CHAPTER 2.5
CHAPTER 2 ANALYTICAL TRIGONOMETRY
PART 5 –Multiple-Angle and Product-to-Sum Formulas
TRIGONOMETRY MATHEMATICS CONTENT STANDARDS:

9.0
- Students compute, by hand, the values of the trigonometric functions and
the inverse trigonometric functions at various standard points.

10.0
- Students demonstrate an understanding of the addition formulas for
sines and cosines and their proofs and can use those formulas to prove
and/or simplify other trigonometric identities .

11.0
- Students demonstrate an understanding of half-angle and double-angle formulas for sines and cosines and can use those formulas to prove and/or simplify other trigonometric identities.

19.0
- Students are adept at using trigonometry in a variety of applications
and word problems.
OBJECTIVE(S):

Students will learn the double-angle formulas.

Students will learn how to solve a multiple-angle equation.

Students will learn how to evaluate functions involving double angles.

Students will learn how to derive a multiple-angle formula.

Students will learn the power-reducing formulas.

Students will learn how to reduce a power.

Students will learn the half-angle formulas.

Students will learn how to solve a trigonometric equation using the half-angle formulas.

Students will learn the product-to-sum and sum-to-product formulas.

Students will learn gain further practice applying trigonometry to the real-world.
Multiple-Angle Formulas
In this section you will study four other categories of trigonometric identities.
1.
The first category involves functions of multiple angles such as sin ku and cos ku .
2.
The second category involves squares of trigonometric functions such as sin
2
3.
The third category involves functions of half-angles such as sin
F u
H
J
. u .
4.
The fourth category involves products of trigonometric functions such as sin cos v .

CHAPTER 2.5
The most commonly used multiple-angle formulas are the double-angle formulas .
Double-Angle Formulas sin 2 u

2 sin cos u cos2 u = cos
2 u
 sin
2
2 cos
2 u

1
2
1 2 sin u u tan 2 u

2 tan u
1
 tan
2 u
Note that sin 2 u

2 sin u , cos 2 u

2 cos u , and tan 2 u

2 tan u .
PROOF: sin 2 u = sin b g
=
=
EXAMPLE 1: Solving a Multiple-Angle Equation
Solve 2 cos x
 sin 2 x

0 .
Begin by rewriting the equation so that it involves functions of x (rather than 2 x ). Then factor and solve as usual.
2 cos x
 sin 2 x

0 Write original equation.
Double-angle formula.
Factor.
Set factors equal to zero.
Solutions in 0 2
 g
.
So, the general solution is where n is an integer. and General solution.

CHAPTER 2.5
EXAMPLE 2: Using Double-Angle Formulas to Analyze Graphs
Use a double-angle formula to rewrite the equation y

4 cos
2 x

2 .
Using the double-angle formula for _________, you can rewrite the original equation as y

4 cos
2 x

2 Write original equation.
Factor
Use double-angle formula
EXAMPLE 3: Evaluating Functions Involving Double Angles
Use the following to find sin 2

, cos2

, and tan2

. cos
 
5
13
,
3
2
  y sin
  y
 r
Consequently, using each of the double-angle formulas, you can write sin 2

= = = cos2

= tan2

=
=
=
=

5 , x

12


CHAPTER 2.5
1.) Find sec2

.
5
12

CHAPTER 2.5
2.) If cos u
 
2
7 and

2

, find sin 2 u .
y x
The double-angle formulas are not restricted to angles 2

, and

. Other double combinations, such as 4

and 2

, or 6
 and 3

, are also valid. Her are two examples. sin 4
 
2 sin 2
 cos 2
 and cos 6
  cos
2
3
  sin
2
3


CHAPTER 2.5
3.) Solve on the interval 0 2
 g
:
2 x
 cos 2 x
2 g

1
4.) Simplify 6 cos
2 x

3 .

CHAPTER 2.5
DAY 1
By using double-angle formulas together with the sum formulas derived in the previous section, you can form other multiple-angle formulas.
EXAMPLE 4: Deriving a Triple-Angle Formula
Express sin 3 x in terms of sin x . sin 3 x =
= Sum formula.
=
=
Double-angle formula.
Multiply.
=
=
=
Pythagorean identity.
Multiply.
Simplify.

CHAPTER 2.5
Power-Reducing Formulas
The double-angle formulas can be used to obtain the following power-reducing formulas . sin
2 u

1
 cos 2 u
2 cos
2 u

1
 cos 2 u
2 tan
2 u

1
 cos 2 u
1
 cos 2 u
EXAMPLE 5: Reducing a Power
Rewrite sin
4 x as a sum of first powers of the cosines of multiple angles.
Note the repeated use of power-reducing formulas. sin
4 x = Property of exponents.
=
=
Power-reducing formula.
Expand binomial.
=
=
=
=
Power-reducing formula.
Distributive Property.
Factor out a common factor.
Simplify.

CHAPTER 2.5
Half-Angle Formulas
You can derive some useful alternative forms of the power-reducing formulas by replacing u with u
2 sin u
2
 
1
 cos u
2
. The results are called half-angle formulas . cos u
2
 
1
 cos u
2 tan u
2

1
 cos u sin u
 sin u
1
 cos u
The signs of sin
F u
H
J and cos
F u
H
J depend on the quadrant in which u
2 lies.

CHAPTER 2.5
EXAMPLE 6: Using a Half-Angle Formula
Find the exact value of sin105
0
.
y x
Begin by noting that 105
0 sin
F u
H
J is half of __________. Then, using the half-angle formula for and the fact the 105
0 lies in _________________________, you have sin105
0
=
=
=
=
=
The ________________ square root is chosen because sin
 is __________________ in
Quadrant ________.

CHAPTER 2.5
EXAMPLE 7: Solving a Trigonometric Equation
Find all solutions of 2
 sin
2 x

2 cos
2 x
2 in the interval 0 2
 g
.
2
 sin
2 x

2 cos
2 x
2
Write original equation.
Half-angle formula.
Simplify.
Simplify.
Pythagorean identity.
Simplify.
Factor.
By setting the factors _________________ and ________________________ equal to zero, you find that the solutions in the interval 0 2
 g are
DAY 2 x = ________________, x = _________________, and x = ______________.

CHAPTER 2.5
5.) Find sin

2
8
.
15

CHAPTER 2.5
6.) Find the exact value of:
7
 a. cos
12 b. tan .
0

CHAPTER 2.5
7.) If cot u

7 and

3

2
, find cos u
2
. y x

CHAPTER 2.5
8.) Simplify
1
 cos 4 x
.
2

CHAPTER 2.5
9.) Find the exact zeros on 0 2
 g
: bg
 x sin cos
2
 x

1
DAY 3

CHAPTER 2.5
Product-to-Sum Formulas
Each of the following product-to-sum formulas is easily verified using the sum and difference formulas discussed in the preceding section. sin u sin v cos u cos v


1
2
 cos
 u
 v

 cos
 u
 v
 
1
2
 cos
 u
 v
  cos
 u
 v
  sin u cos v cos u sin v


1
2
 sin
 u
 v

 sin
 u
 v
 
1
2
 sin
 u
 v
  sin
 u
 v
 
EXAMPLE 8: Writing Products as Sums
Rewrite the product as a sum or difference. cos cos
5
5 x sin 4 x x sin 4 x =
=
Occasionally, it is useful to reverse the procedure and write a sum of trigonometric functions as a product. This can be accomplished with the following sum-to-product formulas .
Sum-to-Product Formulas sin x
 sin y

2 sin x
2 y cos
 x
2 y sin x
 sin y

2 cos
 x
2 y sin x
2 y cos x
 cos y

2 cos cos x
 cos y
 
2 sin x x
2
2 y y cos
 x
2 y x y sin
2

CHAPTER 2.5
EXAMPLE 9: Using a Sum-to-Product Formula
Find the exact value of cos 195
0  cos 105
0
.
Using the appropriate sum-to-product formula, you obtain cos 195
0  cos 105
0
=
=
=
=

CHAPTER 2.5
EXAMPLE 10: Solving a Trigonometric Equation
Find all solutions of sin 5 x
 sin 3 x

0 in the interval

0 , 2
 
. sin 5 x
 sin 3 x

0 Write original equation.
Sum-to-product formula.
Simplify.
Note that the general solution would be x = __________, where n is an integer.

CHAPTER 2.5
EXAMPLE 11: Verifying a Trigonometric Identity
Verify the identity sin cos t t

 sin cos
3 t
3 t
 tan 2 t .
Using the appropriate sum-to-product formulas, you have sin cos t t
 sin 3 t
 cos 3 t
=
=
=
DAY 4
=
10.) Write as sum or difference:
5 sin 3
 sin 4

=
=
=
=
11.) Write as a product: cos x
 cos 7 x =
=
=

CHAPTER 2.5
12.) Find the zeros on

0 , 2
 
: h
 
 cos 2 x
 cos 6 x
DAY 5
13.) Verify the following trigonometric identities: a. tan u
2
 csc u
 cot u b. cos 4 x
 cos 2 x
2 sin 3 x
  sin x

CHAPTER 2.5
Application
EXAMPLE 12: Projectile Motion
Ignoring air resistance, the range of a projectile fired at an angle
 with the horizontal and with an initial velocity of v feet per second given by
0 r

1
2 v
0
16 sin
 cos
 where r is the horizontal distance (in feet) that the projectile will travel. A place kicker for a football team can kick a football from ground level with an initial velocity of 80 feet per second. a.) Write the projectile motion model in a simpler form.
You can use a double-angle formula to rewrite the projectile motion model as r

1
2 v
0
16 sin
 cos

Rewrite original projectile motion model.
Rewrite model using a double-angle formula.

CHAPTER 2.5 b.) At what angle must the player kick the football so that the football travels 200 feet?
Write projectile motion model.
Substitute ______ for r and
_____ for v .
0
Simplify.
Divide each side by 200.
You know that 2
 
_____, so dividing this result by _____ produces
 
_____.
Because _____ = ______, you can conclude that he player must kick the football at an angle of ______ so that the football will travel 200 feet. c.) For what angle is the horizontal distance the football travels a maximum?
From the model _______________________ you can see that the amplitude is ________.
So the maximum range is r = _________ feet. From part b.), you know that this corresponds to angle of ______. Therefore, kicking the football at an angle of ______ will produce a maximum horizontal distance of ________ feet.