CHAT Pre-Calculus
Section 5.1
Using Fundamental Identities
In this lesson we will use the fundamental trig identities to
do the following:
1. Evaluate trig functions
2. Simplify trig expressions
3. Develop additional trig identities
4. Solve trig equations
Fundamental Trigonometric Identities
Reciprocal Identities
1
sin u 
csc u
1
csc u 
sin u
1
sec u
1
sec u 
cos u
cos u 
1
cot u
1
cot u 
tan u
tan u 
Quotient Identities
tan u 
sin u
cos u
cos u
cot u 
sin u
1
CHAT Pre-Calculus
Section 5.1
Pythagorean Identities
sin 2 u  cos 2 u  1
1  tan 2 u  sec2 u
1  cot 2 u  csc 2 u
Cofuntion Identities*


sin   u   cos u
2



cos  u   sin u
2



tan  u   cot u
2



cot   u   tan u
2



sec  u   csc u
2



csc  u   sec u
2


* 90° can be substituted for 2 in the cofunction identities.
Even/Odd Identities
sin(u )   sin u
csc( u )   csc u
cos(u )  cos u
sec( u )  sec u
tan(u )   tan u
cot(u )   cot u
Note: Sometimes variations of these are used, such as
cos2u = 1 – sin2 u.
2
CHAT Pre-Calculus
Section 5.1
Using Fundamental Identities
Example: If csc u = -5/3, and cos u > 0, find the values of
the other five trig functions.
(By allsintancos we know we are in Quadrant IV.)
3
sin u 
5
(csc and sin are reciprocals)
16
  3
cos u  1  sin u  1  
 
25
 5 
4
cos u 
5
5
sec u 
(sec and cos are reciprocals)
4
3
sin u
3 5 3
5
tan u 


 
4
cos u
5 4
4
5
4
cot u 
(cot and tan are reciprocals)
3
2
2
2
3
CHAT Pre-Calculus
Section 5.1
2
csc
x cot x  cot x .
Example: Simplify
csc 2 x cot x  cot x  cot x(csc2 x  1)
 cot x(cot 2 x)
 cot 3 x
Example: Simplify tan x sin x  cos x .
sin x
tan x sin x  cos x 
 sin x  cos x
cos x
sin 2 x

 cos x
cos x
sin 2 x
 cos x 

 cos x

cos x
 cos x 
sin 2 x cos 2 x


cos x cos x
sin 2 x  cos 2 x

cos x
1

cos x
 sec x
4
CHAT Pre-Calculus
Section 5.1
Note: Sometimes it is helpful to write everything in terms of
sine and cosine. Other times you will see forms of
the trig identities, as in the following example.
sec t
tan t

Example: Simplify tan t 1  sec t .
sec t
tan t
sec t (1  sec t )
tan t (tan t )





tan t 1  sec t tan t (1  sec t ) 1  sec t (tan t )
sec t  sec 2 t
tan 2 t


tan t (1  sec t ) tan t (1  sec t )
sec t  sec 2 t  tan 2 t

tan t (1  sec t )
sec t  1

tan t (1  sec t )
(1  sec t )

tan t (1  sec t )
1

tan t
 cot t
5
CHAT Pre-Calculus
Section 5.1
Example: Factor cos x  1 .
2
cos 2 x  1
(cos x  1)(cos x  1)
This is the difference of squares.
Example: Factor sin u  3sin u  10 .
2
sin 2 u  3 sin u  10
(sin u  2)(sin u  5)
This is backwards FOIL.
Example: Factor sec t  tan t  3 .
2
Write first in terms of one trig function.
sec 2 t  tan t  3  (tan 2 t  1)  tan t  3
 tan 2 t  tan t  2
 (tan t  1)(tan t  2)
6
CHAT Pre-Calculus
Section 5.1
1
Example: Rewrite sec x  1 so that it is not a fraction.
Multiply by 1 using the conjugate of the denominator.
1
1
(sec x  1)


sec x  1 sec x  1 (sec x  1)
sec x  1

sec 2 x  1
sec x  1

tan 2 x
sec x
1


tan 2 x tan 2 x
1
 1 
 sec x

2
2
 tan x  tan x
 sec x cot 2 x  cot 2 x
2
 1  cos x 
2



cot
x

2

 cos x  sin x 
 cos x 
2


cot
x

2
 sin x 
 1  cos x 
2


  cot x
 sin x  sin x 
 csc x cot x  cot
7
CHAT Pre-Calculus
Section 5.1
Example: Perform the addition and simplify.
sin x
cos x

1  cos x sin x
sin x
(sin x) cos x (1  cos x)




(1  cos x) (sin x) sin x (1  cos x)
sin 2 x
cos x  cos 2 x


(1  cos x)(sin x) (1  cos x)(sin x)
sin 2 x  cos x  cos 2 x

(1  cos x)(sin x)
[sin 2 x  cos 2 x]  cos x

(1  cos x)(sin x)
1  cos x

(1  cos x)(sin x)
(1  cos x)

(1  cos x)(sin x)
1

sin x
 csc x
8