Transcript Thinking Mathematically by Robert Blitzer

Chapter 5
Verifying
Trigonometric
Identities
Basic Trigonometric Identities
Reciprocal Identities
csc x =
1
sinŹx
sec x =
1
cosŹx
cot x =
1
tanŹx
Quotient Identities
sinŹx
tan x = cosŹx
cosŹx
cot x = sinŹx
Pythagorean Identities
sin2x + cos2x = 1
tan 2x + 1 = sec 2x
1 + cot 2x = csc2x
Even-Odd Identities
sin(-x) = -sin x
csc(-x) = -csc x
cos(-x) = cos x
sec(-x) = sec x
tan(-x) = -tan x
cot(-x) = -cot x
Text Example
Verify the identity: sec x cot x = csc x.
Solution
The left side of the equation contains the more complicated
expression. Thus, we work with the left side. Let us express this side of the
identity in terms of sines and cosines. Perhaps this strategy will enable us to
transform the left side into csc x, the expression on the right.
1
cosx
sec x cotx =

cosx sin x
Apply a reciprocal identity: sec x = 1/cos x
and a quotient identity: cot x = cos x/sin x.
1
=
= csc x
sin x
Divide both the numerator and the
denominator by cos x, the common factor.
Text Example
Verify the identity: cosx - cosxsin2x = cos3x..
Solution
We start with the more complicated side, the left side. Factor out
the greatest common factor, cos x, from each of the two terms.
cos x - cos x sin2 x = cos x(1 - sin2 x)
= cos x ·
= cos3 x
cos2
x
Factor cos x from the two terms.
Use a variation of sin2 x + cos2 x = 1.
Solving for cos2 x, we obtain cos2 x =
1 – sin2 x.
Multiply.
We worked with the left and arrived at the right side. Thus, the identity is
verified.
Guidelines for Verifying
Trigonometric Identities
1.
Work with each side of the equation independently of the other
side. Start with the more complicated side and transform it in a
step-by-step fashion until it looks exactly like the other side.
2.
Analyze the identity and look for opportunities to apply the
fundamental identities. Rewriting the more complicated side
of the equation in terms of sines and cosines is often helpful.
3.
If sums or differences of fractions appear on one side, use the
least common denominator and combine the fractions.
4.
Don't be afraid to stop and start over again if you are not
getting anywhere. Creative puzzle solvers know that
strategies leading to dead ends often provide good problemsolving ideas.
Example
• Verify the identity:
csc(x) / cot (x) = sec (x)
Solution:
csc x
= sec x
cot x
1
sin x = 1
cos x cos x
sin x
1
sin x
1

=
sin x cos x cos x
Example
• Verify the identity:
cos x = cos x  cos x sin x
3
2
Solution:
cos x = cos3 x  cos x sin 2 x
cos x = cos x(cos x  sin x)
2
cos x = cos x(1)
2
Example
• Verify the following identity:
Solution:
tan2 x - cot2 x
= tan x - cot x
tan x  cot x
sin 2 x cos2 x
tan2 x - cot2 x cos2 x sin 2 x
=
sin x cos x
tan x  cot x

cos x sin x
sin 4 x - cos4 x
4
4
2
2
sin
x
cos
x cos x sin x
cos
x
sin
x
=
=

2
2
2
2
sin x  cos x
cos x sin x
1
cos x sin x
Example cont.
Solution:
sin 4 x - cos4 x
cos x sin x
(sin 2 x  cos2 x)(sin2 x - cos2 x)
=
sin x cos x
sin 2 x - cos2 x
sin 2 x
cos2 x
=
=
sin x cos x
sin x cos x sin x cos x
sin x cos x
=
= t an x - cot x
cos x sin x
Sum and Difference
Formulas
The Cosine of the Difference of
Two Angles
cos( -  ) = cos cos   sin  sin 
The cosine of the difference of two angles
equals the cosine of the first angle times the
cosine of the second angle plus the sine of
the first angle times the sine of the second
angle.
Text Example
• Find the exact value of cos 15°
Solution
We know exact values for trigonometric functions of 60° and 45°.
Thus, we write 15° as 60° - 45° and use the difference formula for cosines.
cos l5° = cos(60° - 45°)
= cos 60° cos 45°  sin 60° sin 45°
1
2
3
2



2 2
2
2
=
2
6

4
4
=
2 6
4
cos( -) = cos  cos   sin  sin 
Substitute exact values from
memory or use special triangles.
Multiply.
Add.
Text Example
Find the exact value of cos 80° cos 20°  sin 80° sin 20°.
Solution
The given expression is the right side of the formula for cos( - )
with  = 80° and  = 20°.
cos( -) = cos  cos   sin  sin 
cos 80° cos 20°  sin 80° sin 20° = cos (80° - 20°) = cos 60° = 1/2
Example
• Find the exact value of cos(180º-30º)
Solution
cos(180- 30)
= cos180cos30  sin 180sin 30
3
1
= -1*
 0*
2
2
3
=2
Example
• Verify the following identity:
5 
2

cos x -  = (cosx  sin x)
4 
2

Solution
5 

cos x 
4 

 5
= cos x cos
 4

 5 
  sin x sin  

 4 
2
2
cos x  sin x
2
2
2
=(cos x  sin x )
2
=-
Sum and Difference Formulas for
Cosines and Sines
cos(   ) = cos cos  - sin  sin 
cos( -  ) = cos cos   sin  sin 
sin(   ) = sin  cos   cos sin 
sin( -  ) = sin  cos  - cos sin 
Example
• Find the exact value of sin(30º+45º)
Solution
sin(   ) = sin  cos   cos sin 
sin(30  45) = sin 30 cos 45  cos30sin 45
1
2
3
2
= 


2 2
2
2
2 6
=
4
Sum and Difference Formulas for
Tangents
The tangent of the sum of two angles equals the tangent of the first angle
plus the tangent of the second angle divided by 1 minus their product.
tan  tan 
tan(   ) =
1 - tan tan 
tan - tan 
tan( -  ) =
1  tan tan 
The tangent of the difference of two angles equals the tangent of the first
angle minus the tangent of the second angle divided by 1 plus their
product.
Example
• Find the exact value of tan(105º)
Solution
•tan(105º)=tan(60º+45º)
tan  tan 
tan(   ) =
1 - tan tan 
tan60  tan 45
=
1 - tan60 tan 45
3 1 1 3
=
=
1- 3 1- 3
Example
• Write the following expression as the sine, cosine, or
tangent of an angle. Then find the exact value of the
expression.
7

7

sin
cos - cos
sin
12
12
12
12
Solution
7

7

sin
cos - cos sin
12
12
12
12
6
 7  
= sin 
-  = sin
12
 12 12 
= sin

2
=1
Double-Angle and
Half-Angle Formulas
Double-Angle
Identities
sin 2 = 2 sin cos
cos 2 = cos2 Š sin2 = 1 Š 2 sin2 = 2 cos2 Š 1
2tan
tan 2 =
1 - tan2
Three Forms of the DoubleAngle Formula for cos2
cos 2 = cos  - sin 
2
2
cos 2 = 2 cos  - 1
2
cos 2 = 1 - 2 sin 
2
Power-Reducing Formulas
1 - cos 2
sin  =
2
1  cos 2
2
cos  =
2
1 - cos 2
2
tan  =
1  cos 2
2
Example
• Write an equivalent expression for sin4x that does
not contain powers of trigonometric functions
greater than 1.
Solution
 1 - cos 2 x  1 - cos 2 x 
sin 4 x = sin 2 x sin 2 x = 


2
2



1  cos 4 x 

1
2
cos
2
x

2


 1 - 2 cos 2 x  cos 2 x 
2


 = 
4
4


 



 2 - 4 cos 2 x  1  cos 4 x  3 - 4 cos 2 x  cos 4 x 
=
=
8
8


Half-Angle Identities
x
sin 2 = ±
1 – cos x
2
x
cos 2 = ±
1 + cos x
2
x
tan 2 = ±
1 – cos x
sin x
1 – cos x
=
=
1 + cos x 1 + cos x
sin x
x
where the sign is determined by the quadrant in which 2 lies.
Text Example
Find the exact value of cos 112.5°.
Solution
Because 112.5° = 225°/2, we use the half-angle formula for cos /2
with  = 225°. What sign should we use when we apply the formula? Because
112.5° lies in quadrant II, where only the sine and cosecant are positive, cos
112.5° < 0. Thus, we use the - sign in the half-angle formula.
225
cos112.5 = cos
2
- 2

1 
1  cos225
 2 
==
2
2
2- 2
2- 2
==4
2
Half-Angle Formulas for:

1 - cos
t an =
2
sin 

sin 
t an =
2 1  cos
Example
• Verify the following identity:
(sin - cos )2 = 1 - sin 2
Solution
(sin - cos ) 2
= sin 2  - 2 sin  cos  cos2 
1 - cos 2 1  cos 2
=

- 2 sin  cos
2
2
2
= - 2 sin  cos = 1 - sin 2
2
Product-to-Sum and
Sum-to-Product
Formulas
Product-to-Sum Formulas
1
sin  sin  = cos( -  ) - cos(   )
2
1
cos cos  = cos( -  )  cos(   )
2
1
sin  cos  = sin(   )  sin( -  )
2
1
cos sin  = sin(   ) - sin( -  )
2
Example
• Express the following product as a sum or difference:
cos 3x cos 2 x
Solution
1
cos cos  = cos( -  )  cos(   )
2
cos3x cos 2 x
1
= cos(3x - 2 x)  cos(3x  2 x)
2
1
= cos(x)  cos(5 x)
2
Text Example
Express each of the following products as a sum or difference.
a. sin 8x sin 3x
b. sin 4x cos x
Solution
The product-to-sum formula that we are using is shown in each
of the voice balloons.
a.
sin  sin  = 1/2 [cos( - ) - cos( + )]
sin 8x sin 3x = 1/2[cos (8x - 3x) - cos(8x  3x)] = 1/2(cos 5x - cos 11x)
b.
sin  cos  = 1/2[sin( + ) + sin( - )]
sin 4x cos x = 1/2[sin (4x  x)  sin(4x - x)] = 1/2(sin 5x  sin 3x)
Sum-to-Product Formulas
sin   sin  = 2 sin

cos
 -
2
2
-

sin  - sin  = 2 sin
cos
2
2

-
cos  cos  = 2 cos
cos
2
2

 -
cos - cos  = -2 sin
sin
2
2
Example
• Express the difference as a product:
sin 4 x - sin 2 x
Solution
sin  - sin  = 2 sin
-
cos

2
2
4x - 2x
4x  2x
sin 4 x - sin 2 x = 2 sin
cos
2
2
2x
6x
= 2 sin cos = 2 sin x cos3x
2
2
Example
• Express the sum as a product:
sin x  sin 4 x
Solution
sin   sin  = 2 sin

cos
 -
2
2
x  4x
x - 4x
sin x  sin 4 x = 2 sin
cos
2
2
5x
- 3x
= 2 sin cos
2
2
Example
• Verify the following identity:
sin x  sin y
x y
x- y
= tan
cot
sin x - sin y
2
2
Solution
x y
x- y
2 sin
cos
sin x  sin y
2
2
=
sin x - sin y 2 sin x - y cos x  y
2
2
x y
x- y
sin
cos
x y
x- y
2
2
=
= tan
cot
x y
x- y
2
2
cos
sin
2
2
Trigonometric
Equations
Equations Involving a Single
Trigonometric Function
To solve an equation containing a single
trigonometric function:
• Isolate the function on one side of the
equation.
• Solve for the variable.
Trigonometric Equations
y
y = cos x
1
y = 0.5
–4 
–2 
2
4
x
–1
cos x = 0.5 has infinitely many solutions for –  < x < 
y
y = cos x
1
0.5
2
–1
x
cos x = 0.5 has two solutions for 0 < x < 2
Text Example
Solve the equation: 3 sin x - 2 = 5 sin x - 1.
Solution
The equation contains a single trigonometric function, sin x.
Step 1 Isolate the function on one side of the equation. We can solve for
sin x by collecting all terms with sin x on the left side, and all the constant
terms on the right side.
3 sin x - 2 = 5 sin x - 1
3 sin x - 5 sin x - 2 = 5 sin x - 5 sin x – 1
-2 sin x - 2 = -1
-2 sin x = 1
sin x = -1/2
This is the given equation.
Subtract 5 sin x from both sides.
Simplify.
Add 2 to both sides.
Divide both sides by -2 and solve for sin x.
Text Example
Solve the equation:
2 cos2 x  cos x - 1 = 0,
0  x < 2.
Solution
The given equation is in quadratic form 2t2  t - 1 = 0 with t =
cos x. Let us attempt to solve the equation using factoring.
2 cos2 x  cos x - 1 = 0
This is the given equation.
(2 cos x - 1)(cos x  1) = 0
2 cos x - 1= 0
or
Factor. Notice that 2t2 + t – 1 factors as (2t – 1)(2t + 1).
cos x  1 = 0
2 cos x = 1 cos x = -1 cos x = 1/2
Set each factor equal to 0.
Solve for cos x.
x =  x = 2 -=  x = 
The solutions in the interval [0, 2) are /3, , and 5/3.
Example
• Solve the following equation:
7 cos   9 = -2 cos 
Solution:
7 cos  9 = -2 cos
9 cos = -9
cos = -1
 =  ,3 ,5
 =   2n
Example
• Solve the equation on the interval [0,2)

3
tan =
2
3
Solution:

3
tan =
2
3


7
= and
2 6
6

7
 = and
3
3
Example
• Solve the equation on the interval [0,2)
cos x  2 cos x - 3 = 0
2
Solution:
cos2 x  2 cos x - 3 = 0
(cos x  3)(cosx - 1) = 0
cos x  3 = 0 cos x - 1 = 0
cos x = -3 cos x = 1
no solution x = 0
x=0
Example
• Solve the equation on the interval [0,2)
sin 2 x = sin x
Solution:
sin 2 x = sin x
2 sin x cos x = sin x
2 cos x = 1
1
cos x =
2
 5
x= ,
3
3