Identities
• The set of real numbers for which an equation is defined is called the domain of the equation. If an equation is true for all values in its domain it is called an identity .
• Example 1. The equation x x
2


1
 x

1
1 has domain equal to all real numbers except

1.
all real numbers except those of the form Since it is
2
 n
 true for all numbers in its domain, it is an identity.
• Is the equation in Example 1 an identity?

Conditional equations
• If an equation is only true for some values in its domain, it is called a conditional equation .
• Example. The equation 2x +1 = 3 has domain equal to all real numbers, but it is only true for x = 1. Therefore, it is a conditional equation.
• Problem. Which equation is conditional?
( a) cos
2 x
 sin
2 x

1
(b) cos
2 x
 sin
2 x

1

Fundamental identities
• Reciprocal identities sin u

1 csc u cos u

1 sec u t an u

1 cot u
1 csc u
 sin u sec u

1 cos u cot u

1 t an u
• Quotient identities t an u
 sin u cos u cot u
 cos u sin u
• Pythagorean identities sin
2 u
 cos
2 u

1 1

tan
2 u
 sec
2 u 1
 cot
2 u
 csc
2 u

Fundamental identities, continued
• Cofunction identities sin(

2
 u )  cos u cos (

2 tan (

2 sec(

2
 u)
 cot u cot (

2
 u)
 csc u csc(

2
 u)
 u )
 u)
 sin u

tan u
 sec u
• Even/Odd identities sin (
 u)
  sin u cos (
 u)
 cos u tan(
 u)
  tan u csc (
 u)
  csc u sec (
 u)
 sec u cot (
 u)
  cot u

Two possible ways to describe a right triangle
4
 x
2 arctan(x/2 )
2 x
2
 sec


2
2
 tan
 x is independen t variable ,
0
 x
 
 is independen
0
   
2 t variable ,
x

2
 tan

,
4
 x
2 
4 ( 1
 tan
2

)

4
 sec
2
 
2
 sec


Using fundamental identities--an example
• Use fundamental identities to verify the trig identity
2
 sin t

3
 sin(
 t)

4
 cos(
π
2
 t)

9
 sin t.
• We simplify the left-side of the identity.
2
 sin t


2
2


 sin t sin t
3


 sin(
3

3

 t) sin t sin t



4
 cos(
π
2
 t)
4
 cos(
π
2
4
 sin t
 t) using since a sine is odd cofunction identity

9
 sin t using algebra.

Guidelines for verifying trigonometric identities
1.
Work with one side of the equation at a time.
2.
Look to factor an expression, add fractions, square a binomial, or create a monomial denominator.
3.
Look to use the fundamental identities. Note which functions are in the final expression you want. Sines and cosines pair up well, as do secants and tangents, and cosecants and cotangents.
4.
If the preceding guidelines do not help, then try converting all terms to sines and cosines.
5.
Always try something.

Practice problems--verify the identity
1.
tan x
 cot x

1
2.
sec x
 cos x

1
3.
(cot 2 x)(sec 2 x

1 )

1
4 .
( 1
 sin x)(1
 sin x)
 cos
2 x
5 .
cot x sec x
 csc x
 sin x
6.
tan(sin

1 x)
 x
1
 x
2

Solving trigonometric equations
• If you are given a trigonometric equation, your task is to manipulate it using algebra and the fundamental trigonometric identities until you can write an equation of the form trig function
 number.
• Example. Solve 2∙cos x = 1. Clearly, this is equivalent to cos x = 1/2. To solve for x, note that there are two solutions in [0, 2

), namely

/3 and 5

/3. Also, because cos x has a period of 2

, there are infinitely many other solutions, which can be written as x

π
3

2 nπ and x

5π
3

2 nπ where n is an integer.
• Since the latter solution gives all possible solutions of the equation, it is called the general solution .

Graphical approaches to solving cos x = 1/2
• The graph indicates how an infinite number of solutions can occur.
x
 π
3 x

5π
3 y

1
2 x
 π
3

2π
      y
 cos x x

5π
3

4π x
 π
3

2π x

5π
3

2π
• Also, the unit circle shows infinitely many solutions occur.
cos
π
3
 
 1
2 cos

5π
3

2 nπ

 1
2

Solving trigonometric equations--an example
• Solve (tan 3x)(tan x) = tan 3x. This equation can be rewritten as (tan 3x)(tan x – 1) = 0. Next, set the factors to zero. We have
(i) tan 3x

0,
(ii) tan x

1.
• To solve (i) for 3x in the interval [0, 
), we have 3x = 0. In general, we have 3x = n
 so that x = n

/3, n an integer.
• To solve (ii) for x in the interval [0, 
), we have x =

/4.
In general, we have x =

/4 + n

, n an integer.

Solving trigonometric equations--another example
• Solve csc x + cot x = 1. First, let's convert to sines and cosines. We obtain
1 sin x
 cos x sin x

1

1
 cos x
 sin x.
• Since there is no obvious way to solve 1 + cos x = sin x directly, we will try squaring both sides in the hope that the
Pythagorean identity will result in a simplification. Of course, we know that squaring may introduce extraneous solutions.
We have
( 1
 cos x)
2  sin
2
x

1

2cos x
 cos
2
x

1
 cos
2
x.
• After canceling 1, the latter equation becomes
2cos x

2cos
2
x

0
 cos x

(1
 cos x)

0.
• cos x = 0 yields 
/2 + 2n

, n an integer, cos x = –1 yields

+ 2n

, n an integer, which is extraneous since csc x and cot x are undefined at x =

.

Solving trigonometric equations--using inverse functions
• Solve sec
2 x

2 tan x

4
1

tan 2 x

2 tan x

4

0 tan
2
x

2 tan x

3

0
(tan x

3 )(tan x

1)

0 x
 arctan 3 and x
 arctan(

1 ) are two solutions in (
 
2
,

2
).
• The general solution is: x
 arctan( 3 )
 nπ and x

(
 
4
)
 nπ

Guidelines for solving trigonometric equations
1.
Try to isolate the trigonometric function on one side of the equation.
2.
Look to use standard techniques such as collecting like terms and factoring (or use the quadratic formula).
3.
Look to use the fundamental identities.
4.
To solve equations that contain forms such as sin kx or cos kx, first solve for kx and then divide by k.
5.
If you can't get a solution using exact values, use inverse trigonometric functions to solve.

Practice problems--solving trigonometric equations
1 .
3
 csc x

2

0 8 .
cos 2 x
 sin 2 x

1
9.
tan
2 x

2 tan x

0 2 .
3
 sec
2 x

4

0
3 .
3
 sin x

1
 sin x
4.
sin
2 x

3
 cos
2 x
5.
(sin 3x)(sin x

1 )

0
6 .
2
 cos 2 x
 cos x

1

0
7 .
tan 3x

1

0

Application of sum and difference formulas
• Sum and Difference Formulas sin(u
 v )  sin u
 cos v
 cos u
 sin v sin(u
 v )  sin u
 cos v
 cos u
 sin v cos(u
 v )  cos u
 cos v
 sin u
 sin v cos(u
 v )  cos u
 cos v
 sin u
 sin v t an(u
 v)
 t an u
 t an v
1
 t an u
 t an v t an u
 t an v t an(u
 v)

1
 t an u
 t an v

Application of sum and difference formulas--two examples
• Find the exact value of cos
π
12
.
cos
π
12
 cos(
π
3
 π
4
)
 cos
π
3
 cos
π
4
 sin
π
3
 sin
π
4

1
2


2

4
2
2

6
2
3

2
2
• Verify that
2 sin( x
 π
4
)
 sin x
 cos x.
2 sin( x
 π
4
)

2 (sin x
 cos
π
4
 cos x
 sin
π
4
)


2 ( sin x sin x
 1
 cos x
2
 cos x
 1
2
)

Double-angle and power-reducing formulas
• Double-Angle Formulas sin(2u ) 
2sin u
 cos u cos(2u )  cos
2 u
 sin
2 u

2
 cos
2 u

1

1

2
 sin
2 u
2
 t an u t an(2u)

1
 t an
2 u
• Power-Reducing Formulas (double-angle formulas restated) sin
2 u

1
 cos 2u cos
2
2 u

1
 cos 2u
2
t an
2 u

1
 cos 2u
1
 cos 2u

Using double-angle formulas to solve an equation
• Solve cos 2x + cos x = 0. First use the double-angle formula for cos 2x.
( 2
 cos
2
x

1)
 cos(x)

0
2
 cos
2
x
 cos(x)

1

0 quadratic with unknown cos x
(cos x
 cos x
 
1)(cos x
1
 x


π
1
2
)


0
2nπ factor the quadratic, divide by 2 cos x

1
2
 x

π
3

2 nπ , x

5π
3

2 nπ n an integer

Using the power-reducing formulas
• Rewrite sin 4 x in terms of first powers of the cosines of multiple angles.
sin
4 x

(sin
2 x)
2 algebra
2

 cos 2x
2 power reduction

1

1

4

1
4


1

2 cos 2x

2 cos 2x
 cos
2
2 x
 expand cos 4x
2

 power reduction

1
8

3

4 cos 2x
 cos 4x
 algebra

Other formulas not covered
• Half-Angle formulas, Product-to-Sum Formulas, and Sumto-Product Formulas are interesting, but will not be covered in this course.

More practice problems--verify using all available identities
1.
csc 2x

2
 csc x cos x
2 .
1
 cos 10x

2
 cos
2
5 x
3.
cos
4 x
 sin
4 x
 cos 2x
4.
(sin x
 cos x)
2 
1
 sin 2x
5.
cos 3x

4
 cos
3 x

3
 cos x
6.
cos

π
3
 x

 cos

π
3
 x

 cos x

More practice problems--solve for x using all available identities
1 .
sin 2x
 cos x

0
2.
cos x

(sin x)(ta n x)

2
3.
sin(x
 
2
)
 sin( x

3π )
2

1
4 .
tan( x
 π) 
2
 sin(x
 π) 
0
5.
sin(x
 π
2
)
 cos
2 x

0
6.
cos(x
 π)  cos x

1

0
7.
sin( x 2)
 cos x

0

Solution of number 7 from previous slide sin(u)
 cos 2u

0.
• Using a double angle formula, we have sin(u)

(1

2 sin
2 u)

0.
0,
( 2 sin u

1)( sin u

1 )

0.
• We have two cases (i) sin u = −1/2, (ii) sin u = 1.
• (i) yields u = 7π/6, u = 11π/6, (ii) yields u = π/2.
• Thus x = 7π/3, x = 11π/3, x = π , which are all the solutions in
[0, 4π).

S u m an d form u l as sin(u
 v )  sin u
 cos v
 cos u
 sin v sin(u
 v )  sin u
 cos v
 cos u
 sin v cos(u
 v )  cos u
 cos v
 sin u
 sin v cos(u
 v )  cos u
 cos v
 sin u
 sin v t an(u
 v)
 t an u
 t an v
1
 t an u
 t an v t an u
 t an v t an(u
 v)

1
 t an u
 t an v
Dou bl e an gl e form u l as sin(2u ) 
2sin u
 cos u cos(2u )  cos
2 u
 sin
2 u

2
 cos
2 u

1

1

2
 sin
2 u
2
 t an u t an(2u)

1
 t an
2 u
Powe r re du ci n g form u l as sin
2 u

1
 cos 2u cos
2
2 u

1
 cos 2u
2
t an
2 u

1
 cos 2u
1
 cos 2u