Name ________________________________________ Date __________________ Class__________________
LESSON
11-4
Reteach
Sum and Difference Identities
You can use angle addition and subtraction identities to find the exact
value of some trigonometric expressions.
Look for ways to use 30° or
π
π
, and 60° or
π
to make
6
4
3
the sum because exact values are known for these angles.
Sum Identities
, 45° or
Difference Identities
sin (A + B) = sinAcosB + cosAsinB
sin (A – B) = sinAcosB – cosAsinB
cos (A + B) = cosAcosB – sinAsinB
cos (A − B) = cosAcosB + sinAsinB
tan ( A + B ) =
tan A + tan B
1 − tan A tan B
tan ( A − B ) =
Find the exact value of cos105°.
cos105° = cos(60° + 45°)
= cos60°cos45° – sin60°sin45°
1 2
3
2
= ⋅
−
⋅
2 2
2
2
2
6
2− 6
=
−
=
4
4
4
⎛π ⎞
Find the exact value of sin ⎜ ⎟ .
⎝ 12 ⎠
⎛π ⎞
⎛π π ⎞
sin ⎜ ⎟ = sin ⎜ − ⎟
⎝ 12 ⎠
⎝3 4⎠
= sin
π
3
cos
π
4
− cos
π
3
sin
tan A − tan B
1 + tan A tan B
Think: 60° + 45° = 105°.
Use cos(A + B) identity.
Evaluate.
The value is negative. This
makes sense since 105° lies
in Quadrant II where cosine
is negative.
Simplify.
Think:
π
π
3
−
π
4
=
π
12
.
Substitute:
Use sin(A − B) identity.
4
3
2 1 2
=
⋅
− ⋅
2
2
2 2
6
2
6− 2
=
−
=
4
4
4
Substitute:
A = 60° and
B = 45°.
A=
Evaluate.
B=
π
and
3
π
4
.
Simplify.
Use the sum or difference identity to find the exact value of each
expression.
⎛ 7π ⎞
⎛π π ⎞
1. cos(–15°) = cos(30° – 45°)
2. sin ⎜
= sin ⎜ + ⎟
⎟
⎝ 12 ⎠
⎝3 4⎠
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Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
11-30
Holt McDougal Algebra 2
Name ________________________________________ Date __________________ Class__________________
LESSON
11-4
Reteach
Sum and Difference Identities (continued)
You can use angle addition and subtraction identities to prove identities.
Use an identity to make one side of the equation resemble the other side.
Sum Identities
Difference Identities
sin(A + B) = sinAcosB + cosAsinB
sin(A – B) = sinAcosB – cosAsinB
cos(A + B) = cosAcosB – sinAsinB
cos (A − B) = cosAcosB + sinAsinB
tan ( A + B ) =
tan A + tan B
1 − tan A tan B
tan ( A − B ) =
tan A − tan B
1 + tan A tan B
Prove: tan(π + x) = tanx
tan(π + x) = tanx
tan π + tan x
= tan x
1 − tan π tan x
0 + tan x
= tan x
1 − (0) tan x
tan x
= tan x
1
Modify the left side.
Use tan(A + B) identity.
Substitute:
A = π and
B = x.
Evaluate. Think: tanπ = 0.
Simplify.
Prove: sin(π − x) = sinx
sin(π – x) = sinx
sinπ cosx – cosπ sinx = sinx
(0)cosx –(–1) sinx = sinx
0 + sinx = sinx
Modify the left side.
Use sin(A – B) identity.
Evaluate. Think: sinπ = 0 and cosπ = –1.
Simplify.
Write the missing steps or reasons to prove each identity.
⎛π
⎞
3. sin ⎜ + x ⎟ = cos x
⎝2
⎠
Modify the left side.
⎛π ⎞
⎛π ⎞
sin ⎜ ⎟ cos x + cos ⎜ ⎟ sin x = cos x
⎝2⎠
⎝2⎠
_________________________________
_____________________________________ Evaluate.
_____________________________________ Simplify.
4. cos(π + x) = −cosx
Modify the left side.
_____________________________________ Use cos(A + B) identity.
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Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
11-31
Holt McDougal Algebra 2
1 x cosx − 0 x sinx = cosx
⎛ 3π ⎞
⎛ 3π ⎞
− sin ⎜
sin x cos ⎜
⎟
⎟ cos x = cosx
⎝ 2 ⎠
⎝ 2 ⎠
sinx x 0 − (−1)cosx = cosx
cosx = cosx
cosx = cosx
9.
π⎞
⎛
8. cos ⎜ x − ⎟ = sin x
2⎠
⎝
⎛π ⎞
⎛π ⎞
cos x cos ⎜ ⎟ + sin x sin ⎜ ⎟ = sinx
⎝2⎠
⎝2⎠
cosx x 0 + sinx x 1 = sinx
21
221
11. −
21
220
140
13. −
221
10. −
220
221
12. −
171
221
13. −
204
325
14. −
=
3
2 1 2
⋅
+ ⋅
=
2
2
2 2
2. sin
=
253
204
π
3
cos
π
4
+ cos
π
3
sin
3
2 1 2
=
x
+ x
2
2
2 2
6+ 2
4
π
4
6+ 2
4
3. Use sin(A + B) identity
(1)(cosx) + 0(sinx) = cosx
cosx = cosx
Practice C
5.
253
325
1. cos30°cos45° + sin30°sin45°
16. 207 m
2.
2− 6
4
12.
Reteach
171
14.
140
c. A′(0.50, 0.87), B′(5, 8.66),
C′(−4.2, 4.73)
3. −2 − 3
323
36
17. 156.3 mi
⎡0.50 5.00 −4.20 ⎤
b. ⎢
⎥
⎣0.87 8.66 4.73 ⎦
1
2
11. −
36
325
16. F′(−9.66, 2.59), G′(2.59, 9.66),
H′(1.22, 0.71)
⎡cos 60° − sin 60° ⎤ ⎡ 1 10 2 ⎤
15. a. ⎢
⎥⎢
⎥
⎣sin 60° cos 60° ⎦ ⎣0 0 6 ⎦
1.
10. −
15. P′(−0.71, 0.71), Q′(2.83, −2.83),
R′(−2.12, 4.95)
sinx = sinx
9.
323
325
4. cosπcosx − sinπ sinx = −cosx
2− 6
4
(−1)cosx − (0)(sinx) = −cosx
Evaluate.
4. −2 + 3
−cosx = −cosx
6− 2
6. −
4
Simplify.
Challenge
7. tan(π − x) = − tanx
1. Possible answer: Since the triangles are
right triangles and the circle has a radius
of 1, the length of the side opposite angle
x is sinx and the length of the adjacent
side is cosx. In the right triangle with
angle y, the length of the hypotenuse is
cosx so the length of the side opposite
angle y is cosxsiny and the adjacent side
is cosxcosy.
tan π − tan x
= − tan x
1 + tan π tan x
0 − tan x
= − tan x
1 + 0 x tan x
−tanx = −tanx
⎛π
⎞
8. sin ⎜ − x ⎟ = cos x
2
⎝
⎠
2. Possible answer: m∠OAD + a = 90° − x; x
+ y = 90° − m∠OAD; so a = 90° − x −
⎛π ⎞
⎛π ⎞
sin ⎜ ⎟ cos x − cos ⎜ ⎟ sin x = cos x
⎝2⎠
⎝2⎠
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A51
Holt McDougal Algebra 2