Name ________________________________________ Date __________________ Class__________________
LESSON
11-4
Reading Strategy
Follow a Procedure
You can use sum and difference identities, along with the Unit Circle, to
simplify expressions and find solutions.
Sine Sum Identity
Sine Difference Identity
sin(A + B) = sinAcosB + cosAsinB
sin(A − B) = sinAcosB − cosAsinB
To find the value of sin105°:
STEP 1: Determine
whether the angle
is the sum of two
known angles, or
the difference:
105° = 45° + 60°
STEP 2: Substitute the new angles into
the sum or difference identity:
sin(105°) = sin45°cos60° + cos45°sin60°
STEP 3: Solve:
sin(105°) = sin45°cos60° + cos45°sin60°
2 1
2
3
i +
i
=
2 2
2
2
2
6
2+ 6
=
+
=
4
4
4
Identify a pair of angles whose trigonometric values are known and
whose sum or difference is equal to each angle given.
1. 75° _____________________
2. 255°
___________________________
3. 165° _____________________
4. 105°
___________________________
Write the following as sum or difference identities.
5. sin(45° − 30°)
6. sin(240° + 60°)
_________________________________________
7. sin(300° − 210°)
________________________________________
8. sin(180° + 45°)
_________________________________________
9. sin(90° − 60°)
________________________________________
10. sin(225° − 135°)
_________________________________________
________________________________________
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
11-34
Holt McDougal Algebra 2
m∠OAD and y = 90° − m∠OAD − x;
therefore, a = y.
⎡ −1.37
2. ⎢
⎣0.37
3. Possible answer: Triangle ABD is a right
triangle, therefore, AD = AB cosa and BD
= AB sina. But AB = sinx and a = y, so AD
= sinxcosy and BD = sinxsiny.
4. Possible answer: In right triangle OBC,
OB = cosx and BC = OBsiny = cosxsiny.
From the diagram, it is clear why the
length EF is, as the label indicates, equal
to sin(x + y). But EF = AD + BC.
Substitution of the expressions for EF,
AD, and BC gives sin(x + y) = sinxcosy +
cosxsiny.
5. Possible answer: Extend segment AD to
intersect segment OC at a point P.The
segment AP will be perpendicular to
segment OC so it follows that OP = cos(x +
y). But OP + PC = OC, and hence OP = OC
− PC. Since OC = cosxcosy and PC = BD
= sinxsiny, it follows that cos(x + y) =
cosxcosy − sinxsiny.
Reading Strategy
1. 30° + 45°
2. 210° + 45°
3. 135° + 30°
4. 150° + 45°
5. sin45°cos30° − cos45°sin30°
6. sin240°cos60° + cos240°sin60°
7. sin300°cos210° − cos300°sin210°
8. sin180°cos45° + cos180°sin45°
9. sin90°cos60° − cos90°sin60°
10. sin225°cos135° − cos225°sin135°
11-5 DOUBLE-ANGLE AND HALF-ANGLE
IDENTITIES
Practice A
7
25
b.
336
625
c. −
527
625
d.
24
7
e. −
336
527
5.
3 4 3
, ,
5 5 4
2. −
3.
− sin 60° ⎤
cos 60° ⎥⎦
4
0
−3
1
1.73
24
7 24
;−
;
25 25 7
5 39 7 5 39
;
;
32 32
7
4. −
⎡ −0.37
b. ⎢
⎣1.37
5. H
1. a. −
2
1
4. A
1. a.
⎡1
⎢1
⎣
−1.96 −0.73 ⎤
4.60 2.73 ⎥⎦
P''(−1.37, −0.37), Q''(−1.73, 1), R''(−1.96,
4.60), and S''(−0.73, 2.73)
Problem Solving
⎡cos 60°
⎢ sin 60°
⎣
− 1.73
35 17
35
; ;−
18 18
17
6. a. sin2θ + 2sinθ cosθ + cos2θ
b. 2sinθ cosθ + sin2θ + cos2θ
2 ⎤
−2⎥⎦
4.60
1.96
c. sin 2θ + sin2θ + cos2θ
d. sin 2θ + 1
2.73 ⎤
0.73 ⎥⎦
7.
sin 2θ
2 sinθ cosθ
= tanθ ;
= tanθ ;
2
2 cos θ
2cos2 θ
sinθ
= tanθ ; tanθ = tanθ
cos θ
c. P'(−0.37, 1.37), Q'(1, 1.73), R'(4.60,
1.96), and S'(2.73, 0.73)
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
A52
Holt McDougal Algebra 2