 2sin^{2}θ  sinθ  1 = 0 algebraically.
 Let's review factoring by decomposition by factoring 3x^{2} + 7x – 6.
 Note: There is more than one way to factor trinomials, and often guessandcheck is faster than decomposition. If you use a different factoring method, no problem. Use that instead.
 A trinomial can be represented with the form ax^{2} + bx + c.
 In this trinomial, a = 3, b = 7, and c = 6.
 In step one, determine the value of a×c.
 3×6 = 18.
 In step two, find two numbers that have a product of a×c and a sum of b.
 Let's draw a chart that we can use to find the two numbers. The first column will hold the numbers we are testing. The second column will hold the product of these numbers. The third column will hold the sum of the two numbers, and the fourth column will indicate success or failure. Remember, we are looking for two numbers with a product of 18 and a sum of +7.
 Let's try the numbers 2 and 9.
 The product is 18.
 The sum is 7.
 This doesn't work, since we need a sum of +7.
 Now test 2 and 9.
 The product is 18.
 The sum is 7.
 This works, since we have a product of 18 and a sum of 7.
 In step three, rewrite the middle term using the numbers we just found, 2 and 9.
 This gives us 3x^{2}  2x + 9x  6.
 The middle terms return 7x, so we haven't actually changed the trinomial. We've just written it in a more useful form.
 In step four, factor the first two terms, factor the second two terms, then simplify.
 x(3x  2) + 3(3x  2)
 The binomial (3x 2) is common to each term. Factor it out to get (3x 2)(x + 3).
 We have now factored the trinomial.
 Now we'll return to the original question. Solve 2sin^{2}θ  sinθ  1 = 0 algebraically.
 We can factor 2sin^{2}θ  sinθ – 1 using decomposition. For simplicity, rewrite the trinomial as 2a^{2}  a  1, where a = sinθ.
 We need two numbers that multiply to 2, the product of a×c, and add to 1, the bvalue.
 The two numbers are 2 and +1.
 Rewrite the middle term using the two numbers. This gives us 2a^{2}  2a + a – 1.
 Factor 2a out of the first two terms. There is nothing to factor from the second two terms, so just factor out 1.
 The factored form is (a  1)(2a + 1).
 Now that we have factored, restore the a's to sinθ. This gives us (sinθ  1)(2sinθ + 1).
 Now we'll find the values of θ that satisfy the equation.
 Set each factor equal to zero.
 Isolate sinθ by adding 1 to each side of the equation.
 Bring up the unit circle.
 sinθ = 1 when θ = π/2.
 Moving on to the next factor, subtract 1 from each side of the equation.
 Divide both sides of the equation by 2.
 This gives us sinθ = 1/2.
 sinθ = 1/2 when θ = 7π/6, 11π/6.
 The solution over the domain where 0 ≤ θ ≤ 2π is π/2, 7π/6, and 11π/6.
 Let's see how this looks graphically.
 Graph the original function, Y_{1} = 2sin^{2}θ  sinθ  1, in a calculator.
 The θintercepts of the factors will give us back the roots of the original equation.
 Graph the first factor, Y_{2} = sinθ – 1. This gives us back the root of π/2.
 Now graph the second factor, Y_{3} = 2sinθ + 1. This gives us back the roots of 7π/6 and 11π/6.
 In Part B, solve csc^{2}θ – 3cscθ + 2 = 0 algebraically.
 Rewrite csc^{2}θ  3cscθ + 2 as a^{2}  3a + 2 for simplicity, where a = cscθ.
 Now factor a^{2}  3a + 2.
 We need two numbers that multiply to 2, the product of a×c, and add to 3, the bvalue.
 The two numbers are 2 and 1.
 Rewrite the middle term using the two numbers. This gives us a^{2}  2a  a + 2.
 Factor a out of the first two terms. Factor 1 out of the second two terms.
 The factored form is (a – 2)(a – 1).
 Now that we have factored, restore the a's to cscθ. This gives us (cscθ  2)(cscθ  1).
 Now we'll solve the equation (cscθ  2)(cscθ  1) = 0.
 Set each factor equal to zero.
 Add 2 to each side of the equation to get cscθ = 2.
 Take the reciprocal of each side to get sinθ = 1/2.
 Bring up the unit circle.
 sinθ = 1/2 when θ = π/6 and 5π/6.
 Moving on to the next factor, add 1 to each side to get cscθ = 1.
 Take the reciprocal of each side to get sinθ = 1.
 sinθ = 1 when θ = π/2.
 The solution over the domain where 0 ≤ θ ≤ 2π is π/6, π/2, and 5π/6.
 Let's see how this looks graphically.
 Graph the original function, Y_{1} = csc^{2}θ  3cscθ + 2, in a calculator.
 Now graph the first factor, Y_{2} = cscθ – 2. This gives us the roots π/6 and 5π/6.
 And graph the second factor, Y_{3} = cscθ – 1. This gives us the root π/2.
 In Part C, solve 2sin^{2}θ – 5sin^{2}θ + 2sinθ = 0 algebraically.
 Rewrite 2sin^{2}θ  5sin^{2}θ + 2sinθ as 2a^{2}  5a^{2} + 2a, where a = sinθ.
 Now factor 2a^{2}  5a^{2} + 2a.
 Factor out a to get a(2a^{2}  5a + 2).
 We need two numbers that multiply to 4 (product of a×c) and add to 5 (b).
 The two numbers are 4 and 1.
 Replace the middle term with the two numbers. This gives us a[2a^{2}  4a  a + 2].
 Factor out 2a from the first two terms, and 1 from the second two terms.
 This gives us a(a  2)(2a 1).
 Replace a with sinθ. This gives us sinθ(sinθ  2)(2sinθ  1).
 Now we'll solve the equation.
 Set each factor equal to zero.
 Bring up the unit circle.
 sinθ = 0 when θ = 0, π, and 2π.
 Moving on to the next factor, add 2 to both sides to isolate sinθ.
 There is no solution.
 Moving on to the last factor, add 1 to both sides to isolate sinθ.
 Divide both sides by 2 to get sinθ = 1/2.
 sinθ = 1/2 when θ = π/6 and 5π/6.
 The solution over the domain where 0 ≤ θ ≤ 2π is 0, π/6, 5π/6, π, and 2π.
 Let's see how this looks graphically.
 Graph the original function, Y_{1} = 2sin^{3}θ  5sin^{2}θ + 2sinθ, in a calculator.
 Now graph the first factor, Y_{2} = sinθ. This gives us the roots 0, π, and 2π.
 Graph the second factor, Y_{3} = sinθ – 2.
 And graph the third factor, Y_{4} = 2sinθ – 1. This gives us the roots π/6 and 5π/6.

Section 4:
DAY FOUR: Examples 17  20 (37 minutes) 

Lecture 17 

11:21 

Example 17: Double and Triple Angle Equations  Solve each trigonometric equation. In Part A, we'll solve sin2θ = root(3)/2 graphically and nongraphically.
 We'll begin by solving this equation graphically.
 Draw the graph of y = sin2θ. The period is π. The steps required to find the period are shown.
 Now draw the line y = root(3)/2.
 The solution for the domain where 0 ≤ θ ≤ 2π is 2π/3, 5π/6, 5π/3, and 11π/6.
 The first equation of the general solution is θ = 2π/3 + nπ, nεI.
 The second equation of the general solution is θ = 5π/6 + nπ, nεI.
 Now we'll solve the equation nongraphically.
 Even though we're going to solve this equation algebraically, we should be thinking about the graph as we work so the algebra makes sense.
 In Step 1, solve the more simple equation, sinθ = root(3)/2.
 Take the inverse sine of both sides of the equation to isolate θ.
 Bring up the unit circle.
 sinθ = root(3)/2 when θ = 4π/3 and 5π/3.
 We can see these solutions on the graph.
 In Step 2, we can get the graph of y = sin2θ with a horizontal stretch by a scale factor of 1/2.
 Stretch the graph horizontally by a scale factor of 1/2.
 We need to multiply both original angles by 1/2 since the width of the graph was halved.
 Multiply the first angle, 4π/3, by 1/2.
 This multiplies to 4π/6.
 Reduce the fraction to 2π/3.
 Now multiply the second angle, 5π/3, by 1/2.
 This multiplies to 5π/6.
 In Step 3, add the period to each angle to get the other solutions.
 Add the period, π, to 2π/3.
 Get a common denominator of 3.
 This adds to 5π/3.
 Now add the period to 5π/6.
 Get a common denominator of 6.
 This adds to 11π/6.
 The solution for the domain where 0 ≤ θ ≤ 2π is 2π/3, 5π/6, 5π/3, and 11π/6. This is the same result we obtained earlier.
 In Part B, we'll solve cos3θ = root(2)/2 graphically and nongraphically.
 We'll begin by solving this equation graphically.
 It takes 6 ticks to get to 90° (π/2 radians).
 90°/6 = 15°. The ticks are spaced 15° (or π/12 radians) apart.
 Graph y = cos3θ and the line y = root(2)/2. The period is 2π/3.
 Write the radian values of the solutions.
 This is the first tick, so it is 1π/12.
 This is the seventh tick, so it is 7π/12.
 This is the ninth tick, so it is 9π/12. This reduces to 3π/4.
 The solution for the domain where 0 ≤ θ ≤ 2π is π/12, 7π/12, 3π/4, 5π/4, 17π/12, and 23π/12.
 The first equation of the general solution is θ = π/12 + n(2π/3), nεI.
 The second equation of the general solution is θ = 7π/12 + n(2π/3), nεI.
 Now we'll solve the equation nongraphically.
 Even though we're going to solve this equation algebraically, we should be thinking about the graph as we work so the algebra makes sense.
 In Step 1, solve the more simple equation, cosθ = root(2)/2.
 Take the inverse cosine of both sides of the equation to isolate θ.
 Bring up the unit circle.
 cosθ = root(2)/2 when θ = π/4 and 7π/4.
 We can see these solutions on the graph.
 In Step 2, we can get the graph of y = cos3θ with a horizontal stretch by a scale factor of 1/3.
 Stretch the graph horizontally by a scale factor of 1/3.
 We need to multiply both original angles by 1/3 since the width of the graph is now onethird the original.
 Multiply the first angle, π/4, by 1/3. This gives us π/12.
 Multiply the second angle, 7π/4, by 1/3. This gives us 7π/12.
 In Step 3, add the period to each angle to get the other solutions.
 Recall that the period is 2π/3.
 Add the period to 2π/3.
 Get a common denominator of 12.
 This adds to 9π/12.
 Reduce the fraction to 3π/4.
 Now add the period to 7π/12.
 Get a common denominator of 12.
 This adds to 15π/12.
 Reduce the fraction to 5π/4.
 Add the period again to get the next two solutions.
 Add the period to 3π/4.
 Get a common denominator of 12.
 This adds to 17π/12.
 Now add the period to 5π/4.
 Get a common denominator of 12.
 This adds to 23π/12.
 The solution for the domain where 0 ≤ θ ≤ 2π is π/12, 7π/12, 3π/4, 5π/4, 17π/12, and 23π/12. This is the same result we obtained earlier.


Lecture 18 

06:37 

Example 18: Half and Quarter Angle Equations  Solve each trigonometric equation. In Part A, we'll solve cos1/2θ = 1/2 graphically and nongraphically.
 We'll begin by solving this equation graphically.
 Draw the graph of y = cos1/2θ. The period is 4π. The steps required to find the period are shown.
 Now draw the line y = 1/2.
 The solution for the domain where 0 ≤ θ ≤ 4π is 2π/3 and 10π/3.
 How do we get 10π/3? We know that each tick is spaced π/6 radians apart. This tick is the 20th tick, so we get 20π/6, or 10π/3 when we reduce the fraction.
 The first equation of the general solution is θ = 2π/3 + n(4π), nεI.
 The second equation of the general solution is θ = 10π/3 + n(4π), nεI.
 Now we'll solve the equation nongraphically.
 Even though we're going to solve this equation algebraically, we should be thinking about the graph as we work so the algebra makes sense.
 In Step 1, solve the more simple equation, cosθ = 1/2.
 Take the inverse cosine of both sides of the equation to isolate θ.
 Bring up the unit circle.
 cosθ = 1/2 when θ = π/3 and 5π/3.
 We can see these solutions on the graph.
 In Step 2, we can get the graph of y = cos1/2θ with a horizontal stretch by a scale factor of 2.
 Stretch the graph horizontally by a scale factor of 2.
 We need to multiply both original angles by 2 since the width of the graph was doubled.
 Multiply the first angle, π/3, by 2 to get 2π/3.
 Now multiply the second angle, 5π/3, by 2 to get 10π/3.
 The solution for the domain where 0 ≤ θ ≤ 4π is 2π/3 and 10π/3.
 In Part B, we'll solve sin1/4θ = 1 graphically and nongraphically.
 We'll begin by solving this equation graphically.
 Graph y = sin1/4θ. The period is 8π.
 Now draw the line y = 1.
 The solution for the domain where 0 ≤ θ ≤ 8π is 6π.
 The general solution is θ = 6π + n(8π), nεI.
 Now we'll solve the equation nongraphically.
 Even though we're going to solve this equation algebraically, we should be thinking about the graph as we work so the algebra makes sense.
 In Step 1, solve the more simple equation, sinθ = 1.
 Take the inverse sine of both sides of the equation to isolate θ.
 Bring up the unit circle.
 sinθ = 1 when θ = 3π/2.
 We can see this solution on the graph.
 In Step 2, we can get the graph of y = cos1/4θ with a horizontal stretch by a scale factor of 4.
 Stretch the graph horizontally by a scale factor of 4.
 We need to multiply the original angle by 4 since the width of the graph was quadrupled.
 Multiply the original angle, 3π/2, by 4. This gives us 12π/2, which reduces to 6π.
 The solution for the domain where 0 ≤ θ ≤ 8π is 6π.


Lecture 19 

04:02 

Example 19: Problem Solving (phases of the moon)  It takes the moon approximately 28 days to go through all of its phases. In Part A, write a function, P(t), that expresses the visible percentage of the moon as a function of time. Draw the graph.
 In the diagram, write the moon's visible percentage for each phase.
 The moon cycle begins with a visibility of zero. Halfway through the cycle it reaches a maximum visibility of 1.00. The visibility then returns to zero.
 The goal is to create a function with the form y = acosb(θ  c) + d. We need to determine the a, b, c, and d parameters.
 We'll begin by finding the amplitude.
 The minimum of the graph is 0, and the maximum is 1.
 This gives us a = 0.50. Make sure you remember to include a negative with the function since the cosine shape is upsidedown.
 Create a box to store the graph data.
 Now we'll find the bparameter.
 The period is 28. Plug this in for P.
 This reduces to π/14.
 There is no phase shift, so c = 0.
 Now we'll find the vertical displacement.
 The minimum of the graph is 0, and the maximum is 1.
 This gives us d = 0.50.
 The function is P(t) = 0.50cosπ/14t + 0.50.
 Now we'll move on to Part B. In one cycle, for how many days is 60% or more of the moon's surface visible?
 Bring up the function and its graph.
 Draw the line y = 0.60. The distance between the points of intersection is the number of days where more than 60% of the moon is visible.
 Using a graphing calculator, graph Y_{1} = 0.50cosπ/14t + 0.50 and Y_{2} = 0.60.
 Use the intersect feature to get 7.90 and 20.10 days.
 Subtract the two values to get 12.20 days.
 60% or more of the moon's surface is visible for approximately 12 days each cycle.


Lecture 20 

03:41 

Example 20: Problem Solving (lawn sprinkler)  A rotating sprinkler is positioned 4 m away from the wall of a house. The wall is 8 m long. As the sprinkler rotates, the stream of water splashes the house d meters from point P. Note: north of point P is a positive distance, and south of point P is a negative distance. In Part A, write a tangent function, d(θ), that expresses the distance where the water splashes the wall as a function of the rotation angle θ.
 We have a triangle with side lengths of 4 and d.
 This can be represented with the trigonometric ratio tanθ = d/4.
 Cross multiply to get d = 4tanθ.
 We can write this as the function d(θ) = 4tanθ.
 In Part B, graph the function for one complete rotation of the sprinkler. Draw only the portion of the graph that actually corresponds to the wall being splashed.
 Rewrite the function d(θ) = 4tanθ.
 Let's redraw the sprinkler and wall so we can animate it.
 Now set up a coordinate grid to draw the graph. The length of the wall is 8 m, and the wall is centered at 0 m.
 Just so we're perfectly clear on the wall's boundary, draw the lines y = 4 and y = 4.
 Rotate the sprinkler and draw the graph of y = 4tanθ.
 Now erase the portions of the graph shouldn't be there.
 Erase the regions shown.
 If we replay the animation, we can see that this graph is the one we actually need.
 Now we'll move on to Part C. If the water splashes the wall 2.0 m north of point P, what is the angle of rotation (in degrees)?
 Redraw the sprinkler and graph from Part B.
 Rotate the sprinkler so it contacts the wall 2.0 m north of point P. Also, draw the line y = 2.
 Using a graphing calculator, find the point of intersection. This gives us θ = 0.4636 rad.
 In degrees, the angle of rotation is 26.6°. We can get this by multiplying the radian angle by 180°/π.


Lecture 21 

11:05 

Example 21: Inverse Trigonometric Functions  When we solve a trigonometric equation like cosx = 1, one possible way to write the solution is shown. In this example, we will explore the inverse functions of sine and cosine to learn why taking an inverse actually yields the solution. In Part A, when we draw the inverse of trigonometric graphs, it is helpful to use a grid that is labeled with both radians and integers. Briefly explain how this is helpful.
 When we graph ordinary trigonometric functions, there are two things we should make note of:
 First, we use angles on the xaxis and integers on the yaxis.
 Second, we can use nonsquare grids.
 This kind of graph won't work for trigonometric inverses!
 An inverse graph requires integers on the xaxis and angles on the yaxis. This is swapped from the original orientation.
 If we want to graph a trigonometric graph and its inverse simultaneously, we require a perfectly square grid.
 Draw a perfectly square grid. Since the grid is a square, we need to put the exact same value at each tick mark.
 Since trigonometric graphs also use integers, add tick marks at the integral values.
 Now that we have both angles and integers labeled, we can draw an original trigonometric graph and its inverse on the same grid.
 Now we'll move on to Part B. Draw the inverse function of each graph. State the domain and range of the original and inverse graphs (after restricting the domain of the original so the inverse is a function).
 We will be graphing y = sinx and its inverse, y = sin^{1}x.
 When drawing y = sinx, make sure the maximum and minimum are correctly aligned with the integer ticks, rather than the grid lines.
 Recall that an inverse is reflected across the line y = x.
 We have a problem. The inverse does not pass the vertical line test, so it is not a function!
 Let's draw a box around a portion of the inverse graph that would pass the vertical line test.
 The domain of the inverse is 1 ≤ x ≤ 1, and the range is –π/2 ≤ y ≤ π/2.
 We don't need the rest of the inverse graph, so erase it.
 Now draw a rectangle around the portion of the original graph required to make the inverse segment.
 The domain of the original is –π/2 ≤ x ≤ π/2, and the range is –1 ≤ y ≤ 1.
 Now return the inverse segment back to its original position and erase the portions of the sine graph we don't need.
 If the graph of y = sinx is restricted to the domain –π/2 ≤ x ≤ π/2, then the graph of y = sin^{1}x is a function that exists over the domain 1 ≤ x ≤ 1.
 Now we'll graph y = cosx and its inverse, y = cos^{1}x.
 When drawing y = cosx, make sure the maximum and minimum are correctly aligned with the integer ticks, rather than the grid lines.
 Recall that an inverse is reflected across the line y = x.
 We have a problem. The inverse does not pass the vertical line test, so it is not a function!
 Let's draw a box around a portion of the inverse graph that would pass the vertical line test.
 We don't need the rest of the inverse graph, so erase it.
 The domain of the inverse is –1 ≤ x ≤ 1, and the range is 0 ≤ y ≤ π.
 Now draw a rectangle around the portion of the original graph required to make the inverse segment.
 The domain of the original is 0 ≤ x ≤ π, and the range is –1 ≤ y ≤ 1.
 Now return the inverse segment back to its original position and erase the portions of the sine graph we don't need.
 If the graph of y = cosx is restricted to the domain 0 ≤ x ≤ π, then the graph of y = cos^{1}x is a function that exists over the domain 1 ≤ x ≤ 1.
 Now we'll move on to Part C. Is there more than one way to restrict the domain of the original graph so the inverse is a function? If there is, generalize the rule in a sentence.
 Bring up the graph of y = sinx and its inverse.
 Scan the graph of the original and its inverse to find intervals where the inverse passes the vertical line test.
 The first potential domain restriction occurs when 3π/2 ≤ x ≤ π/2.
 The next potential domain restriction occurs when π/2 ≤ x ≤ π/2.
 And we have another potential domain restriction when π/2 ≤ x ≤ 3π/2.
 We can generalize that the inverse is a function for all intervals where the original is decreasing or increasing.
 Now bring up the graph of y = cosx and its inverse.
 Scan the graph of the original and its inverse to find intervals where the inverse passes the vertical line test.
 When the cosine graph is decreasing, the inverse is a function. The first potential domain restriction occurs when 2π ≤ x ≤ π.
 When the cosine graph is increasing, the inverse is a function. The second potential domain restriction occurs when π ≤ x ≤ 0.
 When the cosine graph is decreasing, the inverse is a function. The third potential domain restriction occurs when 0 ≤ x ≤ π.
 When the cosine graph is increasing, the inverse is a function. And finally, we have a potential domain restriction that occurs when 2π ≤ x ≤ 2π.
 We can generalize that the inverse is a function for all intervals where the original is decreasing or increasing.
 Now we'll move on to Part D. Using the inverse graphs from part (b), evaluate each of the following:
 We'll begin by evaluating sin^{1}(1).
 Bring up the graph of y = sin^{1}x.
 We can evaluate sin^{1}(1) by locating 1 on the xaxis, then finding the corresponding yvalue.
 The answer is π/2.
 Now we'll evaluate arccos(1).
 Note that arcsin means the same thing as sin^{1}, and arccos means the same thing as cos^{1}.
 Bring up the graph of y = cos^{1}x.
 We can evaluate arccos(1) by locating 1 on the xaxis, then finding the corresponding yvalue.
 The answer is π.
 Recall from the beginning of this lesson that we solved the equation cosx = 1.
 We get an answer of π because the function y = cos^{1}x has the ordered pair (1, π), so cos^{1}(1) = π.

Section 5:
Other Lesson Materials 

Lecture 22 

22 pages 

Lecture 23 

00:52 

Trigonometric Equations: Introduction  Welcome to Trigonometric Equations. This topic will take four days to complete.
 In Day 1 we will learn how to solve simple trigonometric equations with the primary trigonometric ratios.
 In Day 2 we will learn how to solve simple trigonometric equations with the reciprocal trigonometric ratios.
 In Day 3 we will study firstdegree and seconddegree trigonometric equations.
 In Day 4 we will study double and half angle equations. We will also solve problems with trigonometric equations.


Lecture 24 

00:52 

Trigonometric Equations: Summary  You have now completed Trigonometric Equations. Over the past four days you have learned:
 How to solve trigonometric equations using the unit circle, points of intersection, a calculator, and θintercepts.
 how to solve equations involving both primary and reciprocal trigonometric ratios.
 how to factor and solve firstdegree and seconddegree trigonometric equations.
 how to model realworld scenarios with trigonometric equations.


Quiz 1 

18 questions 
Full curriculum

Simply Great
A really great course with understandable material and a good instructor. Worth learning from!
Thanks!
Learn something new through this course!
very well explained