Download Double-Angle Formulas

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 sin2 u .
u
3. The third category involves functions of half-angles such as sin
.
2
4. The fourth category involves products of trigonometric functions such as sin u cos v .
F
I
G
HJ
K
CHAPTER 2.5
The most commonly used multiple-angle formulas are the double-angle formulas.
Double-Angle Formulas
sin 2u  2 sin u cos u
cos2 u  sin2 u
2 cos2 u  1
1  2 sin2 u
cos2u =
tan 2u 
2 tan u
1  tan 2 u
Note that sin 2u  2 sin u , cos 2u  2 cos u , and tan 2u  2 tan u .
PROOF:
sin 2u =
b g
sin u  u
=
=
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 2x). 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.
g
Solutions in 0,2 .
So, the general solution is
and
where n is an integer.
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 sin2 , cos2 , and tan2 .
cos 
5
,
13
y
3
   2
2
x
sin  
y

r
Consequently, using each of the double-angle formulas, you can write
sin2 =
=
=
cos2 =
=
=
tan2 =
=
5,12
CHAPTER 2.5
1.) Find sec2 .
5
12
CHAPTER 2.5
2.) If cosu  
2

and  u   , find sin 2u .
7
2
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  cos2 3  sin 2 3
CHAPTER 2.5
g
3.) Solve on the interval 0,2 :
bsin 2 x  cos 2 xg 1
2
4.) Simplify 6 cos2 x  3 .
CHAPTER 2.5
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 3x in terms of sin x .
sin 3x =
DAY 1
=
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 2u
2
cos2 u 
1  cos 2u
2
tan 2 u 
1  cos 2u
1  cos 2u
EXAMPLE 5: Reducing a Power
Rewrite sin4 x as a sum of first powers of the cosines of multiple angles.
Note the repeated use of power-reducing formulas.
sin4 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
u
replacing u with . The results are called half-angle formulas.
2
sin
u
1  cos u

2
2
u
1  cos u
cos  
2
2
tan
u 1  cos u
sin u


2
sin u
1  cos u
The signs of sin
u
uI
F
Fu I
and cosGJdepend on the quadrant in which lies.
G
J
H2 K H2 K
2
CHAPTER 2.5
EXAMPLE 6: Using a Half-Angle Formula
Find the exact value of sin1050 .
y
x
Begin by noting that 1050 is half of __________. Then, using the half-angle formula for
u
sin
and the fact the 1050 lies in _________________________, you have
2
F
I
G
HJ
K
sin1050
=
=
=
=
=
The ________________ square root is chosen because sin  is __________________ in
Quadrant ________.
CHAPTER 2.5
EXAMPLE 7: Solving a Trigonometric Equation
x
Find all solutions of 2  sin 2 x  2 cos2 in the interval 0,2 .
2
g
2  sin 2 x  2 cos2
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 are
g
x = ________________, x = _________________, and x = ______________.
DAY 2
CHAPTER 2.5
5.) Find sin

2
.
8
15
CHAPTER 2.5
6.) Find the exact value of:
7
a. cos
12
b. tan 202.50
CHAPTER 2.5
7.) If cot u  7 and   u 
3
u
, find cos .
2
2
y
x
CHAPTER 2.5
8.) Simplify
1  cos 4 x
.
2
CHAPTER 2.5
g
9.) Find the exact zeros on 0,2 :
x
f x  sin  cos x  1
2
bg
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.
1
cosu  v   cosu  v 
2
1
cos u cos v  cosu  v   cosu  v 
2
1
sin u cos v  sin u  v   sin u  v 
2
1
cos u sin v  sin u  v   sin u  v 
2
sin u sin v 
EXAMPLE 8: Writing Products as Sums
Rewrite the product as a sum or difference.
cos 5x sin 4x
cos 5x sin 4x =
=
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
 x y  x y
sin x  sin y  2 sin 
 cos

 2   2 
x y x y
sin x  sin y  2 cos
 sin 

 2   2 
x y x y
cos x  cos y  2 cos
 cos

 2   2 
x y x y
cos x  cos y  2 sin 
 sin 

 2   2 
CHAPTER 2.5
EXAMPLE 9: Using a Sum-to-Product Formula
Find the exact value of cos 1950  cos 1050 .
Using the appropriate sum-to-product formula, you obtain
cos 1950  cos 1050
=
=
=
=
CHAPTER 2.5
EXAMPLE 10: Solving a Trigonometric Equation
Find all solutions of sin 5x  sin 3x  0 in the interval 0,2  .
sin 5x  sin 3x  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
sin t  sin 3t
 tan 2t .
Verify the identity
cos t  cos 3t
Using the appropriate sum-to-product formulas, you have
sin t  sin 3t
=
cos t  cos 3t
=
=
=
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x  cos 2x  cos 6x
13.) Verify the following trigonometric identities:
u
cos 4 x  cos 2 x
  sin x
a. tan  csc u  cot u
b.
2
2 sin 3x
DAY 5
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 v0 feet per second given by
r
1 2
v0 sin  cos 
16
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
v0 sin  cos 
16
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 v0 .
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.