1. Basic Trigonometric Identities

2. Example Verify the identity: sec x cot x = csc x.
Solution The left side of the equation contains the more complicated expression. Thus, we work with the left side. Let us express this side of the identity in terms of sines and cosines. Perhaps this strategy will enable us to transform the left side into csc x, the expression on the right.
Apply a reciprocal identity: sec x = 1/cos x
and a quotient identity: cot x = cos x/sin x.
Divide both the numerator and the
denominator by cos x, the common factor.

3. Example Verify the identity: cosx - cosxsin2x = cos3x..
Solution We start with the more complicated side, the left side. Factor out the greatest common factor, cos x, from each of the two terms.
cos x - cos x sin2 x = cos x(1 - sin2 x) Factor cos x from the two terms.
Use a variation of sin2 x + cos2 x = 1. Solving for cos2 x, we obtain cos2 x = 1 – sin2 x.
= cos x · cos2 x
Multiply.
= cos3 x
We worked with the left and arrived at the right side. Thus, the identity is verified.

4. Guidelines for Verifying Trigonometric Identities
Work with each side of the equation independently of the other side. Start with the more complicated side and transform it in a step-by-step fashion until it looks exactly like the other side.
Analyze the identity and look for opportunities to apply the fundamental identities. Rewriting the more complicated side of the equation in terms of sines and cosines is often helpful.
If sums or differences of fractions appear on one side, use the least common denominator and combine the fractions.
Don't be afraid to stop and start over again if you are not getting anywhere. Creative puzzle solvers know that strategies leading to dead ends often provide good problem-solving ideas.

5. Example
Verify the identity: csc(x) / cot (x) = sec (x)
Solution:

6. Example
Verify the identity:
Solution:

7. Example
Verify the following identity:
Solution:

8. 7.2 Trigonometric Equations

9. Equations Involving a Single Trigonometric Function
To solve an equation containing a single trigonometric function:
• Isolate the function on one side of the equation.
• Solve for the variable.

10. Trigonometric Equationsy y
= cos x 1 y
= 0.5 x –4  –2  2  4  –1
cos x = 0.5 has infinitely many solutions for –  < x <  y y
= cos x 1
0.5 x 2 
cos x = 0.5 has two solutions for 0 < x < 2 –1

11. Example Solve the equation: 3 sin x - 2 = 5 sin x - 1, 0 ≤ x < 360°
Solution The equation contains a single trigonometric function, sin x.
Step 1 Isolate the function on one side of the equation. We can solve for sin x by collecting all terms with sin x on the left side, and all the constant terms on the right side.
3 sin x - 2 = 5 sin x - 1
This is the given equation.
3 sin x - 5 sin x - 2 = 5 sin x - 5 sin x – 1
Subtract 5 sin x from both sides.
-2 sin x - 2 = -1
Simplify.
-2 sin x = 1
Add 2 to both sides.
sin x = -1/2
Divide both sides by -2 and solve for sin x.
x = 210° or x = 330°

12. Example Solve the equation: 2 cos2 x + cos x - 1 = 0, 0 £ x < 2p.
Solution The given equation is in quadratic form 2t2 + t - 1 = 0 with t = cos x. Let us attempt to solve the equation using factoring.
2 cos2 x + cos x - 1 = 0
This is the given equation.
(2 cos x - 1)(cos x + 1) = 0
Factor. Notice that 2t2 + t – 1 factors as (2t – 1)(2t + 1).
2 cos x - 1= or cos x + 1 = 0
Set each factor equal to 0.
2 cos x = 1 cos x = -1
Solve for cos x.
cos x = 1/2
x = p x = 2ppp x = p
The solutions in the interval [0, 2p) are p/3, p, and 5p/3.

13. Example Solve the following equation for θ is any real number.
Solution:

14. Example
Solve the equation on the interval [0,2)
Solution:

15. Example
Solve the equation on the interval [0,2)
Solution:

16. 7.3 Sum and Difference Formulas

17. The Cosine of the Difference of Two Angles
The cosine of the difference of two angles equals the cosine of the first angle times the cosine of the second angle plus the sine of the first angle times the sine of the second angle.

18. Example Find the exact value of cos 15°
Solution We know exact values for trigonometric functions of 60° and 45°. Thus, we write 15° as 60° - 45° and use the difference formula for cosines.
cos l5° = cos(60° - 45°)
cos( -) = cos  cos  + sin  sin 
= cos 60° cos 45° + sin 60° sin 45°
Substitute exact values from
memory or use special triangles.
Multiply.
Add.

19. cos 80° cos 20° + sin 80° sin 20° = cos (80° - 20°) = cos 60° = 1/2
Example
Find the exact value of cos 80° cos 20° + sin 80° sin 20°.
Solution The given expression is the right side of the formula for cos( - ) with  = 80° and  = 20°.
cos( -) = cos  cos  + sin  sin 
cos 80° cos 20° + sin 80° sin 20° = cos (80° - 20°) = cos 60° = 1/2

20. Example
Find the exact value of cos(180º-30º)
Solution

21. Example
Verify the following identity:
Solution

22. Sum and Difference Formulas for Cosines and Sines

23. Example
Find the exact value of sin(30º+45º)
Solution

24. Sum and Difference Formulas for Tangents
The tangent of the sum of two angles equals the tangent of the first angle plus the tangent of the second angle divided by 1 minus their product.
The tangent of the difference of two angles equals the tangent of the first angle minus the tangent of the second angle divided by 1 plus their product.

25. Example Find the exact value of tan(105º) tan(105º)=tan(60º+45º)
Solution
tan(105º)=tan(60º+45º)

26. Example
Write the following expression as the sine, cosine, or tangent of an angle. Then find the exact value of the expression.
Solution

27. Review Quiz Answers

28. 7.4 Double-Angle and Half-Angle Formulas

29. Double-Angle Identities

30. Three Forms of the Double-Angle Formula for cos2

31. Power-Reducing Formulas

32. Example
If sin α = 4/5 and α is an acute angle, find the exact values of sin 2 α and cos 2 α.
Solution
If we regard α as an acute angle of a right triangle, we obtain
cos α = 3/5. We next substitute in double-angle formulas:
Sin 2 α = 2 sin α cos α = 2 (4/5)(3/5) = 24/25.
Cos 2 α = cos2 α – sin2 α = (3/5)2 – (4/5)2 = 9/25 – 16/25 = -7/25.

33. Half-Angle Identities

34. Example Find the exact value of cos 112.5°.
Solution Because 112.5° = 225°/2, we use the half-angle formula for cos /2 with  = 225°. What sign should we use when we apply the formula? Because 112.5° lies in quadrant II, where only the sine and cosecant are positive, cos 112.5° < 0. Thus, we use the - sign in the half-angle formula.

35. Half-Angle Formulas for:

36. Example
Verify the following identity:
Solution

37. 7.5 Product-to-Sum and Sum-to-Product Formulas

38. Product-to-Sum Formulas

39. Example
Express the following product as a sum or difference:
Solution

40. Text Example
Express each of the following products as a sum or difference.
a. sin 8x sin 3x b. sin 4x cos x
Solution The product-to-sum formula that we are using is shown in each of the voice balloons. a.
sin 8x sin 3x = 1/2[cos (8x - 3x) - cos(8x + 3x)] = 1/2(cos 5x - cos 11x)
sin  sin  = 1/2 [cos( - ) - cos( + )]
sin  cos  = 1/2[sin( + ) + sin( - )] b.
sin 4x cos x = 1/2[sin (4x + x) + sin(4x - x)] = 1/2(sin 5x + sin 3x)

41. Sum-to-Product Formulas

42. Example
Express the difference as a product:
Solution

43. Example
Express the sum as a product:
Solution

44. Example
Verify the following identity:
Solution

45. Inverse Trig Functions
7.6
Inverse Trig Functions
Objective: In this section, we will look at the definitions and
properties of the inverse trigonometric functions. We will recall
that to define an inverse function, it is essential that the function
be one-to-one.

46. 7.6 Inverse Trigonometric Functions and Trig Equations
Domain: [–1, 1]
Range:
Domain:
Range:
Domain: [–1, 1]
Range: [0, π] 46

47. Let us begin with a simple question:
What is the first pair of inverse functions that pop into YOUR mind?
This may not be your pair but this is a famous pair. But something is not quite right with this pair. Do you know what is wrong?
Congratulations if you guessed that the top function does not really have an inverse because it is not 1-1 and therefore, the graph will not pass the horizontal line test.

48. Consider the graph of
Note the two points on the graph and also on the line y=4.
f(2) = 4 and f(-2) = 4 so what is an inverse function supposed to do with 4?
By definition, a function cannot generate two different outputs for the same input, so the sad truth is that this function, as is, does not have an inverse.

49. So how is it that we arrange for this function to have an inverse?4
y=x 2
We consider only one half of the graph: x > 0.
The graph now passes the horizontal line test and we do have an inverse:
Note how each graph reflects across the line y = x onto its inverse.

50. A similar restriction on the domain is necessary to create an inverse function for each trig function.
Consider the sine function.
You can see right away that the sine function does not pass the horizontal line test.
But we can come up with a valid inverse function if we restrict the domain as we did with the previous function.
How would YOU restrict the domain?

51. Take a look at the piece of the graph in the red frame.
We are going to build the inverse function from this section of the sine curve because:
This section picks up all the outputs of the sine from –1 to 1.
This section includes the origin. Quadrant I angles generate the positive ratios and negative angles in Quadrant IV generate the negative ratios.
Lets zoom in and look at some key points in this section.

52. I have plotted the special angles on the curve and the table.

53. The new table generates the graph of the inverse.
The domain of the chosen section of the sine is
So the range of the arcsin is
To get a good look at the graph of the inverse function, we will “turn the tables” on the sine function.
The range of the chosen section of the sine is [-1 ,1] so the domain of the arcsin is [-1, 1].

54. Note how each point on the original graph gets “reflected” onto the graph of the inverse.
etc.
You will see the inverse listed as both:

55. In the tradition of inverse functions then we have:
Unless you are instructed to use degrees, you should assume that inverse trig functions will generate outputs of real numbers (in radians).
The thing to remember is that for the trig function the input is the angle and the output is the ratio, but for the inverse trig function the input is the ratio and the output is the angle.

56. The other inverse trig functions are generated by using similar restrictions on the domain of the trig function. Consider the cosine function:
What do you think would be a good domain restriction for the cosine?
Congratulations if you realized that the restriction we used on the sine is not going to work on the cosine.

57. The chosen section for the cosine is in the red frame
The chosen section for the cosine is in the red frame. This section includes all outputs from –1 to 1 and all inputs in the first and second quadrants.
Since the domain and range for the cosine are the domain and range for the inverse cosine are

58. The other trig functions require similar restrictions on their domains in order to generate an inverse.
Like the sine function, the domain of the section of the
tangent that generates the arctan is
y=arctan(x)
y=tan(x)

59. arcsin(x) arccos(x) arctan(x) Domain Range
The table below will summarize the parameters we have so far. Remember, the angle is the input for a trig function and the ratio is the output. For the inverse trig functions the ratio is the input and the angle is the output.
arcsin(x)
arccos(x)
arctan(x)
Domain
Range
When x<0, y=arcsin(x) will be in which quadrant?
y<0 in IV
When x<0, y=arccos(x) will be in which quadrant?
y>0 in II
y<0 in IV
When x<0, y=arctan(x) will be in which quadrant?

60. The graphs give you the big picture concerning the behavior of the inverse trig functions. Calculators are helpful with calculations (later for that). But special triangles can be very helpful with respect to the basics. 2 1 2
Use the special triangles above to answer the following. Try to figure it out yourself before you click.

61. OK, lets try a few more. Try them before you peek.2 1 2

62. Negative inputs for the arccos can be a little tricky.x -1 2 1 2
From the triangle you can see that arccos(1/2) = 60 degrees. But negative inputs for the arccos generate angles in Quadrant II so we have to use 60 degrees as a reference angle in the second quadrant.

63. You should be able to do inverse trig calculations without a calculator when special angles from the special triangles are involved. You should also be able to do inverse trig calculations without a calculator for quadrantal angles.
Its not that bad. Quadrantal angles are the angles between the quadrants—angles like
To solve arccos(-1) for example, you could draw a quick sketch of the cosine section:
And observe that arccos(-1) =

64. But a lot of people feel comfortable using the following sketch and the definitions of the trig ratios.
r = 1
For arccos(-1) for example, you can observe that, since
the point (-1, 0) is the one we want. That point is on the terminal side of
So, since

65. Finally, we encounter the composition of trig functions with inverse trig functions. The following are pretty straightforward compositions. Try them yourself before you click to the answer. so
First, what do we know about
We know that is an angle whose sine is
Did you suspect from the beginning that this was the answer because that is the way inverse functions are SUPPOSED to behave? If so, good instincts but….

66. Consider a slightly different setup:
This is also the composition of two inverse functions but…
Did you suspect the answer was going to be 120 degrees? This problem behaved differently because the first angle, 120 degrees, was outside the range of the arcsin. So use some caution when evaluating the composition of inverse trig functions.
The remainder of this presentation consists of practice problems, their answers and a few complete solutions.

67. Find the exact value of each expression without using a calculator
Find the exact value of each expression without using a calculator. When your answer is an angle, express it in radians. Work out the answers yourself before you click.

68. Answers for problems 1 – 9.
Negative ratios for arccos generate angles in Quadrant II. y x 1 2
The reference angle is
so the answer is

69. y x -1 2
14. x 1 2 y
15.

70. Review Answers for Test