Download Reteach 11.5

Name ________________________________________ Date __________________ Class__________________
Reteach
LESSON
11-5
Double-Angle and Half-Angle Identities
There are special double-angle formulas for sin2θ, cos2θ, and tan2θ.
Double-Angle Identities
The double angle is twiceθ, or 2θ.
sin2θ = 2sinθ cosθ
cos2θ = cos2θ − sin2θ
There are three equivalent forms of
the identity for cos2θ. If you know one
form, you can use sin2θ + cos2θ = 1
to derive the other forms.
cos2θ = 2cos2θ − 1
cos2θ = 1 − 2sin2θ
tan 2θ =
2 tanθ
1 − tan2 θ
Find sin2θ and cos2θ if cos θ = −
Step 1 Find sinθ.
sin2θ = 1 – cos2θ
⎛ 2⎞
sin2 θ = 1 − ⎜ − ⎟
⎝ 3⎠
5
2
sin θ =
9
sin θ = −
2
Pythagorean identity
2
Substitute cos θ = − .
3
Notice the interval
in which θ lies. Cosine and
sine are both negative in
Quadrant III.
Evaluate.
5
3
Use the negative
value since θ lies in
Quadrant III.
Solve for sinθ.
Step 2 Find sin2θ.
sin2θ = 2sinθ cosθ
⎛
5 ⎞⎛ 2 ⎞
−
sin2θ = 2 ⎜ −
⎜ 3 ⎟⎟ ⎜⎝ 3 ⎟⎠
⎝
⎠
sin2θ =
2
and 180° < θ < 270°.
3
4 5
9
Write the formula.
Substitute values for sinθ and cosθ.
Simplify.
Step 3 Find cos2θ.
cos2θ = 2cos2θ – 1
2
⎛ 2⎞
cos 2θ = 2 ⎜ − ⎟ − 1
⎝ 3⎠
8
1
cos 2θ = − 1 = −
9
9
Choose a formula.
Substitute the value for cosθ.
Simplify.
Evaluate each expression if cos θ =
1. sinθ
Use sin2θ = 1 − cos2θ.
1
and 270° < θ < 360°.
4
2. cos2θ
Use cos2θ = 1 − 2sin2θ.
_________________________________________
3. tanθ
________________________________________
4. tan2θ
sin θ
.
Use tan θ =
cos θ
Use tan 2θ =
_________________________________________
2 tanθ
.
1 − tan2 θ
________________________________________
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
11-38
Holt McDougal Algebra 2
Name ________________________________________ Date __________________ Class__________________
LESSON
11-5
Reteach
Double-Angle and Half-Angle Identities (continued)
You can use half-angle identities to find the exact value of some
trigonometric expressions.
Half�Angle Identities
1 − cos θ
θ
θ
⎛θ ⎞
sin ⎜ ⎟ = ±
: (+) for
in Quadrants I, II; (−) for
in Quadrants III, IV
2
2
2
2
⎝ ⎠
1 + cos θ
θ
θ
⎛θ ⎞
cos ⎜ ⎟ = ±
: (+) for
in Quadrants I, IV; (−) for
in Quadrants II, III
2
2
2
⎝2⎠
1 − cosθ
θ
θ
⎛θ ⎞
tan ⎜ ⎟ = ±
: (+) for
in Quadrants I, III; (−) for
in Quadrants II, IV
2
2
1 + cos θ
⎝2⎠
Think: 225° has a
reference angle of 45°.
Find the exact value of cos 112.5°.
225°
2
1 + cos θ
=−
2
1 + cos 225°
=−
2
cos112.5° = cos
⎛
2⎞
1 + ⎜⎜ −
⎟
2 ⎟⎠
⎝
=−
2
⎛ 2 − 2 ⎞⎛ 1⎞
=− ⎜
⎜ 2 ⎟⎟ ⎜⎝ 2 ⎟⎠
⎝
⎠
=−
Find the angle: θ = 225°.
Choose the half-angle identity.
Substitute θ = 225°.
Evaluate.
Simplify.
112.5° is in QII where
cosine is negative.
cos225° = −cos45°
2
=−
2
2− 2
2− 2
=−
4
2
Use the half-angle identity to find the exact value of each expression.
315°
30°
6. sin15° = sin
5. tan 157.5° = tan
2
2
1 − cos 315°
⎛ 315° ⎞
tan ⎜
=−
⎟
1 + cos 315°
⎝ 2 ⎠
⎛ 30° ⎞
sin ⎜
⎟ = _____________________
⎝ 2 ⎠
_________________________________________
________________________________________
_________________________________________
________________________________________
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
11-39
Holt McDougal Algebra 2
Reteach
1.
− 15
4
3. − 15
2. −
7
8
4.
15
7
5. tan
2
2− 2
2 ;−
2 ;
5. −
2
2+ 2
1+
2
2
6.
7. tan
sin A
1 − cos A
sin A
1. a. sin 2θ = 2 sinθ cosθ
b. d (θ ) =
c. 35 =
3v 0 2 sin 2θ
32
3(20)2 sin 2θ
32
d. sin2θ = 0.933
Challenge
e. 34.5°
sin 2 A
cos 2 A
2. a.
2 sin A cos A
b. tan 2 A =
cos2 A − sin2 A
Angle of
Elevation (°)
Horizontal
Distance (yd)
15
18.75
25
28.73
35
35.24
45
37.50
55
35.24
2 sin A cos A
cos2 A
c. tan 2 A =
2
cos A sin2 A
−
cos2 A cos2 A
2 tan A
1 − tan2 A
2. tan2A = tan(A + A) =
tan A + tan A
2 tan A
=
1 − tan A tan A 1 − tan2 A
3. a. tan 2 A =
=
Problem Solving
2− 3 2− 3
;
4
2
d. tan 2 A =
2
2
3
1−
2 ;
2
1. a. tan 2 A =
A
1 − cos A 1 − cos A
=
⋅
2
1 + cos A 1 − cos A
(1 − cos A )
2+ 2
2+ 2
2− 2 2− 2 2− 2
2
⋅
=
⋅
= 2 −1
2+ 2 2− 2
2
2
b.
2− 2
1 − cos 30°
;
2
2− 2
6. a. tan 22.5° =
1−
−
A
1 − cos A
=±
2
1 + cos A
b. 45°
3. 26.5°
4. Possible answer: For this initial velocity
and target distance, sin2θ ≈ 1.32, which
is outside the possible range of values for
the sine of an angle; so he is incorrect.
24
≈ 3.43
7
b. m∠A = 127°, tan254° ≈ 3.49
5. C
1 − tan2 A
4. a. 2 cot 2 A =
tan A
b. cot A − tan A =
6. G
1 − tan2 A
tan A
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
A54
Holt McDougal Algebra 2