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Chapter 5: Analytic
Trigonometry
Section 5.1a:
Fundamental Identities
HW: p. 451-452 1-7 odd,
27-49 odd
Is this statement true?
x 1
 x 1
x 1
2
This identity is a true sentence, but only
with the qualification that x must be in the
domain of both expressions.
If either side of the equality is undefined (i.e., at x = –1), then
the entire expression is meaningless!!!
The statement tan   sin  cos is a trigonometric identity
because it is true for all values of the variable for which both
sides of the equation are defined.
The set of all such values is called the domain of validity of
the identity.
Basic Trigonometric Identities
Reciprocal Identities
1
1
1
csc  
sec  
cot  
sin 
cos 
tan 
1
1
1
sin  
cos  
tan  
csc 
sec 
cot 
Quotient Identities
sin 
tan  
cos 
 0
cos 
cot  
sin 
is in the domain of validity of exactly three of the basic
identities. Which three?
Basic Trigonometric Identities
Reciprocal Identities
1
1
1
csc  
sec  
cot  
sin 
cos 
tan 
1
1
1
sin  
cos  
tan  
csc 
sec 
cot 
Quotient Identities
sin 
tan  
cos 
cos 
cot  
sin 
For exactly two of the basic identities, one side of the equation
is defined at   0 and the other side is not. Which two?
Basic Trigonometric Identities
Reciprocal Identities
1
1
1
csc  
sec  
cot  
sin 
cos 
tan 
1
1
1
sin  
cos  
tan  
csc 
sec 
cot 
Quotient Identities
sin 
tan  
cos 
cos 
cot  
sin 
For exactly three of the basic identities, both sides of the
equation are undefined at   0 . Which three?
Pythagorean Identities
Recall our unit circle:
sint
P
cost
What are the coordinates of P?
(1,0)
P  cos t ,sin t 
So by the Pythagorean Theorem:
 cos t    sin t 
2
Divide by
 cos t 
2
1
:
 cos t    sin t   1
2
2
2
 cos t   cos t   cos t 
2
2
2
 1   tan t    sec t 
2
2
Pythagorean Identities
Recall our unit circle:
sint
P
What are the coordinates of P?
P  cos t ,sin t 
(1,0)
cost
So by the Pythagorean Theorem:
 cos t    sin t 
2
Divide by
1
 sin t  :
2
 cos t    sin t   1
2
2
2
 sin t   sin t   sin t 
2
2
2
  cot t   1   csc t 
2
2
Pythagorean Identities
cos   sin   1
1  tan   sec 
2
2
cot   1  csc 
2
Given
2
tan   5
2
and
cos  0, find sin 
2
and cos .
sec2   1  tan 2   1  52  26  sec   26
1
cos  
26
We only take the positive answer…why?
sin 
tan   5
 5 sin   5cos
cos 
5
 1 
sin   5 
 sin  
26
 26 
Cofunction Identities


sin      cos 
2



cos      sin 
2



tan      cot 
2



cot      tan 
2



sec      csc 
2



csc      sec 
2

Can you explain why each of these is true???
Odd-Even Identities
If
sin   x    sin x
cos   x   cos x
csc   x    csc x
sec   x   sec x
tan   x    tan x
cot   x    cot x
cos  0.34 , find sin    2.




Sine is odd  sin       sin    
2

2

Cofunction Identity 
  cos  0.34
Simplifying Trigonometric Expressions
Simplify the given expression.
sin x  sin x cos x  sin x  sin 2 x  cos 2 x 
3
2
 sin x 1
 sin x
How can we support this answer graphically???
Simplifying Trigonometric Expressions
Simplify the given expression.
2
2
sec
x

1
sec
x

1


  sec x  1  tan x
2
2
2
sin x
sin x
sin x
2
sin x 1

2
2
cos x sin x
1
2

 sec x
2
cos x
Graphical support?
Simplifying Trigonometric Expressions
Simplify the given expressions to either a constant or a basic
trigonometric function. Support your result graphically.
sin x cos x  sin x
1

1  tan x
cos x 
cos x

cos x sin x  cos x
1  cot x
1
sin x
sin x
sin x
cos x  sin x
sin x
 tan x


cos x
sin x  cos x cos x
Simplifying Trigonometric Expressions
Simplify the given expressions to either a constant or a basic
trigonometric function. Support your result graphically.
sec2 u  tan 2 u 1
 1
2
2
cos v  sin v 1
Simplifying Trigonometric Expressions
Use the basic identities to change the given expressions to ones
involving only sines and cosines. Then simplify to a basic
trigonometric function.
 sec y  tan y sec y  tan y 
sec y
 1
sin y  1
sin y 
 cos y  cos y  cos y  cos y 




1
cos y
 1  sin y  1  sin y   cos y 




 cos y  cos y   1 
Simplifying Trigonometric Expressions
Use the basic identities to change the given expressions to ones
involving only sines and cosines. Then simplify to a basic
trigonometric function.
 1  sin y  1  sin y   cos y 




 cos y  cos y   1 
1  sin y  sin y  sin y 1  sin y cos y



cos y
cos y
cos y
2
2
2
 cos y
Simplifying Trigonometric Expressions
Use the basic identities to change the given expressions to ones
involving only sines and cosines. Then simplify to a basic
trigonometric function.
1
1
2
sec2 x csc x
cos x sin x

1
1
sec2 x  csc2 x
 2
2
cos x sin x
1
cos 2 x sin 2 x
sin x


2
2
2
2
2
cos x sin x sin x  cos x sin x  cos x
sin x

1
 sin x
Let’s start with a practice problem…
Simplify the expression
cos x
sin x
cos x cos x sin x 1  sin x



1  sin x cos x 1  sin x cos x cos x 1  sin x
cos x  cos x    sin x 1  sin x 


1  sin x  cos x 
cos x  sin x  sin x

How about some
1  sin x  cos x  graphical support?
1  sin x
1


 sec x
1  sin x  cos x  cos x
2
2
Combine the fractions and simplify to a multiple of a power of a
basic trigonometric function.
1
1

1  sin x 1  sin x
1  sin x
1  sin x


1  sin x 1  sin x  1  sin x 1  sin x 
2
2
1  sin x  1  sin x



2
2
1

sin
x
cos
x
1  sin x 1  sin x 
 2sec x
2
Combine the fractions and simplify to a multiple of a power of a
basic trigonometric function.
sec x  1   sec x  1

1
1


sec x  1 sec x  1
 sec x  1 sec x  1
2

2
sec x  1
2

tan 2 x
 2 cot x
2
Combine the fractions and simplify to a multiple of a power of a
basic trigonometric function.
sin x
1  cos x sin x  1  cos x 


1  cos x
sin x
sin x 1  cos x 
2
2
sin x  cos x  2 cos x  1

sin x 1  cos x 
2 1  cos x 
2  2cos x


sin x 1  cos x  sin x 1  cos x 
2
2
2


2csc
x
sin x
Quick check of your algebra skills!!!
Factor the following expression (without any guessing!!!)
12 x  8 x  15
a c  180 b  8
2
What two numbers have a product of –180 and a sum of 8?
10,18
12 x  10 x  18 x  15
Group terms and factor: 2 x  6 x  5 3  6 x  5
Rewrite middle term:
Divide out common factor:
2
 6x  5 2x  3
Write each expression in factored form as an algebraic
expression of a single trigonometric function.
e.g.,
sin x  2sin x  1
2
Substitute:
Let
 2sin x  3sin x 1
u  sin x
sin x  2sin x  1  u  2u  1
2
2
Factor:
“Re”substitute for your answer:
  u  1
2
  sin x  1
2
Write each expression in factored form as an algebraic
expression of a single trigonometric function.
e.g.,
 2sin x  3sin x 1
2
2
2
1
 cos x  1  2 cos x  cos x
sec x
2
 1  cos x 
Write each expression in factored form as an algebraic
expression of a single trigonometric function.
e.g.,
 2sin x  3sin x 1
sin x  cos x  1  sin x  1  sin x   1
2
2
 sin x  sin x  2
2
Let
u  sin x
 u  u  2   u 1 u  2
2
  sin x 1 sin x  2
Write each expression in factored form as an algebraic
expression of a single trigonometric function.
e.g.,
 2sin x  3sin x 1
sec x  sec x  tan x
2
2
 sec2 x  sec x   sec2 x  1
 2sec x  sec x  1
2
  2sec x  1 sec x 1
Write each expression as an algebraic expression of a single
trigonometric function.
e.g., 2sin x  3
tan   1  tan   1 tan   1

 tan  1
1  tan 
1  tan 
2
tan x
sec x  1  sec x  1 sec x  1


sec x  1
sec x  1 sec x  1
2
2
 sec x 1