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Chapter 19
Trigonometric Equations and
Identities
Table of Contents




Basic Trigonometric Identities
Types of Expressions
Solving Trigonometric Equations
Solving Second Degree (Quadratic)
Trigonometric Equations
Slide 3
Slide 7
Slide 11
Slide 16
Basic Trigonometric
Identities
Notes
Basic Trigonometric Identities
Reciprocal
Identities
Quotient
Identities
1
sec 
cos
tan 
sin
cos
cos 2   sin 2   1
1
csc 
sin
cot 
cos
sin
1  tan 2   sec2 
cot 
1
tan
Pythagorean
Identities
1  cot 2   csc2 
Basic Trigonometric Identities
Basic Identity
Alternative Form
cos2   sin 2   1
cos2   1  sin 2 
sin 2   1  cos2 
1  tan 2   sec2 
1  cot 2   csc2 
1  sec2   tan 2 
tan 2   sec2   1
1  csc2   cot 2 
cot 2   csc2   1
Basic Trigonometric Identities
Double Angle Identities
sin 2  2 sin cos
cos 2  1  2 sin 
2
2 22 cos 2   1
cos 2  cos 2 cos
 sin
cos 2  cos 2   sin 2 
Types of Expressions
Notes
Types of Expressions
 An equation has a unique solution.
It is an expression that is only true
for certain replacement values.
 An identity is true for all
replacement values.
Types of Expressions
Equation
Identity
2x + 1 = 7
x=3
3x + 4x = 7x
Only true when x = 3.
This is always true no
matter what values
are substituted for x.
Types of Expressions
Example:
Rewrite each expression in
terms of sin θ, cos θ, or a
constant.
1)
tan  cos 
sin 
 cos   sin 
cos 
2)
1  cos 2 
1  cos 2   sin 2 
sec  cot  sin 
1
cos 

 sin   1
cos  sin 
3)
Solving Trigonometric
Equations
Notes
Solving Trigonometric Equations
Example:
Solve for x.
0 ≤ x ≤ 360
2cosx – 1 = 0
2cosx = 1
cosx = ½
x = 60°
Solving Trigonometric Equations
Remember:
 The answer to the problem is your reference angle.
 In the first quadrant, the answer equals the reference
angle.
 In the second quadrant, the answer equals 180 –
reference angle.
 In the third quadrant, the answer equals 180 +
reference angle.
 In the fourth quadrant, the answer equals 360 –
reference angle.
Solving Trigonometric Equations
Example:
Solve for x to the nearest
ten minutes.
0 ≤ x ≤ 360
2tanx + 3 = 2
2tanx = -1
tanx = -½
x = 26°34’
Solving Trigonometric Equations
 When using the calculator, do not enter the
negative sign when pressing 2ND TAN.
 The negative sign is used to determine the
quadrant that the angle lies in (Unit Circle).
 Therefore, the answer 26°34’ is the
reference angle.
 The angle lies in the second and fourth
quadrants, therefore, the angles are
153°30’ and 333°30.
Solving Second Degree
(Quadratic) Trigonometric
Equations
Notes
Solving Second Degree Equations
Example:
Solve. 0° ≤ x ≤ 360°
tan2x – 3tanx – 4 = 0
let x = tan x
x2 – 3x – 4 = 0
(x – 4)(x + 1) = 0
x–4=0
x=4
x+1=0
x = -1
tan x = 4
tan x = -1
x = 76°
x = 256°
x = 135°
x = 315°
Solving Second Degree Equations
Example:
Solve. 0° ≤ x ≤ 360°
3cos2x – 5cosx = 4
let x = cos x
3x2 – 5x – 4 = 0
x
 b  b 2  4 ac
x
2a
  5 
5  73
6
x  2.26
x
x   0.59
 52  4 3  4 
2 3
Solving Second Degree Equations
x = 2.26
x = -0.59
cos x = 2.26 cos x = -0.59
x=Ø
x = 126°
x = 234°
Solving Second Degree Equations
Example:
Solve. 0° ≤ x ≤ 360°
2cos2x = cosx
let x = cos x
2x2 = x
2x2 – x = 0
x(2x – 1) = 0
x=0
2x – 1 = 0
x = .5
cos x = 0
cos x = .5
x = 90°
x = 270°
x = 270°
Solving Second Degree Equations
Example:
Solve. 0° ≤ x ≤ 360°
3sinx + 4 = 1/sinx
let x = sin x
3x + 4 = 1/x
x(3x + 4) = x(1/x)
3x2 + 4x = 1
(3x – 1)(x + 1) = 0
Solving Second Degree Equations
3x - 1= 0
x = .333
x+1=0
x = -1
sin x = .333
sin x = -1
x = 19°
x = 161°
x = 270°
Solving Second Degree Equations
Example:
Solve. 0° ≤ x ≤ 360°
2cos2x – sinx = 1
Use the identity cos2x = 1 – sin2x
2 (1 – sin2x) – sinx = 1
2 – 2sin2x – sinx = 1
– 2sin2x – sinx + 1 = 0
2sin2x + sinx - 1 = 0
Solving Second Degree Equations
2sin2x + sinx - 1 = 0
let x = sin x
2x2 + x - 1 = 0
(2x - 1)(x + 1) = 0
2x - 1 = 0
x=½
x+1=0
x = -1
sin x = ½
x = 30°
x = 150°
sin x = -1
x = 270°
Chapter 19
End of Chapter