Pre calculus Problems of the Day
Simplify the following:
2
2
  

 
a ) sin   cos 


6 
6 
2
2
  

 
b ) sin   cos 


4 
4 
2
2
 4  

4  
c ) sin
  cos



3 
3 
Trigonometric Identities - a statement of equality that is
true for all values where the function is defined.
Reciprocal Identities:
sin  
1
csc 
csc  

1
sin 
cos  
sec  
1
sec 
1
cos 
tan  
cot  
Quotient Identities:

tan  
sin 
cos 
cot  
cos 
sin 
1
cot 
1
tan 
Pythagorean Identities:
2
sin     sin 
2
sin   cos   1
2
2
1  cot   csc 
2
2
tan   1  sec 
Even/Odd Identities:
cos    cos 



tan     tan 
Verifying Trigonometric Identities
To verify a trigonometric identity we must show that one side of
the identity can be simplified so that it is identical to the other
side or each side can be simplified independently until they are
identical.
Never treat an identity like an equation. We are not solving
for the variable.
WE VERIFY (OR PROVE) IDENTITIES
BY DOING THE FOLLOWING:
•
•
•
•
Work with one side at a time.
We want both sides to be exactly the same.
Start with the more complicated side
Use algebraic manipulations and/or the basic
trigonometric identities until you have the same
expression as on the other side.
Techniques for verifying trigonometric identities.
1) Rename using the fundamental identities.
2) Rewrite a more complicated side in terms of sines and
cosines.
3) Factor.
4) Combine fractional expressions using an LCD.
5) Separate a single-term quotient into two terms.
6) Multiply the numerator and denominator on one side by a
binomial factor that appears on the other side of the
identity.
Verifying Trigonometric Identities
tan x cos x  1  cos x
2
2
2
 sin 2 x 
2
2
cos
x

1

cos
x


2
 cos x 
2
2
sin x  1  cos x
1  cos x  1  cos x
2
2
tan  csc   sec 
 sin    1 


  sec 
 cos    sin  
1
cos 
 sec 
sec   sec 
s in x s e c x c o t x  1
 1   cos x 
sin x 

1
 cos x   sin x 
1 1
ta n A  co t A  csc A se c A
sin A  sin A  cos A  cos A 



  csc A sec A
cos A  sin A 
s i n A  cos A 
sin A  cos A
2
2
 csc A sec A
sin A cos A
1
 csc A sec A
sin A cos A
 1  1 


  csc A sec A
 sin A   cos A 
csc A se c A  csc A se c A
csc x
cot x  tan x
 cos x
1
s in x
 cos x
cos x
s in x

s in x
cos x
1
sin x
 co s x
2
2
co s x  sin x
sin x co s x
1

sin x
sin x cos x
cos x  sin x
1
sin x
2

2
sin x cos x
 cos x
 cos x
1
cos x  cos x
1
sin y  1

1
  2 se c y
2
sin y  1
 sin y  1 
 sin y  1 
1
2



2
sec
y




sin y  1  sin y  1  sin y  1  sin y  1 
1
 sin y  1   sin y  1
 sin y  1  sin y  1
2
2
  2 sec y
2
sin y  1
2
2

  2 sec y
  2 se c y
2
 1  sin y  1
2

2
2
  2 sec y
2
cos y
 2 sec y   2 sec y
2
2
cos x  1  2 sin x  sin x
4

2
cos x
2

2
2
4
 1  2 sin x  sin x
 1  sin x   1  2 sin
x   1  sin x   1  2 sin
2
2
 1  sin
2
2
4
2
x  sin x
2
x  sin x
4
4
1  2 sin x  sin x  1  2 sin x  sin x
2
4
2
4
tan x 
sin x
cos x

cos x
1  sin x
cos x
1  sin x
sin x  1  sin x   cos x cos x
cos x  1  sin x 
sin x  sin x  co s x
2
 sec x
 sec x
 sec x
2
co s x  1  sin x 
sin x  1
cos x  1  sin x 
1
 se c x
 sec x
 sec x
cos x
sec x  sec x
sin x  1
1  cos x  sin x
2
sin x  1
sin x  sin x
2
sin x  1
sin x  1  sin x 
1
 csc x
 csc x
 csc x
 csc x
sin x
csc x  csc x
cot x  cot x  tan x   csc x
2
cot x  1  csc x
2
2
csc x  csc x
2
2
cot x  csc x
sin x  tan x
co s x
sin x

sin x 
co s x

 cot x csc x
1
sin x  co t x csc x
sin x
co s x
1
sin x
sin x
 co t x csc x
sin x co s x
sin x

co s x
co s x
cos x  1
sin x

cos x
sin x  cos x  1
1
sin x

cos x
 cot x csc x
 cot x csc x
sin x
cot x csc x  cot x csc x