The document discusses trigonometric identities and product-sum formulas. It shows that the product of sine and cosine, sin(u)cos(v), can be written as a sum of trigonometric functions using the addition formulas for sine. The four product-to-sum formulas are derived from the addition formulas. The formulas can also be used in the reverse as sum-to-product formulas. Examples are provided to demonstrate using the formulas to prove identities and solve equations.

1. 7.1 Trigonometric Identities

2. 2

3. 3
Product-Sum Formulas
It is possible to write the product sin u cos as a sum of
trigonometric functions. To see this, consider the Addition
and Subtraction Formulas for Sine:
sin(u +) = sin u cos + cos u sin
sin(u –) = sin u cos – cos u sin
Adding the left- and right-hand sides of these formulas
gives
sin(u +) = sin(u –) = 2 sin u cos

4. 4
Product-Sum Formulas
Dividing by 2 gives the formula
sin u cos = [sin(u +) + sin(u –)]
The other three Product-to-Sum Formulas follow from the
Addition Formulas in a similar way.

5. 5
Product-Sum Formulas
The Product-to-Sum Formulas can also be used as
Sum-to-Product Formulas. This is possible because the
right-hand side of each Product-to-Sum Formula is a sum
and the left side is a product. For example, if we let
in the first Product-to-Sum Formula, we get
so

6. 6
Product-Sum Formulas
The remaining three of the following Sum-to-Product
Formulas are obtained in a similar manner.

7. 7
Example 5 – Proving an Identity
Verify the identity .
Solution:
We apply the second Sum-to-Product Formula to the
numerator and the third formula to the denominator.
Sum-to-Product
Formulas

8. 8
Example 5 – Solution
Simplify
Cancel
cont’d

9. 9

10. 10
7.4 Basic Trigonometric Equations

11. 11
Basic Trigonometric Equations
An equation that contains trigonometric functions is called a
trigonometric equation. For example, the following are
trigonometric equations:
sin2  + cos2  = 1 2 sin – 1 = 0 tan 2 – 1 = 0
The first equation is an identity—that is, it is true for every
value of the variable . The other two equations are true
only for certain values of .
To solve a trigonometric equation, we find all the values of
the variable that make the equation true.

12. 12
Basic Trigonometric Equations
Solving any trigonometric equation always reduces to
solving a basic trigonometric equation—an equation of
the form T( ) = c, where T is a trigonometric function and c
is a constant.
In the next examples we solve such basic equations.

13. 13
Example 1 – Solving a Basic Trigonometric Equation
Solve the equation
Solution:
Find the solutions in one period. Because sine has
period 2, we first find the
solutions in any interval of
length 2. To find these
solutions, we look at the
unit circle in Figure 1.
Figure 1

14. 14
Example 1 – Solution
We see that sin  = in Quadrants I and II, so the solutions
in the interval [0, 2) are
Find all solutions. Because the sine function repeats its
values every 2 units, we get all solutions of the equation
by adding integer multiples of 2 to these solutions:
where k is any integer.
cont’d

15. 15
Example 1 – Solution
Figure 2 gives a graphical representation of the solutions.
Figure 2
cont’d

16. 16
Example 2 – Solving a Basic Trigonometric Equation
Solve the equation tan  = 2.
Solution:
Find the solutions in one period. We first find one
solution by taking tan–1 of each side of the equation.
tan  = 2
 = tan–1(2)
  1.12
Given equation
Take tan–1 of each side
Calculator (in radian mode)

17. 17
Example 2 – Solution
By the definition of tan–1 the solution that we obtained is the
only solution in the interval (–/2, /2) (which is an interval
of length ).
Find all solutions. Since tangent has period , we get all
solutions of the equation by adding integer multiples of  :
  1.12 + k
where k is any integer.
cont’d

18. 18
Example 2 – Solution
A graphical representation of the solutions is shown in
Figure 6.
You can check that the solutions shown in the graph
correspond to k = –1, 0, 1, 2, 3.
Figure 6
cont’d

19. 19
Basic Trigonometric Equations
In the next example we solve trigonometric equations that
are algebraically equivalent to basic trigonometric
equations.

20. 20
Example 3 – Solving Trigonometric Equations
Find all solutions of the equation.
(a) 2 sin  – 1 = 0 (b) tan2  – 3 = 0
Solution:
(a) We start by isolating sin  .
2 sin  – 1 = 0
2 sin  = 1
sin  =
Given equation
Add 1
Divide by 2

21. 21
Example 3 – Solution
This last equation is the same as that in Example 1. The
solutions are
 = + 2k  = + 2k
where k is any integer.
(b) We start by isolating tan .
tan2  – 3 = 0
tan2  = 3
tan  =
Given equation
Add 3
Take the square root
cont’d

22. 22
Example 3 – Solution
Because tangent has period , we first find the solutions
in any interval of length . In the interval (–/2, /2) the
solutions are  = /3 and  = –/3.
To get all solutions, we add integer multiples of  to
these solutions:
 = + k  = – + k
where k is any integer.
cont’d

23. 23
Solving Trigonometric Equations
by Factoring

24. 24
Solving Trigonometric Equations by Factoring
Factoring is one of the most useful techniques for solving
equations, including trigonometric equations.
The idea is to move all terms to one side of the equation,
factor, and then use the Zero-Product Property.

25. 25
Example 4 – A Trigonometric Equation of Quadratic Type
Solve the equation 2 cos2  – 7 cos  + 3 = 0.
Solution:
We factor the left-hand side of the equation.
2 cos2  – 7 cos  + 3 = 0
(2 cos  – 1)(cos  – 3) = 0
2 cos  – 1 = 0 or cos  – 3 = 0
cos  = or cos  = 3
Factor
Given equation
Solve for cos 
Set each factor equal to 0

26. 26
Example 4 – Solution
Because cosine has period 2, we first find the solutions in
the interval [0, 2). For the first equation the solutions are
 = /3 and  = 5/3 (see Figure 7).
Figure 7
cont’d

27. 27
Example 4 – Solution
The second equation has no solution because cos  is
never greater than 1.
Thus the solutions are
 = + 2k  = + 2k
where k is any integer.
cont’d

28. 28
Example 5 – Solving a Trigonometric Equation by Factoring
Solve the equation 5 sin  cos  + 4 cos  = 0.
Solution:
We factor the left-hand side of the equation.
5 sin  cos  + 2 cos  = 0
cos  (5 sin  + 2) = 0
cos  = 0 or 5 sin  + 4 = 0
sin  = –0.8
Given equation
Factor
Set each factor equal to 0
Solve for sin 

29. 29
Example 5 – Solution
Because sine and cosine have period 2, we first find the
solutions of these equations in an interval of length 2.
For the first equation the solutions in the interval [0, 2) are
 = /2 and  = 3/2 . To solve the second equation, we
take sin–1 of each side.
sin  = –0.80
 = sin–1(–0.80)
Second equation
Take sin–1 of each side
cont’d

30. 30
Example 5 – Solution
  –0.93
So the solutions in an interval of length 2 are  = –0.93
and  =  + 0.93  4.07 (see Figure 8).
Figure 8
Calculator (in radian mode)
cont’d

31. 31
Example 5 – Solution
We get all the solutions of the equation by adding integer
multiples of 2 to these solutions.
 = + 2k  = + 2k
  –0.93 + 2k   4.07 + 2k
where k is any integer.
cont’d

32. 32
7.5 More Trigonometric Equations

33. 33
More Trigonometric Equations
In this section we solve trigonometric equations by first
using identities to simplify the equation. We also solve
trigonometric equations in which the terms contain
multiples of angles.

34. 34
Solving Trigonometric Equations
by Using Identities

35. 35
Solving Trigonometric Equations by Using Identities
In the next example we use trigonometric identities to
express a trigonometric equation in a form in which it can
be factored.

36. 36
Example 1 – Using a Trigonometric Identity
Solve the equation 1 + sin = 2 cos2.
Solution:
We first need to rewrite this equation so that it contains
only one trigonometric function. To do this, we use a
trigonometric identity:
1 + sin = 2 cos2
1 + sin = 2(1 – sin2)
2 sin2 + sin – 1 = 0
(2 sin – 1)(sin + 1) = 0
Given equation
Pythagorean identity
Put all terms on one side
Factor

37. 37
Example 1 – Solution
2 sin – 1 = 0 or sin + 1 = 0
sin = or sin = –1
 = or  =
Because sine has period 2, we get all the solutions of the
equation by adding integer multiples of 2 to these
solutions.
cont’d
Set each factor
equal to 0
Solve for sin
Solve for  in the
interval [0, 2)

38. 38
Example 1 – Solution
Thus the solutions are
 = + 2k  = + 2k  = + 2k
where k is any integer.
cont’d

39. 39
Example 2 – Squaring and Using an Identity
Solve the equation cos + 1 = sin in the interval [0, 2).
Solution:
To get an equation that involves either sine only or cosine
only, we square both sides and use a Pythagorean identity.
cos + 1 = sin
cos2 + 2 cos + 1 = sin2
cos2 + 2 cos + 1 = 1 – cos2
2 cos2 + 2 cos = 0
Given equation
Pythagorean identity
Square both sides
Simplify

40. 40
Example 2 – Solution
2 cos  (cos  + 1) = 0
2 cos = 0 or cos + 1 = 0
cos = 0 or cos = –1
 = or  = 
Because we squared both sides, we need to check for
extraneous solutions. From Check Your Answers we see
that the solutions of the given equation are /2 and .
Set each factor
equal to 0
Solve for cos
Solve for  in [0, 2)
cont’d
Factor

41. 41
Example 2 – Solution
Check Your Answers:
 =  =  = 
cos + 1 = sin cos + 1 = sin cos  + 1 = sin 
0 + 1 = 1 0 + 1 ≟ –1 –1 + 1 = 0
cont’d

42. 42
Example 3 – Finding Intersection Points
Find the values of x for which the graphs of f(x) = sin x and
g(x) = cos x intersect.
Solution 1: Graphical
The graphs intersect where f(x) = g(x). In Figure 1 we
graph y1 = sin x and y2 = cos x on the same screen, for x
between 0 and 2.
(a) (b)
Figure 1

43. 43
Example 3 – Solution
Using or the intersect command on the graphing
calculator, we see that the two points of intersection in this
interval occur where x  0.785 and x  3.927.
Since sine and cosine are periodic with period 2, the
intersection points occur where
x  0.785 + 2k and x  3.927 + 2k
where k is any integer.
cont’d

44. 44
Example 3 – Solution
Solution 2: Algebraic
To find the exact solution, we set f(x) = g(x) and solve the
resulting equation algebraically:
sin x = cos x
Since the numbers x for which cos x = 0 are not solutions
of the equation, we can divide both sides by cos x:
= 1
tan x = 1
Equate functions
Divide by cos x
Reciprocal identity
cont’d

45. 45
Example 3 – Solution
The only solution of this equation in the interval (–/2, /2)
is x = /4. Since tangent has period , we get all solutions
of the equation by adding integer multiples of :
x = + k
where k is any integer. The graphs intersect for these
values of x.
You should use your calculator to check that, rounded to
three decimals, these are the same values that we
obtained in Solution 1.
cont’d

46. 46
Equations with Trigonometric Functions of Multiples of Angles
When solving trigonometric equations that involve functions
of multiples of angles, we first solve for the multiple of the
angle, then divide to solve for the angle.

47. 47
Example 4 – A Trigonometric Equation Involving a Multiple of an Angle
Consider the equation 2 sin 3 – 1 = 0.
(a) Find all solutions of the equation.
(b) Find the solutions in the interval [0, 2).
Solution:
(a) We first isolate sin 3 and then solve for the angle 3.
2 sin 3 – 1 = 0
2 sin 3 = 1
sin 3 =
Given equation
Add 1
Divide by 2

48. 48
Example 4 – Solution
3 =
Solve for 3 in the interval
[0, 2) (see Figure 2)
Figure 2
cont’d

49. 49
Example 4 – Solution
To get all solutions, we add integer multiples of 2 to
these solutions. So the solutions are of the form
3 = + 2k 3 = + 2k
To solve for , we divide by 3 to get the solutions
where k is any integer.
cont’d

50. 50
Example 4 – Solution
(b) The solutions from part (a) that are in the interval [0, 2)
correspond to k = 0, 1, and 2. For all other values of k
the corresponding values of  lie outside this interval.
So the solutions in the interval [0, 2) are
cont’d

51. 51
Example 5 – A Trigonometric Equation Involving a Half Angle
Consider the equation
(a) Find all solutions of the equation.
(b) Find the solutions in the interval [0, 4).
Solution:
(a) We start by isolating tan .
Given equation
Add 1

52. 52
Example 5 – Solution
Since tangent has period , to get all solutions, we add
integer multiples of  to this solution. So the solutions
are of the form
Divide by
Solve for in the interval
cont’d

53. 53
Example 5 – Solution
Multiplying by 2, we get the solutions
 = + 2k
where k is any integer.
(b) The solutions from part (a) that are in the interval
[0, 4) correspond to k = 0 and k = 1. For all other
values of k the corresponding values of x lie outside
this interval. Thus the solutions in the interval [0, 4)
are
cont’d