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Using
Fundamental
Identities
Objectives:
1. Recognize and write the fundamental
trigonometric identities
2. Use the fundamental trigonometric
identities to evaluate trigonometric
functions, simplify trigonometric expressions,
and rewrite trigonometric expressions
WHY???
Fundamental trigonometric
identities can be used to
simplify trigonometric
expressions.
Fundamental Trigonometric
Identities
Reciprocal Identities
1
sin u 
csc u
1
csc u 
sin u
1
cos u 
sec u
1
sec u 
cos u
1
tan u 
cot
1
cot u 
tan u
Quotient Identities
sin u
tan u 
cosu
cosu
cot u 
sin u
Fundamental Trigonometric
Identities
Pythagorean Identities
sin 2 u  cos 2 u  1
1 tan 2 u  sec 2 u
1 cot 2 u  csc 2 u
Even/Odd Identities
sin( u)  sin u
csc(u)  csc u
cos(u)  cosu
sec(u)  sec u
tan(u)  tan u
cot(u)  cot(u)
Fundamental Trigonometric
Identities
Cofunction Identities


sin   u= cos u
2



tan  u cot u
2



sec  u csc u
2



cos  u sin u
2



cot  u tan u
2



csc  u sec u
2

Example: If
and Ө is in quadrant II,
find each function value.
a) sec Ө
To find the value of this
function, look for an
identity that relates
tangent and secant.

Tip: Use Pythagorean Identities.
Example: If
and Ө is in quadrant II,
find each function value. (Cont.)

b) sin Ө
Tip: Use Quotient Identities.

c) cot ( Ө )
Tip: Use Reciprocal and
Negative-Angle Identities.
7
Example:
1
Use the values sin x  and
2
cos x > 0 and identities to find
the values of all six
trigonometric functions.
What quadrant will you use?
1st quadrant
1
1

csc x 
2
sin x 1 / 2
sin x  cos x  1
2
2
2
1
1 3
cos x  1     1  
4 4
2
2
3
cos x 
2
2 2 3
1


sec x 
3
cos x
3
1
sin x
3
1
2
tan x 



cos x
3
3
3
2
1
3
cot x 

 3
tan x 1
Your Turn: Using Identities to
Evaluate a Function
 Use the given values to evaluate the remaining trigonometric functions
 (You can also draw a right triangle)
3
1. sec u   , tan u  0
2
2. csc  5,cos  0
3
3
3. tan x 
,cos x  
3
2
Solution: #1
Quadrant III
2
cos   
3
5
sin   
3
5
tan  
2
3 5
csc   
5
2 5
cot  
5
Solution: #2
Negative y-axis
1
sin   5
tan   undefined
sec  undefined
cot   0
Solution: #3
Quadrant III
1
sin x  
2
csc x  2
2 3
sec x  
3
cot x  3
Simplify an Expression
 Simplify cot x cos x + sin x to a single
trigonometric function.
cos x
cot x 
sin x
cos x
cos 2 x
cos x  sin x 
 sin x 
sin x
sin x
cos 2 x  sin 2 x
1

 csc x
sin x
sin x
Example: Simplify
Simplify cos x csc x  csc x
1. Factor csc x out of the
expression.
2

csc x cos x  1
2


csc x cos x  1
2

2. Use Pythagorean identities
to simplify the expression
in the parentheses.
sin x  cos x  1
2
2
 sin x  cos x  1
2
2

csc x  sin x
2


csc x  sin x
2

3. Use Reciprocal identities
to simplify the expression.
1
2
 sin x
sin x


 sin x
  sin x
sin x
2
Your Turn: Simplifying a
Trigonometric Expression
1. sin x cos x  sin x
2
2. sec 2 x(1  sin 2 x)
2
tan x
3.
2
sec x
Solutions:
1. sin x cos x  sin x   sin x
2
2. sec 2 x(1  sin 2 x)  1
2
tan x
2
3.
 sin x
2
sec x
3
Factoring Trigonometric
Expressions
2
sec  1
-Factor the same way you would factor
any quadratic.
- If it helps replace the “trig” word with x
-Factor

sec  1 the same way you
2
would factor x 1
2
x 1  (x 1)(x  1) so sec   (sec  1)(sec   1)


2
2
Example: 2csc x  7 csc x  6
2
Make it an easier problem.
Let a = csc x
2a2 – 7a + 6
(2a – 3)(a – 2)
Now substitute csc x for a.
2csc x  3 csc x  2
Example: Factor sec x  3tanx  1.
2
1. Use Pythagorean identities
to get one trigonometric
function in the expression.
2
2
sec x  tan x  1.
 tan
2

x  1  3tanx  1
tan x  3tanx  2
2
2. Now factor.
 tan x  2 tan x  1 
Your Turn: Factoring
Trigonometric Expressions
1. 4 tan   tan   3
2
2. csc x  cot x  3
2
Solutions:
1. 4 tan 2   tan   3   4 tan   3 tan   1
2. csc2 x  cot x  3   cot x  2  cot x  1
Your Turn: Factor and
simplify
1. sin x csc x  sin x
2
2
2
2. 1  2cos x  cos x
2
4
Solutions:
1. sin x csc x  sin x  cos x
2
2
2
2
2. 1  2cos x  cos x  sin x
2
4
4
Adding Trigonometric Expressions
(Common Denominator)
sin 
cos

1 cos sin 
sin
2
  cos 2   1
sin 
sin 
cos (1 cos )



 
sin  1 cos  sin  (1 cos )

(sin  )(sin  )  (cos  )(1 cos ) sin 2   cos  cos2 

(1 cos )(sin  )
(1 cos )(sin  )
1 cos 

(1 cos  )(sin  )
1

sin 
 csc
Your Turn: Adding
Trigonometric Expressions
1
1
1.

sec x  1 sec x  1
2
sec x
2. tan x 
tan x
Solutions:
1
1
2
1.

 2cot x
sec x  1 sec x  1
2
sec x
2. tan x 
  cot x
tan x
Rewriting a Trigonometric
Expression so it is not in
Fractional Form
1
1  sin x 1  sin x
1



2
1  sin x 1  sin x 1  sin x 1  sin x
1  sin x
1
sin x



2
2
cos x cos x cos 2 x
1
sin x
1



2
cos x cos x cos x
 sec x  tan x sec x
2
Your Turn: Rewriting a
Trigonometric Expression
5
1.
tan x  sec x
Solution:
5
1.
 5sec x  5tan x
tan x  sec x
Trigonometric Substitution
4 x
x  2 tan 
2
4   2 tan  
2
4  4tan 2 
4(1  tan 2  )
4sec2 
2sec
64  16 x 2
x  2cos
64  16  2cos 
2
64  16  4cos 2  
64  64cos2 
64 1  cos 2  
64sin 2 
8sin 
Your Turn:
1.
x2  4
x  2sec
2.
x 2  100
x  10 tan 
Solutions:
1.
x 2  4  2 tan 
x  2sec
2.
x 2  100  10sec
x  10 tan 
Assignment:
 Sec 5.1 pg. 357 – 359: #1 – 13 odd, 15 – 26 all, 27
71 odd, 81 -91 odd