109
Section 7.6 – Double and Half Angle Formulas
To derive the double-angles formulas, we will use the sum of two angles
formulas that we developed in the last section. We will let α = θ and β = θ:
cos(2θ) = cos(θ + θ) = cos(θ)cos(θ) – sin(θ)sin(θ) = cos2(θ) – sin2(θ)
sin(2θ) = sin(θ + θ) = sin(θ)cos(θ) + cos(θ)sin(θ) = 2sin(θ)cos(θ)
tan(2θ) = tan(θ + θ) =
tan(θ) +tan(θ)
1− tan(θ) tan(θ)
=
2 tan(θ)
1− tan2 (θ)
For the double angle formula for the cosine function, we can derive several
variations of it using the Pythagorean Formula sin2(θ) + cos2(θ) = 1.
Solving for cos2(θ), we get cos2(θ) = 1 – sin2(θ). Thus,
cos2(θ) – sin2(θ) = (1 – sin2(θ)) – sin2(θ) = 1 – 2sin2(θ)
Now, solving for sin2(θ), we get sin2(θ) = 1 – cos2(θ). Hence
cos2(θ) – sin2(θ) = cos2(θ) – [1 – cos2(θ)]
= cos2(θ) – 1 + cos2(θ) = 2cos2(θ) – 1
Double-Angle Formulas:
1)
cos(2θ) = cos2(θ) – sin2(θ) or 1 – 2sin2(θ) or 2cos2(θ) – 1
2)
sin(2θ) = 2sin(θ)cos(θ)
3)
tan(2θ) =
Objective 1:
2 tan(θ)
1− tan2 (θ)
Use Double-angle Formulas to Find Exact Values.
Find the Exact Value of the following:
Ex. 1
Given tan(θ) = –
2
2
,
π
2
< θ < π, find
a)
tan(2θ)
b)
sin(2θ)
c)
cos(2θ)
Solution:
a)
Plug the value of tan(θ) into our double-angle formula:
tan(2θ) =
2 tan(θ)
2
1− tan (θ)
=
2(−
1−(−
2
)
2
2 2
)
2
=
− 2
2
1−( )
4
=
− 2
1
2
=–2
2
110
b)
Before we can apply the double-angle formula for the sine
function, we need to find sin(θ) and cos(θ). Since θ is in the
second quadrant, y will be positive and x will be negative.
2
2
tan(θ) = –
2
−2
=
r2 = x 2 + y 2
r2 = (– 2)2 + (
r2 = 4 + 2
r2 = 6 or r = ±
Thus, sin(θ) =
x
r
cos(θ) =
=
2
, so x = – 2 and y =
2
. Now, find r:
)2
6
y
r
=
2
−2
=
−2
6
y
x
=
6
6
(reject the negative answer)
2
=
•
6
6
•
=
6
6
6
=
−2 6
6
12
6
=–
2 3
6
=
3
3
=
and
6
3
Now, we can find sin(2θ):
sin(2θ) = 2sin(θ)cos(θ) = 2
=–
c)
2 2
3
(
3
3
)(–
6
3
)=–
2
2
(
cos(2θ) = cos (θ) – sin (θ) = –
Objective 2:
=–
6 2
9
6
9
3
9
.
Using the results form part b, we get:
=
2 18
9
1
.
3
6
3
2
) –(
3
3
2
)
=
–
=
3
9
Use the double-angle formulas to establish identities.
Establish the following identities:
Ex. 2a
cos4(θ) – sin4(θ) = cos(2θ)
€
€
Ex. 2b
cos(2θ)
1+sin(2θ)
Ex. 2c
cot(2θ) =
=
cot(θ)− 1
cot(θ)+ 1
1
(cot(θ)
2
– tan(θ))
Solution:
a)
We will start with the left side:
cos4(θ) – sin4(θ)
(factor)
2
2
= [cos (θ) – sin (θ)][cos2(θ) + sin2(θ)]
(cos2(θ) + sin2(θ) = 1)
= [cos2(θ) – sin2(θ)][1]
= cos2(θ) – sin2(θ)
(apply the double-angle identity)
= cos(2θ)
111
b)
We will start with the right side;
cot(θ)− 1
cot(θ)+ 1
=
=
=
(write in terms of sine and cosine)
cos(θ )
−1
sin(θ )
(multiply
cos(θ )
+1
sin(θ )
cos(θ )
•sin(θ)−1•sin(θ)
sin(θ )
cos(θ )
•sin(θ)+1•sin(θ)
sin(θ )
the top & bottom by sin(θ))
(simplify)
cos(θ) −sin(θ)
cos(θ) +sin(θ)
If we examine the left side for a moment, we have cos(2θ) in the
numerator which is equal to cos2(θ) – sin2(θ). This suggests that we
need to multiply top and bottom by the conjugate of the numerator:
=
=
=
=
=
c)
cos(θ) −sin(θ)
cos(θ) +sin(θ)
•
cos(θ) +sin(θ)
cos(θ) +sin(θ)
(expand)
cos2 (θ)−sin2 (θ)
(regroup the denominator)
cos2 (θ) +2cos(θ) sin(θ)+ sin2 (θ)
cos2 (θ)−sin2 (θ)
2
(cos2(θ) + sin2(θ) = 1)
2
cos (θ) +sin (θ)+ 2cos(θ)sin(θ)
cos2 (θ) −sin2 (θ)
1+2cos(θ) sin(θ)
cos(2θ)
1+sin(2θ)
(apply the double-angle formulas)
The identity has been established
Let's start with the left side:
1
tan(2θ)
2 tan(θ)
cot(2θ) =
(apply double-angle formula for tangent)
=1÷
(invert and multiply)
=
=
=
=
1− tan2 (θ)
1− tan2 (θ)
2 tan(θ)
(multiply top and bottom by
1
1
−tan2 (θ)•
tan(θ )
tan(θ )
1
2tan(θ)•
tan(θ )
1
−tan(θ)
tan(θ )
1•
2
1
(cot(θ)
2
(simplify)
(cot(θ) =
– tan(θ))
1
)
tan(θ)
1
)
tan(θ)
112
Solve for all values in [0, 2π):
Ex. 3
sin(2θ)sin(θ) = cos(θ)
Solution:
sin(2θ)sin(θ) = cos(θ)
(get zero on one side)
sin(2θ)sin(θ) – cos(θ) = 0
(sin(2θ) = 2sin(θ)cos(θ))
2sin(θ)cos(θ)sin(θ) – cos(θ) = 0
(factor out cos(θ))
2
cos(θ)[2sin (θ) – 1] = 0
(solve)
2
cos(θ) = 0 or
2sin (θ) – 1 = 0
θ=
π
2
or
3π
2
or
1
2
sin2(θ) =
1
sin(θ) = ±
€
2
=±
The reference
angle is
€
€
π
4
π
3π
π
1
2
π
,
4
•
2
2
2
2
=±
so the four angles are
5π
,π–
=
,π+
=
, 2π –
4
4
€
€ 4
€4
€
π
π 3π 5π 3π 7π
So, the solution is { , ,
,
,
,
}.
4
2
4
2
4
€4
π
4
Ex. 4
€
€ 2(θ) € €
€ €
cos(2θ)
= 2€– 2sin
Solution:
€ €
€
2
cos(2θ) = 2€– 2sin
(θ) € € (cos(2θ)
= 1 – 2sin2(θ))
1 – 2sin2(θ) = 2 – 2sin2(θ)
(add 2sin2(θ) to both sides)
1=2
There is no solution.
Ex. 5
tan(2θ) + 2cos(θ) = 0
Solution:
The tangent function is undefined when the cos(2θ) = 0
or 2θ =
π
2
or
k=1
€
Thus, our
€
€
7π
4
. Writing the general form of the solution, our
π
3π
+ 2kπ or
+ 2kπ (solve for θ)
2
2
π
3π
+ kπ or
+ kπ, k is an integer
4
4
π
π
3π
3π
θ≠
+ (0)π =
or
+ (0)π =
4
4
€ 4
€4
π
5π
3π
7π
θ≠
+ (1)π =
or
+ (1)π =
4
4
4
€ 4
π 3π 5π
7π
restriction is θ ≠ ,
,
, or
.
4
4
4
4€
€
€
restriction is
θ≠
€
€
k=0
3π
2
=
2θ ≠
€
€ €
€
€
€
€
113
(tan(θ) =
sin(2θ)
+ 2cos(θ)
cos(2θ)
2sin(θ)cos(θ)
(use the double angle identities)
2
1−2sin (θ)
2sin(θ)cos(θ)
€
€
sin(2θ)
)
cos(2θ)
tan(2θ) + 2cos(θ) = 0
=0
+ 2cos(θ) = 0
(multiply by 1 – 2sin2(θ))
€
•(1 – 2sin2(θ)) + 2cos(θ)•(1 – 2sin2(θ)) = 0•(1 – 2sin2(θ))
2
1−2sin (θ)
2sin(θ)cos(θ) + 2cos(θ)(1 – 2sin2(θ)) = 0
(factor out 2cos(θ))
€
2
2cos(θ)[sin(θ) + (1 – 2sin (θ)] = 0 (simplify inside the bracket)
2cos(θ)[– 2sin2(θ) + sin(θ) + 1] = 0 (factor out – 1)
– 2cos(θ)[2sin2(θ) – sin(θ) – 1] = 0 (2x2 – x – 1 = (2x + 1)(x – 1)
– 2cos(θ)[2sin(θ)+ 1][sin(θ) – 1] = 0
(solve)
– 2cos(θ) = 0, 2sin(θ) + 1 = 0, or sin(θ) – 1 = 0
1
,
2
π
+
6
cos(θ) = 0, sin(θ) = –
θ=
π
2
or
3π
2
, θ=π
or sin(θ) = 1
=
7π
6
or 2π –
π
6
=
11π
6
, or θ =
π
2
None of these values match our restrictions, so the solution is
π 7π 3π €11π
{ ,
,
,
}.
2
6
2
6
€
€
€ €
€ €
€
We will first derive the half angle formula for the sine function. Recall that
cos(2θ) = 1 – 2sin2(θ). We will now solve this for sin(θ):
€ € 1
€ – 2sin
€ 2(θ) = cos(2θ)
(subtract 1)
2
– 2sin (θ) = cos(2θ) – 1
(divide by – 2)
1−cos(2θ)
2
1−cos(2θ)
sin(θ) = ±
2
1−cos(α)
α
sin( ) = ±
€
2
2
sin2(θ) =
(use the square root property)
(Let
α
2
= θ, then α = 2θ)
Now, we will derive the half angle formula for cos(θ):
€ (add 1)
€ 2(θ) – 1 = cos(2θ)
2cos
2cos2(θ) = 1 + cos(2θ)
(divide by 2)
€
€
1+cos(2θ)
cos2(θ) =
(use the square root property)
cos(θ) = ±
cos(
€
€
€
€
α
2
)=±
2
1+cos(2θ)
2
1+cos(α)
2
(Let
€
α
2
= θ, then α = 2θ)
114
For the half angle formula for the tangent function, we will use the quotient
formula:
2
tan (
α
2
)=
sin2 (
α
)
2
α
cos2 ( )
2
= sin2(
1−cos(α)
2
•
2
1+cos(α)
1−cos(α)
2 α
So, tan ( ) = €
2
1+cos(α)
€
=
€
€tan(
α
2
) =€±
α
2
=
) ÷ cos2(
α
2
)=
1−cos(α)
2
÷
1+cos(α)
2
1−cos(α)
.
1+cos(α)
€(use €
the square €
root property)
1−cos(α)
€
1+cos(α)
Sometimes, it might be better to use the formula you get before you use the
€
€ root property.
square
€
€ angle formulas
Half
α
)=±
2
α
cos( ) = ±
2
α
tan(
€2)=±
sin(
€
€
€
1−cos(α)
2
1+cos(α)
2
or
1−cos(α)
1+cos(α)
or
α
)=
2
α
cos2( ) =
2
α
tan2( ) =
2
sin2(
or
€
1−cos(α)
2
1+cos(α)
2
1−cos(α)
1+cos(α)
Sign equals the sign of trigonometric function in the quadrant of α/2.
€
€
Find the following:
€
€
Ex.€6
Write sin4(θ) as an equivalent expression with the powers of the
sine and cosine equal to one.
Solution:
sin4(θ)
(write as a perfect square and θ =
=
=
=
=
)
2θ 2
)]
(apply the 2nd half-angle formula for sine)
2
2
1−cos(2θ)
(expand)
2
1−cos(2θ) 1−cos(2θ)
1−2cos(2θ) +cos2 (2θ)
•
=
2
2
4
1
1
1
4θ
2
– cos(2θ) + cos (2θ)
(write 2θ =
in the last term)
4
2
4
2
1
1
1
4θ
– cos(2θ) + cos2( )
(apply the 2nd formula for cosine)
4
2
4
2
1
1
1 1+cos(4θ)
– cos(2θ) +
(simplify)
4
2
4
2
= [sin2(
=
2θ
2
[
]
(
)
115
=
=
1
4
3
8
–
–
1
cos(2θ) +
2
1
cos(2θ) +
2
1
1
+ cos(4θ)
8
8
1
cos(4θ)
8
(simplify)
Establish the identity:
Ex. 7a
tan(
α
2
1−cos(α)
sin(α)
)=
Ex. 7b
tan(
α
2
)=
sin(α)
1+cos(α)
Solution:
a) We will start with the right side. Since sin2(
2sin2(
Thus,
α
2
) = 1 – cos(α). Also, sin(α) = sin[2(
1−cos(α)
sin(α)
€
=
b)
α
)
2
α
cos( )
2
sin(
=
2sin2 (
2sin(
α
2
= tan(
α
2
α
)
2
1−cos(α)
€
sin(α)
1−cos(α)
sin(α)
tan(
=
=
=
=
=
α
)
2
) cos(
)
α
2
)
α
2
€
(reduce)
€
α
2
)=
)] = 2
1−cos(α)
, then
2
α
α
sin( )cos( ).
2
2
€
€
The identity has been established.
(use the result from part a)
(multiply top & bottom by the conjugate of 1 – cos(α))
•
1+cos(α)
1+cos(α)
1−cos2 (α)
sin(α)(1+cos(α))
sin2 (α )
sin(α)(1+cos(α ))
sin(α)
The
1+cos(α)
(expand the numerator)
(but, sin2(α) = 1 – cos2(α))
(reduce)
identity has been established.
The advantage of these two formulas for the half-angle of tangent is you do
not have to worry about determining the sign.
Other half-angle formulas for tangent:
€
tan(
α
2
)=
1−cos(α)
sin(α)
tan(
α
2
)=
sin(α)
1+cos(α)
Now, let's apply these formulas to an application problem from the book
(Sullivan's PreCalculus, 9th edition, © 2012, page 493, exercise 95.)
116
Verify the following:
Ex. 8
Given D =
1
2
W
csc(θ)− cot(θ)
, show that W = 2Dtan(
θ
2
)
Solution:
D=
1
2
W
(multiply by 2(csc(θ) – cot(θ))
csc(θ)− cot(θ)
2(csc(θ) – cot(θ))D = 2(csc(θ) – cot(θ))
1
2
W
csc(θ)− cot(θ)
2(csc(θ) – cot(θ))D = W or
W = 2D(csc(θ) – cot(θ)) (rewrite in terms of sine and cosine)
1
cos(θ)
–
sin(θ)
sin(θ)
1−cos(θ)
2D(
)
sin(θ)
θ
2Dtan( )
2
W = 2D(
W=
W=
Objective 3:
)
(combine)
(apply the half-angle formula)
Use half-angle Formulas to Find Exact Values
Find the exact value of the following:
Ex. 9a
sin(15˚)
Ex. 9b
Solution:
a)
sin(15˚) = sin(
30o
2
)=±
cos(– 112.5˚)
1−cos(30o )
2
Since 15˚ is in quadrant I, then the sine function is positive:
=
=
=
=
b)
1−cos(30o )
2
1−
3
2
3
2
)
(multiply the top & bottom by 2)
2
1•2−
(cos(30˚) =
3
2
•2
2•2
2− 3
4
=
(simplify)
2− 3
2
Since the cosine function is even, then
cos(– 112.5˚) = cos(112.5˚). Thus,
cos(112.5˚) = cos(
225o
2
)=±
1+cos(225o )
2
Since 112.5˚ is in quadrant II, then cosine function is negative:
117
1+cos(225o )
2
=–
=–
=–
=–
Ex. 10
2
2
)
2
2
•2
1+(−
(multiply the top & bottom by 2)
2
1•2−
(simplify)
2•2
2− 2
4
2− 2
2
=–
Given cos(θ) =
a)
2
2
(cos(225˚) = – cos(45˚) = –
sin(
θ
2
6
3
)
3π
2
< θ < 2π, find
b)
cos(
θ
2
)
c)
tan(
θ
2
)
Solution:
First. let's figure out what quadrant
3π
2
3π
4
a)
sin(
< θ < 2π
θ
2
<
θ
2
<π
θ
is in
2
1−cos(θ)
2
=
=
€
1−
6
3
(divide by 2)
Thus,
θ
2
is in quadrant II.
quadrant II, then the sine function is positive:
(cos(θ) =
6
3
)
(multiply the top & bottom by 3)
2
1•3−
lies in:
1−cos(θ)
2
)=±
Since
=
θ
2
6
3
2•3
•3
(simplify)
3− 6
.
6
1+cos(θ)
θ
cos( ) = ±
2
2
θ
Since
is in quadrant II, then
2
1+cos(θ)
=–
(cos(θ)
2
=
b)
€
the cosine function is negative:
=
6
3
)
118
1+
=–
1•3+
6
3
•3
2•3
3+ 6
6
=–
tan(
(multiply the top & bottom by 3)
2
=–
c)
6
3
θ
2
Since
=–
=–
.
)=±
1−cos(θ)
1+cos(θ)
θ
is in
2
1−cos(θ)
1+cos(θ)
quadrant II, then the tangent function is negative:
1−
1+
=–
(simplify)
6
3
6
3
(cos(θ) =
6
3
)
(multiply the top & bottom by 3)
3− 6
3+ 6
Solve for all angles in the interval [0, 2π):
Ex. 11
sin3(θ) + cos3(θ) = 0
Solution:
sin3(θ) + cos3(θ) = 0
(F3 + L3 = (F + L)(F2 – FL + L2))
(sin(θ) + cos(θ))(sin2(θ) – sin(θ)cos(θ) + cos2(θ)) = 0
(sin2(θ) + cos2(θ) = 1 in the 2nd parenthesis)
(sin(θ) + cos(θ))(1 – sin(θ)cos(θ)) = 0
(sin(2θ) = 2sin(θ)cos(θ))
(sin(θ) + cos(θ))(1 –
sin(θ) + cos(θ) = 0
or
cos(θ) = – sin(θ)
€ sides)
(square both
2
cos (θ) = sin2(θ)
cos2(θ) – sin2(θ) = 0
cos(2θ) = 0
or
2θ =
2θ =
€
€
1
sin(2θ))
2
π
2
π
2
or
3π
2
+ 2kπ or
€
€
=0
1–
(solve)
1
sin(2θ)
2
=0
sin(2θ) = 2
no solution
(subtract sin2(θ) from both sides)
€
(cos(2θ) = cos2(θ) – sin2(θ))
We will need to write our general solution:
3π
2
+ 2kπ
(solve for θ)
119
θ=
π
4
+ kπ or
k=0
€
k=1
€
€
k is an integer
=
π
4
5π
4
3π
3π
+ (0)π =
4
4
3π
7π
or
+ (1)π =
4
4
or
θ=
=
€
Since we squared the equation, we have to check for the solutions.
π
π
π
θ= €
sin3( €
) + cos€3( ) = 1 + 1€≠ 0
(reject)
θ=
€
θ=
3π
+ kπ,
4
π
+ (0)π
4
π
+ (1)π
4
θ=
θ=
The
4
3π€
4
5π
4 €
7π
4 €
4
4
3π€
3π
3
sin (
) + cos€(
4
4
3 5π
3 5π
sin (
) + cos (
4€
4
3 7π
3 7π
sin (
) + cos (
4€
4
3π 7π
solution is {
,
}
4€ 4
€
3
)=1+€
(– 1) = 0
(keep)
) = – 1 + (– 1) ≠ 0
(reject)
) = (– 1) + 1 = 0
(keep)
€ summarize
€ the formulas
€ that we derived in this section:
Let'
€ Formulas:
€
Double-Angle
1)
cos(2θ) = cos2(θ) – sin2(θ) or 1 – 2sin2(θ) or 2cos2(θ) – 1
2)
sin(2θ) = 2sin(θ)cos(θ)
3)
tan(2θ) =
2 tan(θ)
1− tan2 (θ)
Half angle formulas
4)
5)
6)
€
6b)
€
€
1−cos(α)
α
)=±
2
2
1+cos(α)
α
cos( ) = ±
2
2
α
1−cos(α)
tan(
€ 2 ) = ± 1+cos(α)
α
1−cos(α)
tan(€ ) =
2
sin(α)
sin(
€
or
or
1−cos(α)
α
)=
2
2
1+cos(α)
α
cos2( ) =
2
2
α
1−cos(α)
tan2( ) =
2
1+cos(α)
α
sin(α)
tan( ) =
2
1+cos(α)
sin2(
or
€
6c)
€
€
€