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(x, y)
(- x, y)




(- x, - y)
x  cos 
y  sin  
x  cos  
y  sin   
x  cos 
y   sin  
(x, - y)
sin  
tan   
cos 
sin     sin  
tan    

  tan  
cos  
cos 
Sect 5.1 Verifying Trig identities
Reciprocal
1
sin  
csc 
1
cos  
sec 
1
tan  
cot 
1
cot  
tan 
1
sec  
cos 
1
csc  
sin 
Co-function
Pythagorean
sin  90     cos 
sin 2   cos 2   1
cos  90     sin 
tan  90     cot 
1  tan 2   sec 2 
1  cot 2   csc 2 
cot  90     tan 
sec  90     csc 
csc  90     sec 
Quotient
sin 
tan  
cos 
cos 
cot  
sin 
Even/Odd
sin      sin 
cos     cos 
tan      tan 
cot      cot 
sec     sec 
csc      csc 
If tan    
5
and  is in quadrant II, find each function value.
3
(a) sec 
Negative
answer.
What Trig. Identity
has tan and sec?
sec 2    1  tan 2  
sec2    1   53 
2
sec    1 
2
sec   
2
sec   
25
9
34
9
(b) sin   Positive
answer.
What Trig. Identity
has tan and sin?
sin   y  5
tan   
 
cos  x
3
r 2  x2  y2
r  x2  y2
r  32   5  34
2
34
3
y
5
5 34
sin    

r
34
34
S A
T C
(c) cot    Positive
answer.
What Trig. Identity
has tan and cot?
1
cot   
tan  
1
cot    
tan   
cot    
1
 tan  
1
cot    
  53 
1
3
3
 5  1 5 
5
3
Write cos(x) in terms of tan(x).
sec 2    1  tan 2  
1
1  tan 2  

2
cos  
1
cos 2  
1

1  tan 2  
Secant has a relationship with both
tangent and cosine.
cos   
cos 
1
1  tan 2  
1  tan 2  
 
1  tan 2  
Rationalize the denominator.
1  cot 2 x 
Write
in terms of sin(x) and cos(x), and simplify the expression so
2
1  csc x  that no quotients appear.
csc 2 x   1  cot 2 x 
 cot 2 x   csc 2 x    cot 2 x  csc 2 x 
1  cot 2 x 
1  csc 2 x 
 cot 2 x   1  csc 2 x 
1
2
2
csc 2 x 


sin
x

1
sin
x   1


 2 
2
2
2
 cot x 
x 
cos
cos 2  x 
sin x  cos x 
sin 2  x 
  sec 2 x 
Sect 5.2 Verifying Trig identities
Guidelines to follow.
1. Work with one side of the equation at a time. It is often
better to work on the most complicated.
2.
Look for opportunities to factor, add fractions, square
binomials or multiply a binomial by it’s conjugate to create
a monomial.
3.
Look to use fundamental identities. Look to see what trig
functions are in the answer.
4.
Convert everything to sines and cosines
5. Always try something!
Sect 5.2 Verifying Trig identities
Verify.
cot   1  csc cos   sin  
Work on the right side first.
Distribute the cosecant.
csc cos   csc sin  
1
1
cos  
sin  
sin  
sin  
cos 
1
sin  
cot   1  cot   1
Rewrite to sine and cosine.
Simplify the fractions.
Quotient Identity for cotangent.
Sect 5.2 Verifying Trig identities
Verify. tan x 1  cot x   sec x 
2
2

2

tan 2 x  csc 2 x 
sin x 
1
 2
2
cos x sin x 
2
Work on the left side first.
Pythagorean Identity
1 + cot2x = csc2x
Rewrite to sine and cosine.
Simplify the fractions by canceling .
1
cos 2 x 
Reciprocal Identity for secant.
sec 2 x   sec 2 x 
Sect 5.2 Verifying Trig identities
tan    cot  
 sec 2    csc 2  
Verify.
sin  cos 
tan  
cot  

sin   cos  sin   cos 
sin  
cos 
cos 
sin  

sin   cos  sin   cos 
Work on the left side first.
Rewrite the fraction as
subtraction of two fractions with
the same denominators.
Rewrite to sine and cosine.
Simplify the fractions by multiplying
by the reciprocals and cancel.
sin  
1
cos 
1



cos  sin   cos  sin   sin   cos 
1
1

cos 2   sin 2  
sec 2    csc 2    sec 2    csc 2  
Reciprocal Identity for
secant and cosecant.
Sect 5.2 Verifying Trig identities
Verify.
sec2   1
2

sin

2
sec 
tan 2 
sec 2 
sin 2 
cos 2 
1
cos 2 
sin 2  cos 2 

2
cos 
1
sin 2   sin 2 
Work on the left side first.
Pythagorean Identity
1 + tan2x = sec2x
tan2x = sec2x – 1
Rewrite to sine and cosine.
Rewrite as multiplication.
Cancel and Simplify.
Sect 5.2 Verifying Trig identities
Verify.
2sec2  
1
1

1  sin  1  sin 
Work on the right side first. Two terms
need to be condensed to one term. Find
LCD and combine the fractions.
LCD  1  sin  1  sin  
 1 sin 2 


1
1  sin  
1
1  sin  



1  sin   1  sin   1  sin   1  sin  
1  sin 
1  sin 

2
1  sin  1  sin 2 
2
1  sin 2 
2
1
 2
2
cos 
cos 2 
2 sec 2   2 sec 2 
Pythagorean Identity
sin2x + cos2x = 1
cos2x = 1 – sin2x
Reciprocal of cosine.
Sect 5.2 Verifying Trig identities



2
2
2
Verify.  tan   tan   1 cos   1
sec   sin  
2
2

 1 
2

sin



2
 cos  
sin 2 

cos 2 
 tan 2    tan 2 
Work on the right side first.
Pythagorean Identities.
sin2x + cos2x = 1
cos2x – 1 = – sin2x
1 + tan2x = sec2x

Convert to cosine.
Multiply.
Sect 5.2 Verifying Trig identities
Verify.
tan   cot   sec csc
sin  cos 

cos  sin 
sin  sin  cos  cos 



cos  sin  sin  cos 
sin 2   cos 2 
cos  sin 
1
cos  sin 
1
1

cos  sin 
sec csc  sec csc
Work on the left side first. Try to
combine the two terms into one.
Convert to sine and cosine.
LCD  cos  sin 
Pythagorean Identity
sin2x + cos2x = 1
Rewrite as two fractions
multiplied together.
Reciprocals.
Sect 5.2 Verifying Trig identities
Verify.
cos 
 sec   tan 
1  sin 
1
sin 

cos  cos 
Work on the right side first. Two terms
need to be condensed to one term.
Convert to sine and cosine.
Combine.
1 sin 
cos 
1  sin    1  sin  
1  sin  
cos 
1  sin  
2
cos 1  sin  
cos 2 
cos 1  sin  
cos 
cos 

1  sin   1  sin  
When working with binomials, try
multiplying by the conjugate to create
differences of squares which will
incorporate the Pythagorean Identities.
Pythagorean Identity
sin2x + cos2x = 1
cos2x = 1 – sin2x
Cancel cosine.
Sect 5.2 Verifying Trig identities
Verify.
cot 2 
1  sin 

1  csc 
sin 
csc 2   1
1  csc 
csc   1csc   1
1  csc 
csc  1
1
1
sin 
1
sin 

sin  sin 
1  sin  1  sin 

sin 
sin 
Work on the left side first.
Pythagorean Identity and
convert to sine and cosine.
1 + cot2x = csc2x
cot2x = csc2x – 1
csc2x – 1 is Diff. of Squares.
Factor.
Cancel (csc x + 1)
Convert to sine.
Combine to one term.
cosA  B, sin A  B
Sect 5.3 Sum and Difference Formulas
cos B, sin B
A
cos A, sin A
A– B
Using Distance Formula
A– B
x2  x1 2   y2  y1 2
(1,0)
B
Dist. from (cos(A-B), sin(A-B)) to (1,0) = Dist. from (cosA, sinA) to (cosB,sinB)
cos A  B 12  sin  A  B  02

cos A  cos B2  sin A  sin B2
F.O.I.L.
F.O.I.L.
F.O.I.L.
cos  A  B   2 cos A  B   1  sin  A  B   cos A  2 cos A cos B  cos B  sin 2 A  2 sin A sin B  sin 2 B
2
Pythagorean Identity
Subtract by 2.
2
Pythagorean Identity
Pythagorean Identity
1  2 cos A  B  1 1 2 cos Acos B  1 2 sin Asin B
2  2 cos A B  2  2 cos A cos B  2 sin Asin B
– 2
Divide by –2.
2
2
– 2
 2 cos A  B  2 cos A cos B  2 sin Asin B
2
2
2
cos A  B  cos A cos B  sin A sin B
The Cosine of the Difference of Two Angles
The Cosine of the Difference of Two Angles
cos A  B  cos A cos B  sin A sin B
Substitute (-B) for B in the formula to make the
Cosine of the Sum of Two Angle.
cos A   B  cos A cos B  sin A sin  B
cos (– B) = cos (B)
The Cosine of the Sum of Two Angles
sin (– B) = – sin (B)
cos A  B  cos A cos B  sin A sin B
To make the Sine of the Sum & Difference of Two Angles
we will need the Cofunction Identities for Sine and Cosine.
sin    cos90   
cos   sin 90   
Start with   A B .
sin  A  B  cos90   A  B  cos90  A B  cos90  A  B
cos A  B  cos A cos B  sin A sin B
cos90  A  B  cos90  Acos B  sin 90  Asin B
sin  A  B  sin Acos B  cos Asin B
Substitute (-B) for B in the formula to make the Sine of the Sum of Two Angle.
sin  A   B  sin A cos B  cos Asin  B
cos (– B) = cos (B)
sin (– B) = – sin (B)
sin  A  B  sin A cos B  cos Asin B
To make the Tangent of the Sum & Difference of Two Angles
we will need the Quotient Identities for Tangent.
sin  A  B  sin A cos B  cos A sin B
 cos (A)
cos (B)
tan  A  B  
cos A  B  cos A cos B  sin A sin B
This is what
we need
divide by
cos (A) cos (B) all the
factors.
Tricky manipulation: We want this fraction to have tangents in the formula.
Need to divide by the same factor in both the top and bottom to make tangents.
Start with where we need to divide by cosine.
sin A cos B cos A sin B

tan A  tan B
cos
A
cos
B
cos
A
cos
B

tan  A  B  
cos A cos B sin A sin B
1  tan A tan B

cos A cos B cos A cos B
tan A  tan B
tan  A  B  
1  tan A tan B
sin  A  B  sin A cos B  cos A sin B
tan  A  B  

cos A  B  cos A cos B  sin A sin B
This is what
we need
divide by
cos (A) cos (B) all the
factors.
sin A cos B cos A sin B

 cos A cos B cos A cos B  tan A  tan B
cos A cos B sin A sin B
1  tan A tan B

cos A cos B cos A cos B
tan A  tan B
tan  A  B  
1  tan A tan B
 7
Find the exact value of cos
 12

.

cos105  cos60  45
cos A  B  cos A cos B  sin A sin B
7 7180

 715  105
12
12
Use the special right triangle
angles, 30o, 45o, and 60o. We
may need to use multiples of
these angles.
cos60  45  cos60cos45  sin 60sin 45
2
1
45
1
1 2
3 2
 


2 2
2 2
2
6
2 6



4
4
4
1
60
2
30
3
 5  
 .
Find the exact value of cos
 3 4

2
1
60
2

 3
45
5 5180

 560  300
3
3

4
1
1
30
cos A  B  cos A cos B  sin A sin B
 45
Use the special right triangle
angles, 30o, 45o, and 60o. We
may need to use multiples of
these angles.
cos300  45  cos300cos45  sin 300sin 45
3  2 
 1  2  


  


 2  2   2  2 
2
6
2 6



4
4
4
Suppose that sin    for a Q2 angle  and sin    for a
13
5
Q1 angle  .
Find the exact value of each of the following.
A. cos 
B. cos 
C. cos   
D. cos   
12
3
cos     cos  cos   sin  sin 
12
13

5


4
5
cos  
3
5
13
cos  
4
5
  5  4   12  3 

     
 13  5   13  5 
 20 36  56



65 65 65
cos     cos  cos   sin  sin 
  5  4   12  3 

     
 13  5   13  5 

 20 36 16


65 65 65
Find the exact value of sin 75 .
sin 75  sin 30  45
sin  A  B  sin A cos B  cos Asin B
Use the special right triangle
angles, 30o, 45o, and 60o. We
may need to use multiples of
these angles.
sin 30  45  sin 30cos45  cos30sin 45
2
1
45
1
1 2
3 2
 


2 2
2 2
2
6
2 6



4
4
4
1
60
2
30
3
Find the exact value of
60

2
1
30
 3
2

1
 7 
tan 
 .
 12 
7 7180

 715  105
12
12
150  45  105
45
1
tan A  tan B
tan  A  B  
1  tan A tan B
Use the special right triangle
angles, 30o, 45o, and 60o. We
may need to use multiples of
these angles.
1
1

1

1
tan 150  tan 45
3
3
tan 150  45 


1  tan 150 tan 45
1
 1 
1

1  
1
3
3

1
3

1

3


3
3
1  3 1 1  3
3
1 3





3 1
3 1 1 1  3
3 1
1 3

3
3
3




Find the exact value of
2 30

60
1
3
2

 7 
tan 
 .
 12 
45
7 7180

 715  105
12
12
60  45  105
Use the special right triangle
angles, 30o, 45o, and 60o. We
may need to use multiples of
these angles.
1
1
tan A  tan B
tan  A  B  
1  tan A tan B
tan 60  tan 45
tan  60  45  
1  tan 60 tan 45
Another approach.

3 1
1
 3  1
1 3

1 3
Find the exact value of sin 40cos160  cos40sin 160.
sin  A  B  sin A cos B  cos Asin B
sin 40 160  sin  120
1
60
120
 3
30
2
3

2
Sect 5.5 Dble Angle, Power Reducing, and Half Angle Formulas
Double Angle Formulas: Revise the Sum of Sin, Cos, & Tan Formulas
Substitute A in for B.
sin  A  B  sin A cos B  cos Asin B
sin  A  A  sin A cos A  cos Asin A => sin
cos A  B  cos A cos B  sin A sin B
cos A  A  cos A cos A  sin A sin A


cos2 A  1  sin 2 A  sin 2 A

2 A  2 sin A cos A
 cos2 A  cos 2 A  sin 2 A
 cos2 A  1  2 sin 2 A
  cos2 A  2 cos 2 A  1
cos2 A  cos 2 A  1  cos 2 A
tan A  tan B
tan  A  B  
1  tan A tan B
tan A  tan A
tan  A  A 
1  tan A tan A
2 tan A
 tan 2 A 
1  tan 2 A
Find sin  2  ,cos  2  , tan  2 
5
given cos  
and
13
120
  12  5 

2



 
sin 2   2 sin  cos
13
169

 13 
119
 5    12 
 
cos2   cos 2   sin 2      
169
 13   13 
  12 
2

2 tan 
5

  120
tan 2  

2
119
1  tan 2 
  12 
1 

 5 
2
2
Quadrant 4.
3
   2 .
2

y  132  52  12
 12
sin   
13
tan   
 12
5
5
13
 12
Quadrant 2.
Find the values of the six trigonometric functions of  if cos2   and 90    180.
cos2   cos 2    sin 2  
Choose one of the
double angle identities
to find a value for sine
or cosine.
cos2   1  2 sin 2  
10
1
cos2   2 cos 2    1
cos2   1  2 sin  
Substitute in 4/5.
4
2
 1  2 sin  
5
Subtract by 1.
2
1
  2 sin 2  
5
1
 sin 2  
10
1

 sin  
10
sin   
4
5
1
10

10 10
2
10  12  3
SOH-CAH-TOA
cos  
 3  3 10

10
10
tan   
1
3
Divide by -2.
cot    3
Square root both sides,
but the answer will be
positive, since we are Q2.
sec  
10
3
csc   10

Verify. cot   sin 2   1  cos2 
cos 
 sin 2 
sin 
cos 
 2 sin   cos 
sin  
2 cos 2  
1  cos2   1  cos2 
Work on the left side first.
Convert to sine and cosine
with Quotient Identity.
Double angle identity.
2sin(x) cos(x) = sin(2x)
Cancel
Rewrite the double angle formula.
2cos2x – 1 = cos(2x)
2cos2x = 1 + cos(2x)
sin 2 A  2 sin A cos A
cos    sin    cos2 
2
2
cos 2 7 x   sin 2 7 x   cos2  7 x 
 cos14 x 
1
 2  sin 15 cos15
2
1
 sin 2 15
2
1
1 1 1
 sin 30   
2
2 2 4
Find an identity for cos3  cos  2 
cos A  B  cos A cos B  sin A sin B
Substitute Dble angle Identity.
cos  2   cos ( cos2)  sin  ( sin 2 )


cos  2   cos  (2 cos 2   1)  sin (2 sin  cos  )
cos   2   2cos3   cos  2sin 2  cos
Pythagorean Identity,
rewrite with all cosines.
cos   2   2 cos3   cos   2 1  cos 2   cos 
cos   2   2 cos3   cos   2 cos  1  cos 2  
Distribute
cos   2   2cos3   cos  2cos  2cos3 
 4 cos3   3cos 
cos 3  4 cos3   3cos 
Product to Sum & Sum to Product Formulas
How to create the Product to Sum Formulas. Add and
subtract Sum and Difference formulas for Sine and Cosine.
cos A cos B  sin A sin B  cos A  B


cos A cos B  sin A sin B  cos A  B
2 cos A cos B  cos A  B  cos A  B
cos A cos B 
1
cos A  B   cos A  B 
2
sin A cos B  cos Asin B  sin  A  B


sin A cos B  cos Asin B  sin  A  B
2 sin A cos B  sin  A  B  sin  A  B
1
sin A cos B  sin  A  B   sin  A  B 
2
cos A cos B  sin A sin B  cos A  B


cos A cos B  sin A sin B  cos A  B
2 sin A sin B  cos A  B  cos A  B
sin A sin B 
1
cos A  B   cos A  B 
2
sin A cos B  cos Asin B  sin  A  B


sin A cos B  cos Asin B  sin  A  B
2 cos A sin B  sin  A  B  sin  A  B
1
cos A sin B  sin  A  B   sin  A  B 
2
Product to Sum Formulas
cos A cos B  12 cos A  B   cos A  B  sin A cos B  12 sin  A  B   sin  A  B 
sin A sin B  12 cos A  B   cos A  B  cos A sin B  12 sin  A  B   sin  A  B 
Sum to Product Formulas
cos A cos B  12 cos A  B   cos A  B 
2 cos A cos B  cos A  B  cos A  B
 x y
Let A  

 2 
The reason we choose these two fractions
for A and B is because we need two values
that add up to x and two values that
subtract to be y.
 x y
and B  

 2 
 x  y   x  y 
 x  y   x  y 
 x y  x y
2 cos
cos

cos




cos
 
 


 


 

 2   2 
 2   2 
 2   2 
 x y  x y
2 cos
 cos
  cos y   cos x 
 2   2 
 A B   A B 
cos A  cos B  2 cos
 cos

 2   2 
Product to Sum Formulas
cos A cos B  12 cos A  B   cos A  B  sin A cos B  12 sin  A  B   sin  A  B 
sin A sin B  12 cos A  B   cos A  B  cos A sin B  12 sin  A  B   sin  A  B 
Sum to Product Formulas
 A B   A B 
cos A  cos B  2 cos
 cos

2
2

 

 A B   A B 
sin A  sin B  2 sin 
 cos

 2   2 
 A B   A B 
sin A  sin B  2 sin 
 cos

2
2

 

 A B   A B 
cos A  cos B  2 sin 
 sin 

2
2

 

Rewrite sin  6 x  cos  2 x  as a sum or difference of two functions
sin A cos B  12 sin  A  B   sin  A  B 
sin 6 x  2 x  sin 6 x  2 x
 12 sin 8x   sin 4 x 
sin 6 x  cos2 x  
1
2
Rewrite cos 4x  cos 3x using sums to product identity.
 A B   A B 
cos A  cos B  2 sin 
 sin 

 2   2 
 4 x  3x   4 x  3x 
cos4 x   cos3x   2 sin 
 sin 

2
2

 

 7x   x 
 2 sin   sin  
 2  2
Half Angle Formulas
cos  2 A  1  2sin 2  A
cos  2 A  2cos2  A 1
2sin 2  A  1  cos  2 A
1  cos  2 A
cos  A 
2
1  cos  2 A
sin  A 
2
2
Let A 

2
1  cos  
sin  
2
2

2

   sin  2 
tan   
 2  cos  2 
 
tan    
2
1  cos  2 A
cos  A 
 
2
tan




Let A 
2
2
2
cos

2

1  cos 
2
1 cos  
2
1 cos  
2
1 cos 
2
1 cos 
2
1  cos 
tan  
2
1  cos 

The + symbol in each formula DOES NOT mean there are 2 answers,
instead it indicates that you must determine the sign of the trigonometric
functions based on which quadrant the half angle falls in.
1  cos 
tan  
2
1  cos 



1  cos 1  cos 
1  cos 1  cos 
1  cos 2   1  cos 2
1  cos 2  
sin 2  
1  cos 
tan 
2
sin  

1  cos 
tan  
2
1  cos 



1  cos 1  cos 
1  cos 1  cos 
1  cos    
2
1  cos 2
sin 2  
1  cos 2
sin  
tan 
2 1  cos 

Find the exact value for cos112.5.

1  cos 
 225 
cos


cos112.5  cos

2
2
2


S
A
T
C

2
2
2

1   


2 
225
1  cos225
2

cos


 2 2 
2
2
2
2
2
2 2
2 2


4
2
1  cos 2
sin 2
1  1  2 sin 2 

2 sin  cos 
Verify the identity. tan  

2 sin 2 

2 sin  cos 
tan  
sin 
cos 
