# Trigonometry: Trigonometric Identities I

Part 6 of 7
Instructed by Barry Mabillard
• Lectures 22
• Video 3 Hours
• Skill level expert level
• Languages English
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### Course Description

Trigonometry: Trigonometric Identities I

This lesson consists of a workbook and animated solutions for the workbook.

Workbook: The workbook for Trigonometric Identities I can be found in the introduction. Students should make two printouts of this workbook. Label the first printout "Lecture Notes", and label the second printout "Practice".

Videos: The videos present a fully animated solution for each question in the workbook.

Strategy for Trigonometric Identities I:

This lesson takes four days to complete, with a time commitment of 60-75 minutes each day. Students should work through this lesson as follows:

Watch the videos and fill in your Lecture Notes workbook. (30 - 45 minutes)

In the Practice workbook, do the examples you just learned on your own, without external help. This step is required to reinforce the concepts and gain mastery. (15 - 30 minutes)

Curriculum Fit:

This lesson explores its subject matter in full depth. Teachers are encouraged to compare this lesson with their local standards and assign examples to their students accordingly.

Calculator:

Calculator references in this lesson apply to the TI-83/83+/84 series.

### What are the requirements?

• Trigonometry: Trigonometric Equations

### What am I going to get from this course?

• Over 22 lectures and 3 hours of content!
• set up and complete a trigonometric proof.
• derive the three Pythagorean identities.
• complete proofs involving identity substitution.
• find non-permissible values of complex trigonometric functions.
• interpret trigonometric graphs with asymptotes and holes.
• solve trigonometric equations that involve identity substitutions.

### What is the target audience?

• trigonometry students
• precalculus students

### What you get with this course?

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### Curriculum

Section 1: DAY ONE: Examples 1 - 5 (43 minutes)
06:13
##### Example 1: Trigonometric identities are a special type of trigonometric equation.
1. Why are trigonometric identities considered to be a special type of trigonometric equation?
2. Let's begin by defining a trigonometric identity.
3. A trigonometric identity is an equality that holds true no matter what value is assigned to the variable.
4. Consider the identity secx = 1/cosx.
5. This is a trigonometric identity because the L.S. = R.S. for all values of x.
6. If we try x = 0, we get 1 on both sides. The L.S. = R.S.
7. If we try x = π/3, we get 2 on both sides. The L.S. = R.S.
8. And if we try x = π/2, we get undefined on both sides. The L.S. = R.S.
9. Now draw the graph for both the L.S. and R.S.
10. The graph of the L.S. is identical to the graph of the R.S. This verifies that the statement holds true for all values of x and can be called an identity.
11. So how does a trigonometric identity differ from a trigonometric equation?
12. In general, trigonometric equations can hold true for no values, some values, or all values. In the case where a trigonometric equation holds true for all values, it can be called a trigonometric identity.
13. Consider the equation secx = 1.
14. If we try x = 0, the equation holds true. The L.S. = R.S.
15. But if we try x = π/3, the equation is false. The L.S. ≠ R.S.
16. secx = 1 is not an identity because it is only true for some values, not all values.
17. Now draw the graph for both the L.S. and R.S.
18. We get points of intersection at x = n(2π), nεI. These are the solutions of the equation, and the only values where the left side and right side are equal.
19. A trigonometric equation can be called an identity if it meets three criterion:
20. One: The left side is algebraically equivalent to the right side.
21. Two: The left side and right side have exactly the same non-permissible values.
22. And Three: The left side and right side have exactly the same graph.
23. If a trigonometric equation does not meet the above criterion, it may still be possible to force it to be an identity using variable restrictions. You will learn more about this later.
24. In this lesson (and the next one), we will be doing a lot of algebra that involves trigonometric functions. This is called analytic trigonometry. A common convention in analytic trigonometry is to use x as the variable instead of θ.
25. In Part B, which of the following trigonometric equations are also trigonometric identities?
26. Begin with question i)
27. This is not an identity because the statement holds true only for specific values of x.
28. Now we'll move on to question ii).
29. This is not an identity because the statement holds true only for specific values of x.
30. Now we'll move on to question iii).
31. This equation is an identity because the statement holds true for all values of x.
32. Now we'll move on to question iv).
33. This equation is an identity because the statement holds true for all values of x.
34. Now we'll move on to question v).
35. This is not an identity because the statement holds true only for specific values of x.
11:21
##### Example 2: The Three Pythagorean Identities
1. Using the definition of the unit circle, derive the identity sin2x + cos2x = 1. Why is sin2x + cos2x = 1 called a Pythagorean Identity?
2. Draw a right triangle within the circle as shown.
3. cosx = a/1.
4. This becomes a = cosx.
5. sinx = b/1.
6. This becomes b = sinx.
7. Using the Pythagorean Theorem, we get a2 + b2 = 12.
8. Replace a with cosx and replace b with sinx.
9. This gives us cos2x + sin2x = 1.
10. We call cos2x + sin2x = 1 a Pythagorean identity because it is derived from the Pythagorean Theorem.
11. In Part B, verify that sin2x + cos2x = 1 is an identity using i) x = π/6 and ii) x = π/2.
12. Begin with π/6.
13. Rewrite the identity sin2x + cos2x = 1.
14. Plug in π/6 for x.
15. sin π/6 = 1/2, and cos π/6 = root(3)/2.
16. Square the terms.
17. Combine the fractions.
18. The result is 1 = 1. The L.S. and R.S. are equal.
19. Now we'll try π/2.
20. Rewrite the identity sin2x + cos2x = 1.
21. Plug in π/2 for x.
22. sin π/2 = 1, and cos π/2 = 0.
23. The result is 1 = 1. The L.S. and R.S. are equal.
24. Note that the term verify is used to check that something is true for a specific case. The term prove is used to show that something is true for all cases.
25. In this example, we verified sin2x + cos2x = 1 for the specific cases x = π/6 and x = π/2. However, our results for these two cases make no guarantee that sin2x + cos2x = 1 works for other x-values as well.
26. In Part C, verify that sin2x + cos2x = 1 is an identity using a graphing calculator to draw the graph.
27. Make sure the calculator is in RADIAN mode and use the settings shown.
28. Now draw the graph of y = sin2x + cos2x.
29. Next, draw the graph of y = 1.
30. The graphs of the L.S. and R.S. are identical, so sin2x + cos2x = 1 is an identity.
31. Note that graphing the L.S. and R.S. of a statement is a verification, not a proof. There is always the chance that the graphs just look similar without actually being identical.
32. Now we'll move on to Part D. Using the identity sin2x + cos2x = 1, derive 1 + cot2x = csc2x and tan2x + 1 = sec2x.
33. We'll begin with 1 + cot2x = csc2x.
34. Rewrite the identity sin2x + cos2x = 1.
35. Divide both sides by sin2x.
36. Simplify
37. cos2x/sin2x = cot2x, and 1/sin2x = csc2x. This gives us 1 + cot2x = csc2x, completing the derivation.
38. Now we'll derive tan2x + 1 = sec2x.
39. Rewrite the identity sin2x + cos2x = 1.
40. Divide both sides of the equation by cos2x.
41. sin2x/cos2x = tan2x, and 1/cos2x = sec2x. This gives us tan2x + 1 = sec2x, completing the derivation.
42. In Part E, verify that 1 + cot2x = csc2x and tan2x + 1 = sec2x are identities for x = π/4.
43. Begin by verifying 1 + cot2x = csc2x at x = π/4.
44. Rewrite 1 + cot2x = csc2x.
45. Plug in π/4 for x.
46. Rewrite the cotan and cosecant ratios in terms of sine and cosine.
47. Both cos(π/4) and sin(π/4) equal root(2)/2.
48. For each fraction, multiply the numerator by the reciprocal of the denominator.
49. Simplify
50. Simplify further.
51. The result is 2 = 2, which completes the verification.
52. Now we'll verify tan2x + 1 = sec2x at x = π/4.
53. Rewrite tan2x + 1 = sec2x.
54. Plug in π/4 for x.
55. Rewrite the tan and secant ratios in terms of sine and cosine.
56. Both cos(π/4) and sin(π/4) equal root(2)/2.
57. For each fraction, multiply the numerator by the reciprocal of the denominator.
58. Simplify
59. Simplify further.
60. The result is 2 = 2, which completes the verification.
61. In Part F, verify that 1 + cot2x = csc2x and tan2x + 1 = sec2x are identities graphically.
62. Make sure the calculator is in RADIAN mode and use the settings shown.
63. Draw the graph of y = 1 + cot2x.
64. Now draw the graph of y = csc2x.
65. The graphs are identical, so 1 + cot2x = csc2x is an identity.
66. Now we'll move on to the next identity.
67. Draw the graph of y = tan2x + 1.
68. Now draw the graph of y = sec2x.
69. The graphs are identical, so tan2x + 1 = sec2x is an identity.
10:35
##### Example 3: Proofs with Reciprocal Trigonometric Identities
1. Prove that each trigonometric statement is an identity. State the non-permissible values of x so the identity is true. In Part A, prove that sinxsecx = tanx.
2. In a trigonometric proof, we use identities and algebraic techniques to show that the left side of a statement is algebraically equivalent to the right side.
3. Rewrite the statement and draw a line.
4. When completing a proof, we use the term statement to describe the original equation. This is because we don't know yet if the left side does in fact equal the right side, and the term statement makes no implications about truthfulness. If the proof works out, we can upgrade the description to being an identity, which does imply truthfulness.
5. A common approach to proofing is to draw a line separating the two sides of the statement. Work on one side of the statement until it equals the other side.
6. Do not cross the line! This means no cross-multiplication and now adding or subtracting terms from both sides.
7. Now we'll complete the proof.
8. Rewrite secx as 1/cosx.
9. Multiply to get sinx/cosx.
10. sinx/cosx = tanx.
11. The L.S. equals the R.S., completing the proof.
12. Even though we have shown that the L.S. and R.S. are algebraically equivalent, we still need to specify the non-permissible values of x.
13. Recall that non-permissible values occur whenever a denominator becomes zero.
14. For example, 1/(x2 – 4) has non-permissible values when x = ±2. Both of these integers make the denominator evaluate to zero, which makes the fraction undefined.
15. This step has a denominator, so we need to identify the values of x that make cosx = 0.
16. We can do this by solving the equation cosx = 0.
17. Recall that trigonometric equations can be solved by using the unit circle, finding intersection points, finding x-intercepts, or using a calculator. The animated examples for this lesson provide a graphical solution as a convenient reference. However, when you complete these questions on your own, it is recommended you use the unit circle to find solutions.
18. The non-permissible values occur when cosx = 0, at π/2 + nπ, nεI.
19. Let's explore this in more depth. How do we find the non-permissible values of y = 1/cosx?
20. The first thing we need to do is solve the equation cosx = 0.
21. The solution is x = π/2 + nπ, nεI.
22. The solutions of cosx = 0 become the asymptotes, or non-permissible values, of y = 1/cosx.
23. We write the non-permissible values using a 'does not equal' symbol.
24. Remember: The values of x that make the denominator equal zero become the non-permissible values of the fraction!
26. Use a graphing calculator to inspect the graphs.
27. Graph the left side, Y1 = sinxsecx.
28. Then graph the right side, Y2 = tanx.
29. The graphs of the L.S. and R.S. are identical.
30. In conclusion, sinxsecx = tanx is an identity as written since the graphs are exactly the same. However, since non-permissible values are present, it is considered good form to write the identity with the variable restrictions.
31. In Part B, prove that cotxsinxsecx= 1.
32. Rewrite the statement and draw a line.
33. Rewrite the left side using sine and cosine.
34. Cancel.
35. The result is 1.
36. The L.S. equals the R.S., completing the proof.
37. Even though we have shown that the L.S. and R.S. are algebraically equivalent, we still need to specify the non-permissible values of x.
38. There are two denominators. We need to identify the values of x that make sinx = 0 and cosx = 0.
39. Non-permissible values occur at n(π), nεI.
40. They also occur at π/2 + nπ, nεI.
41. If we merge the results, we get non-permissible values at nπ/2, nεI.
42. Now inspect the graphs of the L.S. and R.S. of the equation.
43. Graph the left side, Y1 = cotxsinxsecx.
44. Now graph the right side, Y2 = 1.
45. Why do we get different graphs even though the identity is true?
46. When we graph y = 1 in a calculator, the calculator has no way of knowing if the "1" is just a number or the result of a lengthy simplification!
47. Therefore, the graph we use to represent the proof should come from the more complex expression, and it must account for the non-permissible values.
48. Technically, it is incorrect to say cotxsinxsecx = 1 is an identity because the L.S. and R.S. have different graphs.
49. However, the two expressions are so close to being an identity that we can create a true statement by including restrictions on the variable.
50. If we write the statement as: cotxsinxsecx = 1, with the restriction that x ≠ nπ/2, nεI, we now have a completely true statement and can call it an identity.
51. The graph of the L.S. includes the restrictions, so this is the graph we should use to represent the identity.
52. Note that the graphing calculator does not show the holes for y = cotxsinxsecx. It's up to you to know they exist based on the denominators from the original expression.
53. If we graph the L.S. in a graphing calculator as shown, then use 2nd -> Trace -> Value to evaluate the function at π/2, the calculator returns no result. This is because the function is undefined when x = π/2.
54. In conclusion, cotxsinxsecx = 1 is not an identity as written since the graphs are not identical. However, we can force this to be an identity by including variable restrictions.
05:45
##### Example 4: More Proofs with Reciprocal Trigonometric Identities
1. Prove that each trigonometric statement is an identity. State the non-permissible values of x so the identity is true. In Part A, prove that sinxsecx/cotx = tan2x.
2. Rewrite the statement and draw a line.
3. Rewrite secx and cotx in terms of sine and cosine.
4. Multiply the terms on the numerator.
5. Multiply the numerator by the reciprocal of the denominator.
6. Multiply to get (sinx/cosx)2
7. This becomes tan2x.
8. The L.S. equals the R.S., completing the proof.
9. Even though we have shown that the L.S. and R.S. are algebraically equivalent, we still need to specify the non-permissible values of x.
10. Inspect the proof to find denominators.
11. Non-permissible values exist when cosx = 0 and sinx = 0.
12. Also, non-permissible values exist when cotx = 0.
13. Now we'll solve these three equations.
14. Non-permissible values occur at π/2 + nπ, nεI.
15. They also occur at n(π), nεI.
16. and at π/2 + nπ, nεI.
17. Merging the solutions yields non-permissible values at nπ/2, nεI.
18. Finally, use a graphing calculator to inspect the L.S. and R.S.
19. First graph the left side, Y1 = sinxsecx/cotx, and draw holes at the non-permissible values.
20. Then graph the right side, Y2 = tan2x.
21. The graph of the L.S. includes all the restrictions, so this is the graph we should use to represent the identity.
22. In conclusion, sinxsecx/cotx = tan2x is not an identity as written since the graphs are not identical. However, we can force this to be an identity by including variable restrictions.
23. In Part B, prove that sin2xsec2x = tan2x.
24. Rewrite the statement and draw a line.
25. sec2x = 1/cos2x.
26. Multiply
27. sin2x/cos2x = tan2x.
28. The L.S. equals the R.S., completing the proof.
29. Non-permissible values exist when cos2x = 0.
30. The non-permissible values occur at π/4 + nπ/2, nεI.
31. Finally, use a graphing calculator to inspect the L.S. and R.S.
32. First graph the left side, Y1 = sin2xsec2x,
33. Then graph the right side, Y2 = tan2x.
34. Both sides have the same graph.
35. In conclusion, sin2xsec2x = tan2x is an identity as written since the graphs are exactly the same. However, since non-permissible values are present, it is considered good form to write the identity with the variable restrictions.
09:23
##### Example 5: Proofs using the Pythagorean Identities
1. Prove that each trigonometric statement is an identity. State the non-permissible values of x so the identity is true. In Part A, prove that sin2x + 1/sec2x = 1.
2. Rewrite the statement and draw a line.
3. sec2x = 1/cos2x.
4. Simplify to get sin2x + cos2x.
5. Use the identity sin2x + cos2x = 1 to get a result of 1.
6. The L.S. equals the R.S., completing the proof.
7. Now identify denominators in the proof that could lead to non-permissible values.
8. Non-permissible values occur when cos2x = 0.
9. Non-permissible values could also occur when sec2x = 0.
10. Find the x-intercepts of cos2x = 0.
11. Take the square root of both sides.
12. This gives us cosx = 0.
13. Draw the graph.
14. Non-permissible values occur at π/2 + nπ, nεI.
15. Now find the x-intercepts of sec2x = 0.
16. Take the square root of both sides.
17. This gives us secx = 0.
18. Draw the graph.
19. The graph of y = secx has no x-intercepts.
20. The non-permissible values occur at the asymptotes, and they are the same values we found in the solution of cosx = 0.
21. Finally, use a graphing calculator to inspect the L.S. and R.S.
22. Graph the left side, Y1 = sin2x + 1/sec2x.
23. Graph the right side, Y2 = 1.
24. The graph of the L.S. includes all the restrictions, so this is the graph we should use to represent the identity.
25. In conclusion, sin2x + 1/sec2x = 1 is not an identity as written since the graphs are not identical. However, we can force this to be an identity by including variable restrictions.
26. In Part B, prove that cosx – cos3x = cosxsin2x.
27. Rewrite the statement and draw a line.
28. Factor out cosx.
29. Rearrange the identity sin2x + cos2x = 1 to get sin2x = 1 – cos2x.
30. Replace 1 – cos2x with sin2x. We now have cosxsin2x.
31. The L.S. equals the R.S., completing the proof.
32. There are no non-permissible values since there are no denominators.
33. Graph the left side, Y1 = cosx - cos3x.
34. Graph the right side, Y2 = cosxsin2x.
35. The left side and right side have identical graphs.
36. In conclusion, cosx - cos3x = cosxsin2x is an identity as written since the graphs are exactly the same. There are no non-permissible values.
37. In Part C, prove that sin3x - sinx = -sinxcos2x.
38. Rewrite the statement and draw a line.
39. Factor out sinx.
40. Rearrange the identity sin2x + cos2x = 1 to get sin2x – 1 = -cos2x.
41. Replace sin2x – 1 with -cos2x.
42. This gives us –sinxcos2x.
43. The L.S. equals the R.S., completing the proof.
44. There are no non-permissible values since there are no denominators.
45. Graph the left side, Y1 = sin3x - sinx.
46. Graph the right side, Y2 = -sinxcos2x.
47. The left side and right side have identical graphs.
48. In conclusion, sin3x - sinx = -sinxcos2x is an identity as written since the graphs are exactly the same. There are no non-permissible values.
49. in Part D, prove that sin2x + sin2xcos2x = sin2x(1 + cos2x).
50. Rewrite the statement and draw a line.
51. Factor out sin2x.
52. Note that there is no way to re-arrange sin2x + cos2x = 1 to get 1 + cos2x.
53. The L.S. equals the R.S., completing the proof.
54. There are no non-permissible values since there are no denominators.
55. Graph the left side, Y1 = sin2x + sin2xcos2x.
56. Graph the right side, Y2 = sin2x(1 + cos2x).
57. The left side and right side have identical graphs.
58. In conclusion, sin2x + sin2xcos2x = sin2x(1 + cos2x) is an identity as written since the graphs are exactly the same. There are no non-permissible values.
Section 2: DAY TWO: Examples 6 - 8 (41 minutes)
13:41
##### Example 6: Proofs with Common Denominators
1. Prove that each trigonometric statement is an identity. State the non-permissible values of x so the identity is true. In Part A, prove that cos2x + tan2xcos2x = 1.
2. Rewrite the statement and draw a line.
3. Factor out cos2x.
4. Use the identity 1 + tan2x = sec2x.
5. Replace 1 + tan2x with sec2x.
6. Rewrite sec2x as 1/cos2x.
7. The cos2x on the numerator and denominator cancel, giving us 1.
8. The L.S. equals the R.S., completing the proof.
9. Non-permissible values occur when cos2x = 0.
10. Find the x-intercepts of cos2x = 0.
11. Take the square root of each side.
12. This gives us cosx = 0.
13. Draw the graph.
14. Non-permissible values occur at π/2 + nπ, nεI.
15. Graph the left side, Y1 = cos2x + tan2xcos2x.
16. Graph the right side, Y2 = 1.
17. The graph of the L.S. includes the restrictions, so this is the graph we should draw.
18. In conclusion, cos2x + tan2xcos2x = 1 is not an identity as written since the graphs are not identical. However, we can force this to be an identity by including variable restrictions.
19. In Part B, prove that sec2x - 1/1+tan2x = sin2x.
20. Rewrite the statement and draw a line.
21. For the numerator, rearrange the identity 1 + tan2x =sec2x to get tan2x = sec2x – 1.
22. For the denominator, use the identity 1 + tan2x =sec2x.
23. Replace sec2x- 1 with tan2x, and replace 1+ tan2x with sec2x.
24. Express the fraction in terms of sine and cosine.
25. Multiply the numerator by the reciprocal of the denominator.
26. Cancel to get sin2x.
27. The L.S. equals the R.S., completing the proof.
28. Non-permissible values occur when cos2x= 0.
29. Non-permissible values could also occur when sec2x = 0.
30. Non-permissible values could also occur when 1 + tan2x = 0.
31. First, find the x-intercepts of cos2x = 0.
32. Take the square root of each side.
33. This gives us cosx = 0.
34. Draw the graph.
35. Non-permissible values occur when cosx = 0, at π/2 + nπ, nεI.
36. Next, find the x-intercepts of sec2x = 0.
37. Take the square root of each side.
38. This gives us secx = 0.
39. Draw the graph.
40. There are no solutions for secx = 0 since there are no x-intercepts. However, non-permissible values do occur at the asymptotes, where x = π/2 + nπ, n ε I.
41. And finally, find the x-intercepts of 1 + tan2x = 0.
42. Draw the graph.
43. There are no x-intercepts. However, non-permissible values do occur at the asymptotes, where x = π/2 + nπ, n ε I.
44. Now we'll draw the graphs of the left side and right side.
45. Graph the left side, Y1 = (sec2x - 1)/1+tan2x.
46. Graph the right side, Y2 = sin2x.
47. The graph of the L.S. includes the restrictions, so this is the graph we should use.
48. In conclusion, sec2x - 1/1+tan2x = sin2x is not an identity as written since the graphs are not identical. However, we can force this to be an identity by including variable restrictions.
49. In Part C, prove that sin2x/(1-cosx) = 1 + cosx.
50. Rewrite the statement and draw a line.
51. Rearrange the identity sin2x + cos2x = 1 to get sin2x = 1 – cos2x.
52. Replace sin2x with 1- cos2x.
53. Factor the numerator as a difference of squares.
54. Cancel to get 1 + cosx.
55. The L.S. equals the R.S., completing the proof.
56. Non-permissible values occur when 1 - cosx = 0.
57. Find the x-intercepts of 1 – cosx = 0.
58. Rearrange to get cosx = 1.
59. Draw the graph.
60. Non-permissible values occur at n(2π), nεI.
61. Graph the left side, Y1 = sin2x/1-cosx.
62. Graph the right side, Y2 = 1 +cosx.
63. The graph of the L.S. includes the restrictions, so this is the graph we should draw.
64. In conclusion, sin2x/(1-cosx) = 1 + cosx is not an identity as written since the graphs are not identical. However, we can force this to be an identity by including variable restrictions.
65. In Part D, prove that (sec2x/csc2x)(csc2x – 1) = 1.
66. Rewrite the statement and draw a line.
67. Rearrange the identity 1 + cot2x = csc2x to get csc2x – 1 = cot2x.
68. Express the fraction in terms of sine and cosine. Also, replace csc2x – 1with cot2x.
69. Multiply the numerator by the reciprocal of the denominator. Also, express cotan in terms of sine and cosine.
70. Cancel to get 1.
71. The L.S. equals the R.S., completing the proof.
72. Non-permissible values occur when cos2x = 0.
73. Non-permissible values also occur when sin2x = 0.
74. First, find the x-intercepts of cos2x = 0.
75. Take the square root of both sides.
76. This gives us cosx=0.
77. Draw the graph.
78. Non-permissible values occur at π/2 + nπ, nεI.
79. Now find the x-intercepts of sin2x = 0.
80. Take the square root of both sides.
81. This gives us sinx=0.
82. Draw the graph.
83. Non-permissible values occur at n(π), nεI.
84. We can merge the solutions to get non-permissible values at nπ/2, nεI.
85. Graph the left side, Y1 = (sec2x/csc2x)(csc2x – 1).
86. Graph the right side, Y2 = 1.
87. The graph of the L.S. includes the restrictions, so this is the graph we should draw.
88. In conclusion, (sec2x/csc2x)(csc2x – 1) = 1 is not an identity as written since the graphs are not identical. However, we can force this to be an identity by including variable restrictions.
12:33
##### Example 7: Proofs with Common Denominators
1. Prove that each trigonometric statement is an identity. State the non-permissible values of x so the identity is true. In Part A, prove that 1 + secx = (cosx + 1)/cosx.
2. Rewrite the statement and draw a line.
3. Rewrite secx as 1/cosx.
4. We need a common denominator of cosx, so multiply 1 by cosx/cosx.
5. Simplify.
6. Add the fractions to get (cosx + 1)/cosx.
7. The L.S. equals the R.S., completing the proof.
8. Non-permissible values exist when cosx = 0.
9. Draw the graph of y = cosx.
10. Non-permissible values occur at π/2 + nπ, nεI.
11. Graph the left side, Y1 = 1 + secx.
12. Graph the right side, Y2 = (cosx + 1)/cosx.
13. The graphs of the L.S. and R.S. are identical.
14. In conclusion, 1 + secx = (cosx + 1)/cosx is an identity as written since the graphs are exactly the same. There are no non-permissible values.
15. In Part B, prove that tan2x – sin2x = sin2xtan2x.
16. Rewrite the statement and draw a line.
17. Rewrite tan2x as sinx/cosx.
18. Get a common denominator of cos2x.
19. Simplify.
20. Subtract the fractions.
21. Factor out sin2x.
22. Rearrange the identity sin2x + cos2x = 1 to get 1 – cos2x = sin2x.
23. Replace 1- cos2x with sin2x.
24. Write sin2x/cos2x as tan2x. This gives us sin2xtan2x.
25. The L.S. equals the R.S., completing the proof.
26. Non-permissible values exist when cos2x = 0.
27. Find the solutions of cos2x = 0.
28. Take the square root of both sides.
29. This gives us cosx = 0.
30. Draw the graph.
31. Non-permissible values occur at π/2 + nπ, nεI.
32. Graph the left side, Y1 = tan2x – sin2x.
33. Graph the right side, Y2 = sin2xtan2x.
34. The graphs of the L.S. and R.S. are identical.
35. In conclusion, tan2x – sin2x = sin2xtan2x is an identity as written since the graphs are exactly the same. There are no non-permissible values.
36. In Part C, prove that cotx + tanx = secxcscx.
37. Rewrite the statement and draw a line.
38. Rewrite cotan and tan in terms of sine and cosine.
39. Get a common denominator.
40. Simplify.
42. Use the identity sin2x + cos2x = 1.
43. Replace sin2x + cos2x with 1.
44. 1/sinx becomes cscx, and 1/cosx becomes secx.This givesus cscxsecx.
45. The L.S. equals the R.S., completing the proof.
46. Non-permissible values exist when sinx = 0 and cosx = 0.
47. Graph y = sinx.
48. Non-permissible values occur at n(π), nεI.
49. Now graph y = cosx.
50. Non-permissible values occur at π/2 + nπ, nεI.
51. We can merge the solutions to get non-permissible values at nπ/2, nεI.
52. Graph the left side, Y1 = cotx + tanx.
53. Graph the right side, Y1 = cscxsecx.
54. The graphs of the L.S. and R.S. are identical.
55. In conclusion, cotx + tanx = secxcscx is an identity as written since the graphs are exactly the same. There are no non-permissible values.
56. In Part D, prove that (1 + tanx)/(1+ cotx) = tanx.
57. Rewrite the statement and draw a line.
58. Rewrite tanx and cotx in terms of sine and cosine.
59. Get a common denominator.
60. Simplify.
61. Add the fractions in the numerator and denominator.
62. Multiply the numerator by the reciprocal of the denominator.
63. Combine the fractions.
64. Rearrange the denominator so it matches the numerator.
65. Cancel.
66. sinx/cosx = tanx.
67. The L.S. equals the R.S., completing the proof.
68. Non-permissible values exist when cosx = 0 and sinx = 0.
69. Non-permissible values also exist when 1 + cotx = 0.
70. Draw the graph of y = cosx. Non-permissible values occur at π/2 + nπ, nεI.
71. Draw the graph of y = sinx. Non-permissible values occur at n(π), nεI.
72. We can merge the solutions to get non-permissible values at nπ/2, nεI.
73. Now find the non-permissible values that occur when 1 + cotx = 0.
74. Rearrange to get cotx= -1.
75. Graph y = cotx and y = -1.
76. Non-permissible values occur at 3π/4 + nπ, nεI.
77. Merge the non-permissible values.
78. Graph the left side, Y1 = (1 + tanx)/(1 + cotx) .
79. Graph the right side, Y2 = tanx .
80. The graph of the L.S. includes the restrictions, so this is the graph we should draw.
81. In conclusion, (1 + tanx)/(1+ cotx) = tanx is an identity as written since the graphs are exactly the same. There are no non-permissible values.
14:17
##### Example 8: More Proofs with Common Denominators
1. Prove that each trigonometric statement is an identity. State the non-permissible values of x so the identity is true. In Part A, prove that sinx/cosx + cosx/(1+sinx) = secx.
2. Rewrite the statement and draw a line.
3. Get a common denominator.
4. Simplify.
6. Use the identity sin2x + cos2x = 1.
7. Replace sin2x + cos2x with 1.
8. Rearrange the denominator so it matches the numerator.
9. Cancel to get 1/cosx.
10. 1/cosx = secx.
11. The L.S. equals the R.S., completing the proof.
12. Non-permissible values occur when cosx = 0 and 1 + sinx = 0.
13. Non-permissible values occur at π/2 + nπ, nεI.
14. Non-permissible values also occur at 3π/2 + n(2π), nεI.
15. We don't require the second solution set since it duplicates non-permissible values already in the first set.
16. Graph the left side, Y1 = sinx/cosx + cosx/(1 + sinx).
17. Graph the right side, Y2 = secx.
18. The graphs of the L.S. and R.S. are identical.
19. In conclusion, sinx/cosx + cosx/(1+sinx) = secx is an identity as written since the graphs are exactly the same. There are no non-permissible values.
20. In Part B, prove that (1 + tan2x)/(1 +cot2x) = tan2x.
21. Rewrite the statement and draw a line.
22. Use the identities 1 + tan2x = sec2x and 1 + cot2x = csc2x.
23. Replace 1 + tan2x with sec2x, and replace 1+ cot2x with csc2x.
24. Rewrite secant and cosecant in terms of cosine and sine.
25. Multiply the numerator by the reciprocal of the denominator.
26. Combine the fractions.
27. sin2x/cos2x = tan2x.
28. The L.S. equals the R.S., completing the proof.
29. Non-permissible values exist when cos2x = 0 and sin2x = 0.
30. Non-permissible values could also exist when 1 + cot2x = 0.
31. Begin with cos2x= 0.
32. Take the square root of both sides.
33. This gives us cosx = 0.
34. Graph y = cosx.
35. Non-permissible values occur at π/2 + nπ, nεI.
36. Let's move on to sin2x = 0.
37. Take the square root of both sides.
38. This gives us sinx = 0.
39. Graph y = sinx.
40. Non-permissible values occur at nπ, nεI.
41. We can merge the solutions to get non-permissible values at nπ/2, nεI.
42. Now we'll solve 1 + cot2x= 0.
43. Isolate cot2x.
44. There is no solution. We can't take the square root of a negative number.
45. Graph the left side, Y1 = (1 + tan2x)/(1 + cot2x).
46. Graph the right side, Y2 = tan2x.
47. The graph of the L.S. includes the restrictions, so this is the graph we should draw.
48. In conclusion, (1 + tan2x)/(1 +cot2x) = tan2x is not an identity as written since the graphs are not identical. However, we can force this to be an identity by including variable restrictions.
49. In Part C, prove that cosx/(1 + sinx) + cosx/(1-sinx) = 2secx.
50. Rewrite the statement and draw a line.
51. Get a common denominator.
52. Multiply the fractions.
54. The numerator becomes 2cosx.
55. Rearrange the identity sin2x + cos2x = 1 to get cos2x = 1 - sin2x.
56. Replace 1 – sin2x with cos2x.
57. Cancel.
58. This becomes 2 times secx.
59. The L.S. equals the R.S., completing the proof.
60. Non-permissible values occur when 1 + sinx = 0 and 1 - sinx = 0.
61. Begin with 1 + sinx=0.
62. This becomes sinx= -1.
63. We also have 1 – sinx= 0.
64. this becomes sinx = 1.
65. We can merge these equations to sinx = ±1.
66. Draw the graph.
67. Non-permissible values occur at at π/2 + nπ, nεI.
68. Graph the left side, Y1 = cosx/(1 + sinx) + cosx/(1-sinx).
69. Graph the right side, Y2 = 2secx.
70. The graphs of the L.S. and R.S. are identical.
71. In conclusion, cosx/(1 + sinx) + cosx/(1-sinx) = 2secx is an identity as written since the graphs are exactly the same. There are no non-permissible values.
72. in Part D, prove that cosx/(1 –sinx) = (1 + sinx)/cosx.
73. Rewrite the statement and draw a line.
74. The conjugate of 1 - sinx is 1 + sinx. It's the same binomial with the opposite sign. We can use conjugates in trigonometric proofs to create opportunities that let us use the Pythagorean identities.
75. Multiply the fraction on the left side by the conjugate of the denominator.
76. Multiply the fractions.
77. Rearrange the identity sin2x + cos2x = 1 to get cos2x = 1 - sin2x.
78. Replace 1 – sin2x with cos2x.
79. Cancel.
80. The L.S. equals the R.S., completing the proof.
81. Non-permissible values occur when 1 + sinx = 0 and 1 - sinx = 0.
82. Begin with 1 + sinx= 0.
83. This becomes sinx= -1.
84. We also have 1 – sinx= 0.
85. this becomes sinx = 1.
86. We can merge these equations to sinx = ±1.
87. Draw the graph.
88. Non-permissible values occur at π/2 + nπ, nεI.
89. Graph the left side, Y1 = cosx/(1 - sinx).
90. Graph the right side, Y2 = (1 + sinx)/cosx.
91. The graph of the right side includes all the restrictions, so this is the graph we should draw.
92. In conclusion, cosx/(1 –sinx) = (1 + sinx)/cosx is not an identity as written since the graphs are not identical. However, we can force this to be an identity by including variable restrictions.
Section 3: DAY THREE: Examples 9 - 14 (34 minutes)
05:12
##### Example 9: Assorted Proofs
1. Prove each identity. For simplicity, ignore NPV's and graphs. In Part A, prove that -4cotx/(1-csc2x) = 4tanx.
2. Rewrite the statement and draw a line.
3. Rearrange the identity 1 + cot2x = csc2x to get –cot2x = 1 – csc2x.
4. Replace 1 – csc2x with –cot2x.
5. Cancel to get 4/cotx.
6. Rewrite cotx as 1/tanx.
7. Multiply the numerator by the reciprocal of the denominator to get 4tanx.
8. The L.S. = R.S., completing the proof.
9. In Part B, prove that sin4x – cos4x = 2sin2x – 1.
10. Rewrite the statement and draw a line.
11. Factor the left side as a difference of squares.
12. Use the identity sin2x + cos2x = 1.
13. Replace sin2x + cos2x with 1.
14. Rearrange the identity sin2x + cos2x = 1 to get cos2x = 1 –sin2x.
15. Replace cos2x with 1 – sin2x.
16. Simplify.
17. The L.S. = R.S., completing the proof.
18. In Part C, prove that cot2x – csc2x = -1.
19. Rewrite the statement and draw a line.
20. Rearrange the identity 1 + cot2x = csc2x to get cot2x = csc2x - 1.
21. Replace cot2x with csc2x – 1.
22. Simplify to get -1.
23. The L.S. = R.S., completing the proof.
24. In Part D, prove that cscx – sinx = cosxcotx.
25. Rewrite the statement and draw a line.
26. Rewrite the left side in terms of sine.
27. Get a common denominator.
28. Subtract the fractions.
29. Rearrange the identity sin2x + cos2x = 1 to get 1 –sin2x = cos2x.
30. Replace 1 – sin2x with cos2x.
31. Write cos2x as cosx × cosx.
32. cosx/sinx = cotx. This gives us cosxcotx.
33. The L.S. = R.S., completing the proof.
05:25
##### Example 10: Assorted Proofs
1. Prove each identity. For simplicity, ignore NPV's and graphs. In Part A, prove that 1/(cscxsinxtanx)= cotx.
2. Rewrite the statement and draw a line.
3. Rewrite cscx and tanx in terms of sine and cosine.
4. Cancel.
5. Multiply the numerator by the reciprocal of the denominator.
6. This becomes cotx.
7. The L.S. = R.S., completing the proof.
8. In Part B, prove that csc2xcosx/tanx = csc3x - cscx.
9. Rewrite the statement and draw a line.
10. Rewrite cscx and tanx in terms of sine and cosine.
11. Multiply the numerator by the reciprocal of the denominator.
12. This becomes cos2x/sin3x.
13. Rearrange the identity sin2x + cos2x = 1 to get cos2x = 1 –sin2x.
14. Replace cos2x with 1 – sin2x.
15. Split the fraction.
16. Cancel.
17. Rewrite each fraction in terms of cosecant. This gives us csc3x – cscx.
18. The L.S. = R.S., completing the proof.
19. In Part C, prove that 1/5(sin2x) + 1/5(cos2x) = 1/5.
20. Rewrite the statement and draw a line.
21. Factor out 1/5.
22. Use the identity sin2x + cos2x = 1.
23. Replace sin2x + cos2x with 1.
24. This gives us 1/5.
25. The L.S. = R.S., completing the proof.
26. In Part D, prove that (secx – cosx)/sinx = tanx.
27. Rewrite the statement and draw a line.
28. Rewrite secx as 1/cosx.
29. Get a common denominator.
30. Subtract
31. Multiply the numerator by the reciprocal of the denominator.
32. This gives us (1 – cos2x)/(cosxsinx)
33. Rearrange the identity sin2x + cos2x = 1 to get sin2x = 1 - cos2x.
34. Replace 1 – cos2x with sin2x.
35. Cancel to get sinx/cosx.
36. This becomes tanx.
37. The L.S. = R.S., completing the proof.
05:34
##### Example 11: Assorted Proofs
1. Prove each identity. For simplicity, ignore NPV's and graphs. In Part A, prove that sinx/(1-cosx) = (1+ cosx)/sinx.
2. Rewrite the statement and draw a line.
3. We need to create an opportunity to use the Pythagorean identities. Multiply the numerator and denominator by the conjugate.
4. Multiply the fractions.
5. The denominator becomes 1 – cos2x.
6. Rearrange the identity sin2x + cos2x = 1 to get 1 - cos2x = sin2x.
7. Replace 1 - cos2x with sin2x.
8. Cancel to get (1 + cosx)/sinx.
9. The L.S. = R.S., completing the proof.
10. In Part B, prove that (1 – cosx)/sinx – sinx/(1 + cosx) = 0.
11. Rewrite the statement and draw a line.
12. Get a common denominator.
13. Simplify
14. Subtract the fractions.
15. Factor out -1 from the Pythagorean identity in the numerator.
16. cos2x + sin2x equals 1.
17. This evaluates to zero.
18. The L.S. = R.S., completing the proof.
19. In Part C, prove that (tanx – 1)2 = (1 – 2sinxcosx)/cos2x.
20. Rewrite the statement and draw a line.
21. In a proof, we can work with either side of the statement. It will be easier to complete this proof by working on the right side.
22. Split the fraction.
23. Simplify.
24. sinx/cosx becomes tanx.
25. Use the identity 1 + tan2x = sec2x.
26. Replace sec2x with 1 + tan2x.
27. Reorder the terms.
28. Factor.
29. The L.S. = R.S., completing the proof.
30. In Part D, prove that (1 + cosx)/(1 – cosx) = ((1 +cosx)/sinx)2.
31. Rewrite the statement and draw a line.
32. Square the right side.
33. Rearrange the identity sin2x + cos2x = 1 to get sin2x = 1 - cos2x.
34. Replace sin2x with 1 – cos2x.
35. The numerator can be written as (1 + cosx)(1+cosx). The denominator can be factored as a difference of squares to get (1 + cosx)(1 – cosx).
36. Cancel
37. The L.S. = R.S., completing the proof.
03:37
##### Example 12: Exploring a Proof Algebraically and Graphically
1. Prove algebraically that sinx = tanxcosx.
2. Rewrite the statement and draw a line.
3. On the right side, tanx becomes sinx/cosx.
4. Cancel to get sinx.
5. The L.S. = R.S., completing the proof.
6. In Part B, verify that sinx = tanxcosx for π/3.
7. Rewrite the statement and draw a line.
8. On the left side, plug in π/3.
9. sin π/3 = root(3)/2.
10. On the right side, plug in π/3.
11. tan π/3 becomes sin π/3/cos π/3.
12. Cancel.
13. sin π/3 = root(3)/2.
14. The L.S. = R.S., completing the verification.
15. In Part C, state the NPV's for sinx = tanxcosx.
16. Bring up the proof we completed in Part A.
17. Non-permissible values occur when cosx = 0.
18. cosx = 0 at π/2 + nπ, nεI.
19. In Part D, show graphically that sinx = tanxcosx. Are the graphs exactly the same?
20. Draw y1 = sinx.
21. Draw y2 = tanxcosx.
22. The graphs are not exactly the same. The graph of the R.S. has holes since tanxcosx has non-permissible values.
23. Reject the graph of y1 = sinx and keep the graph of y2 = tanxcosx. Only the graph of y2 = tanxcosx matches the non-permissible values of the proof.
04:46
##### Example 13: Exploring a Proof Algebraically and Graphically
1. Prove algebraically that cscx + cotx = (1 + cosx)/sinx.
2. Rewrite the statement and draw a line.
3. Rewrite cscx and cotx in terms of sine and cosine.
5. The L.S. = R.S., completing the proof.
6. In Part B, verify that cscx + cotx = (1 + cosx)/sinx for π/3.
7. Rewrite the statement and draw a line.
9. Rewrite cosecant and cotan in terms of sine and cosine.
10. cos π/3 = 1/2, and sin π/3 = root(3)/2.
11. Multiply each numerator by the reciprocal of the denominator.
12. Cancel.
14. Rationalize the denominator.
15. Multiply the fractions.
16. Cancel to get root(3).
17. Now move on to the right side. Plug in π/3.
18. cos π/3 = 1/2, and sin π/3 = root(3)/2.
19. Get a common denominator for the fractions in the numerator.
20. Add the fractions in the numerator, then multiply by the reciprocal of the denominator.
21. Cancel.
22. Rationalize the denominator.
23. Multiply the fractions.
24. Cancel to get root(3).
25. The L.S. = R.S., completing the verification.
26. In Part C, state the NPV's for cscx + cotx = (1 + cosx)/sinx.
27. Bring up the proof we completed in Part A.
28. Non-permissible values occur when sinx = 0.
29. sinx = 0 at nπ, nεI.
30. In Part D, show graphically that cscx + cotx = (1 + cosx)/sinx. Are the graphs exactly the same?
31. Graph y1 = cscx + cotx.
32. Graph y2 = (1 + cosx)/sinx.
33. The graphs are identical.
05:02
##### Example 14: Solving Trigonometric Equations with Identitity Substitutions
1. Prove algebraically that 1/(1-cosx) + 1/(1 + cosx) = 2csc2x.
2. Rewrite the statement and draw a line.
3. Get a common denominator.
4. Simplify.
6. Rearrange the identity sin2x + cos2x = 1 to get 1 - cos2x = sin2x.
7. Replace 1 – cos2x with sin2x.
8. This becomes 2csc2x.
9. The L.S. = R.S., completing the proof.
10. In Part B, verify that 1/(1 – cosx) + 1/(1 + cosx) = 2csc2x for π/2.
11. Rewrite the statement and draw a line.
12. Begin with the left side. Plug in π/2.
13. cos(π/2) = 0.
14. This becomes 1 + 1.
15. Which equals 2.
16. On the right side, plug in π/2.
17. Rewrite cosecant in terms of sine.
18. sin(π/2) = 1.
19. We get a result of 2.
20. The L.S. = R.S., completing the verification.
21. In Part C, state the NPV's for 1/(1 – cosx) + 1/(1 + cosx) = 2csc2x.
22. Bring up the proof we completed in Part A.
23. Non-permissible values occur when 1 - cosx = 0 and 1 + cosx = 0.
24. Begin with 1 – cosx = 0.
25. This becomes cosx = 1.
26. Next we have 1 + cosx = 0.
27. This becomes cosx = -1.
28. Merge the equations to get cosx = ±1.
29. cosx = ±1 at nπ, nεI.
30. In Part D, show graphically that 1/(1-cosx) + 1/(1 + cosx) = 2csc2x. Are the graphs exactly the same?
31. Graph y1 = 1/(1 – cosx) + 1/(1 + cosx).
32. Graph y2 = 2csc2x.
33. The graphs are exactly the same.
Section 4: DAY FOUR: Examples 15 - 19 (39 minutes)
09:45
##### Example 15: Solving Trigonometric Equations with Identitity Substitutions
1. Solve each trigonometric equation over the domain 0 ≤ x ≤ 2π. In Part A, solve 2sin2x – cosx – 1 = 0.
2. So far in this lesson we have been working with trigonometric proofs. In a trigonometric proof, operations across the equals sign is not allowed because we want to show that the L.S. = R.S.
3. We will now work with trigonometric equations. The goal of a trigonometric equation is to solve for the unknown(s) that make the equation true, and operations across the equals sign are allowed.
4. Rewrite the equation.
5. Rearrange sin2x + cos2x = 1 to get sin2x = 1 – cos2x.
6. Replace sin2x with 1 – cos2x.
7. Multiply the 2 through the brackets.
8. Simplify and reorder the terms.
9. Multiply both sides by -1.
10. We have the form 2a2 + a – 1. This factors to (2a – 1)(a + 1).
11. Factor to get (2cosx – 1)(cosx + 1).
12. Solve 2cosx – 1 = 0.
13. Isolate cosx.
14. The solution is x = π/3 and 5π/3.
15. Now solve cosx + 1 = 0.
16. Isolate cosx.
17. The solution is x = π.
18. Combining the solutions, we have π/3, π, and 5π/3.
19. The x-intercepts of the graph match the solutions of the equation.
20. In Part B, solve sinx = secxcotx.
21. Rewrite the equation.
22. Rewrite in terms of sine and cosine.
23. Cancel.
24. Cross multiply.
25. Bring all the terms to the left side of the equation.
26. We have the form a2 – 1, which factors to (a + 1)(a – 1).
27. Factor to get (sinx + 1)(sinx- 1).
28. Solve sinx + 1 = 0.
29. Isolate sinx.
30. The solution is x = 3π/2.
31. Now solve sinx – 1 = 0.
32. Isolate sinx.
33. The solution is x = π/2.
34. Combining the solutions, we have π/2 and 3π/2.
35. The points of intersection match the solutions of the equation.
36. In Part C, solve 2tan2x = -3secx.
37. Rewrite the equation.
38. Bring all the terms to one side of the equation.
39. Rearrange 1 + tan2x = sec2x to get tan2x = sec2x – 1.
40. Replace tan2x with sec2x – 1.
41. Multiply 2 through the brackets.
42. Simplify and reorder the terms.
43. We have the form 2a2 + 3a – 2, which factors to (2a – 1)(a + 2).
44. Factor to get (2secx – 1)(secx + 2).
45. Solve 2secx – 1 = 0.
46. Isolate secx.
47. This is equivalent to cosx = 2.
48. There is no solution to this equation.
49. Now solve secx + 2 = 0.
50. Isolate secx.
51. This is equivalent to cosx =-1/2.
52. The solution is x = 2π/3 and 4π/3.
53. Combining the solutions, we have 2π/3 and 4π/3.
54. The points of intersection match the solutions of the equation.
55. In Part D, solve cos2x = sin2x.
56. Rewrite the equation.
57. Bring all the terms to one side of the equation.
58. Rearrange sin2x + cos2x = 1 to get cos2x = 1 – sin2x.
59. Replace cos2x with 1 – sin2x.
60. Combine like terms.
61. Bring 2sin2x to the other side of the equation.
62. Flip the equation.
63. Divide both sides by 2.
64. Take the square root of each side.
65. Rationalize the denominator.
66. Multiply the fractions.
67. The solution is π/4, 3π/4, 5π/4, 7π/4.
68. The points of intersection match the solutions of the equation.
10:41
##### Example 16: Solving Trigonometric Equations with Identitity Substitutions
1. Solve each trigonometric equation over the domain 0 ≤ x ≤ 2π. In Part A, solve 3 – 3cscx + cot2x = 0.
2. Rewrite the equation.
3. Rearrange 1 + cot2x = csc2x to get cot2x = csc2x – 1.
4. Replace cot2x with csc2x – 1.
5. Simplify and reorder the terms.
6. We have the form a2 – 3a + 2, which factors to (a – 1)(a – 2).
7. Factor to get (cscx – 1)(cscx – 2).
8. Solve cscx – 1 = 0.
9. Isolate cscx.
10. This is equivalent to sinx= 1.
11. The solution is x = π/2.
12. Now solve cscx – 2 = 0.
13. Isolate cscx.
14. This is equivalent to sinx = 1/2.
15. The solution is x = π/6, 5π/6.
16. Combining the solutions, we have π/6, π/2, and 5π/6.
17. The x-intercepts of the graph match the solutions of the equation.
18. In Part B, solve 3sin2x + 3cosx – 4 = sin2x – 2cosx.
19. Rewrite the equation.
20. Collect like terms.
21. Rearrange sin2x + cos2x = 1 to get sin2x = 1 – cos2x.
22. Replace sin2x with 1 – cos2x.
23. Multiply the 2 through the brackets.
24. Reorder the terms.
25. Multiply both sides by -1.
26. We have the form 2a2 – 5a + 2, which factors to (2a – 1)(a – 2).
27. Factor to get (2cosx – 1)(cosx – 2).
28. Solve 2cosx – 1 = 0.
29. Add 1 to both sides.
30. Isolate cosx.
31. The solution is x = π/3 and 5π/3.
32. Now solve cosx – 2 = 0.
33. Isolate cosx.
34. There is no solution.
35. Combining the solutions, we have π/3 and 5π/3.
36. We can check our answer by graphing y = 2sin2x + 5cosx - 4. Recall that we obtained this when we brought all of the terms from the original question to one side of the equation.
37. In Part C, solve sin3x = sinx.
38. Rewrite the equation.
39. Bring all the terms to the left side of the equation.
40. Factor out sinx.
41. Factor sin2x – 1 as a difference of squares to get (sinx + 1) and (sinx – 1).
42. Solve sinx = 0.
43. The solution is x = 0, π, and 2π.
44. Now solve sinx + 1 = 0.
45. Isolate sinx.
46. The solution is x = 3π/2.
47. Now solve sinx – 1 = 0.
48. Isolate sinx.
49. The solution is x = π/2.
50. Combining the solutions, we have 0, π/2, π, 3π/2, and 2π.
51. The points of intersection match the solutions of the equation.
52. In Part D, solve 2sin3x – 2cos2x – sinx + 1 = 0.
53. Rewrite the equation.
54. Rearrange sin2x + cos2x = 1 to get cos2x = 1 – sin2x.
55. Replace cos2x with 1 – sin2x.
56. Multiply -2 through the brackets.
57. Simplify and reorder the terms.
58. Factor 2sin2x out of the first two terms, and factor -1 out of the second two terms.
59. Factor out sinx+ 1. This gives us (sinx + 1)(2sin2x – 1).
60. Solve sinx + 1 = 0.
61. Isolate sinx.
62. The solution is x = 3π/2.
63. Now solve 2sin2x – 1 = 0.
64. Add 1 to both sides.
65. Divide both sides by 2.
66. Square root each side.
67. This gives us sinx = ±1/root(2).
68. Rationalize the denominator.
69. This gives us sinx = ±root(2)/2.
70. The solution is x = π/4, 3π/4, 5π/4, and 7π/4.
71. Combining the solutions, we have π/4, 3π/4, 5π/4, 3π/2, and 7π/4.
72. The x-intercepts of the graph match the solutions of the equation.
09:03
##### Example 17: Solving Trigonometric Equations with Identitity Substitutions
1. Solve each trigonometric equation over the domain 0 ≤ x ≤ 2π. In Part A, solve 2sec2x – tan4x = -1.
2. Rewrite the equation.
3. Use the identity 1 + tan2x = sec2x.
4. Replace sec2x with 1 + tan2x.
5. Multiply 2 through the brackets.
6. Collect like terms.
7. Multiply both sides by -1.
8. Factor to get (tan2x – 3)(tan2x + 1).
9. Solve tan2x – 3 = 0.
10. Isolate tan2x.
11. Take the square root of both sides.
12. This gives us tanx = ±root(3).
13. The solution is x = π/3, 2π/3, 4π/3, and 5π/3.
14. Now solve tan2x + 1 = 0.
15. Isolate tan2x.
16. There is no solution.
17. Combining the solutions, we have x = π/3, 2π/3, 4π/3, and 5π/3.
18. We can check our answer by graphing y = 2sec2x - tan4x + 1. Recall that we obtained this when we brought all of the terms from the original question to one side of the equation.
19. In Part B, solve 2cos3x + 3cosx = 7cos2x.
20. Rewrite the equation.
21. Bring all the terms to the left side of the equation.
22. Factor out cosx.
23. Factor the trinomial to get (2cosx – 1)(cosx – 3).
24. Solve cosx = 0.
25. The solution is x = π/2 and 3π/2.
26. Now solve 2cosx – 1 = 0.
27. Add 1 to both sides of the equation.
28. Divide both sides by 2.
29. The solution is x = π/3 and 5π/3.
30. Now solve cosx – 3 = 0.
31. This becomes cosx = 3. There is no solution.
32. Combining the solutions, we have x = π/3, π/2, 3π/2, and 5π/3.
33. We can check our answer by graphing y = 2cos3x - 7cos2x + 3cosx. Recall that we obtained this when we brought all of the terms from the original question to one side of the equation.
34. In Part C, solve tan2x + 2sec2x – 3 = 0.
35. Rewrite the equation.
36. Use the identity 1 + tan2x = sec2x.
37. Replace sec2x with 1 + tan2x.
38. Multiply 2 through the brackets.
39. Collect like terms.
40. Isolate tan2x.
41. Take the square root of both sides.
42. This gives us tanx = ±1/root(3).
43. Rationalize the denominator.
44. This gives us ±root(3) /3.
45. The solution is x = π/6, 5π/6, 7π/6, and 11π/6.
46. The x-intercepts of the graph match the solutions of the equation.
47. In Part D, solve 4sin2x + 2root(2)sinx + 2root(3)sinx + root(6) = 0.
48. Rewrite the equation.
49. This resembles the expansion of two binomials. Let's see if we can factor (like step 3 of the decomposition method).
50. Factor 2sinx out of the first two terms. Factor root(3) out of the second two terms.
51. Factor out 2sinx + root(2). This gives us (2sinx + root(2))(2sinx + root(3)).
52. Solve 2sinx+ root(2) = 0.
53. Isolate sinx.
54. The solution is x = 5π/4 and 7π/4.
55. Now solve 2sinx + root(3) = 0.
56. Isolate sinx.
57. The solution is x = 4π/3 and 5π/3.
58. Combining the solutions, we have x = 5π/4, 4π/3, 5π/3, and 7π/4.
59. The x-intercepts of the graph match the solutions of the equation.
05:27
##### Example 18: Pythagorean Identities and Unknown Trigonometric Ratios
1. Use the Pythagorean identities to find the indicated value and draw the corresponding triangle. If the value of sinx = 4/7, and 0 ≤ x ≤ π/2, find the value of cosx within the same domain.
2. Begin with the identity sin2x + cos2x= 1.
3. Replace sinx with 4/7.
4. 4/7)2 = 16/49.
5. Isolate cos2x.
6. Get a common denominator.
7. Subtract.
8. Take the square root of both sides.
9. This gives us cosx = ±root(33)/7.
10. We only want the cosine value in quadrant I, which is cosx = root(33)/7.
11. We now have enough information to draw the triangle.
12. Now we'll move on to Part B. If the value of tanA = 3/2, and π ≤ A ≤ 3π/2, find the value of secA within the same domain.
13. Begin with the identity tan2A + 1 = sec2A.
14. Replace tanA with 3/2.
15. (3/2)2 = 9/4.
16. Get a common denominator.
17. This gives us sec2A = 13/4.
18. Take the square root of each side.
19. This gives us secA = ±root(13)/2.
20. We only want the secant value in quadrant III, secA = - root(13)/2.
21. We now have enough information to draw the triangle.
22. Now we'll move on to Part C. If the value of cosθ = root(7)/7, and cotθ < 0, find the value of cosθ within the same domain.
23. Begin with the identity sin2θ + cos2θ= 1.
24. Replace cosθ with root(7)/7.
25. (root(7)/7)2 = 7/49.
26. Reduce 7/49 to 1/7.
27. Subtract 1/7 from both sides.
28. Get a common denominator.
29. Subtract
30. Take the square root of both sides.
31. This gives us sinθ = ±root(6/7).
32. Rationalize the denominator.
33. This gives us sinθ = ±root(42)/7.
34. We only want the sine value in quadrant IV, which is sinθ = -root(42)/7.
35. We now have enough information to draw the triangle.
04:02
##### Example 19: Trigonometric Substitution
1. Using the triangle, show that root(9 – b2)/b2 can be expressed as cosθ/3sin2θ.
2. cosθ = a/3.
3. This gives us a = 3cosθ.
4. sinθ = b/3.
5. This gives us b = 3sinθ.
6. Rewrite root(9 – b2)/b2.
7. Plug in 3sinθ for b.
8. Square 3sinθ in both the numerator and denominator.
9. Factor out 9 in the numerator.
10. 1-sin2θ = cos2θ.
11. Take the square root of the numerator.
12. Reduce the coefficients to get the answer, cosθ/3sin2θ.
13. Now we'll move on to Part B. Using the triangle, show that a2/root(a2 + 16) can be expressed as 4cotθcosθ.
14. tanθ = 4/a.
15. Cross multiply.
16. Divide both sides by tanθ.
17. This gives us a = 4cotθ.
18. Rewrite a2/root(a2 + 16).
19. Plug in 4cotθ for a.
20. Square 4cotθ in both the numerator and denominator.
21. Factor out 16 in the denominator.
22. cot2θ + 1 = csc2θ.
23. Take the square root of the denominator.
24. Divide the coefficients.
25. Rewrite in terms of sine and cosine.
26. Multiply by the reciprocal of the denominator.
27. Cancel to get 4cos2θ/sinθ.
28. Rewrite cosθ as cosθ times cosθ.
29. This gets us to the answer, 4cotθcosθ.
Section 5: Other Lesson Materials
Workbook
24 pages
00:51
##### Trigonometric Identities I: Introduction
1. Welcome to Trigonometric Identities One. This topic will take four days to complete.
2. In Day 1 we will learn how to complete a trigonometric proof. We will also learn about the three Pythagorean identities.
3. In Day 2 we will continue our study of trigonometric proofs.
4. In Day 3 we will continue our study of trigonometric proofs.
5. In Day 4 we will solve trigonometric equations that involve identities.
01:03
##### Trigonometric Identities I: Summary
1. You have now completed Trigonometric Identities One. Over the past four days you have learned:
2. how to set up and complete a trigonometric proof.
3. how to derive the three Pythagorean identities.
4. how to complete proofs involving identitity substitution.
5. how to find non-permissible values.
6. how to interpret trigonometric graphs with asymptotes and holes.
7. and finally, you have learned how to solve trigonometric equations that involve identity substitutions.
Trigonometric Identities One
18 questions

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