13. sum and double half-angle formulas-x

Sum and Difference Angle Formulas
Double-Angle and Half-Angle Formulas

A function f(x) is said to be additive if
f(A ± B) = f(A) ± f(B).

f(A ± B) = f(A) ± f(B).
Trig-functions are not additive.

f(A ± B) = f(A) ± f(B).
Trig-functions are not additive. For example,
sin(π/2 + π/2) = sin(π) = 0 and
sin(π/2 + π/2) ≠ sin(π/2) + sin(π/2) = 1 + 1 = 2.

f(A ± B) = f(A) ± f(B).
sin(π/2 + π/2) = sin(π) = 0 and
sin(π/2 + π/2) ≠ sin(π/2) + sin(π/2) = 1 + 1 = 2.
Hence in general, sin(A ± B) ≠ sin(A) ± sin(B).

f(A ± B) = f(A) ± f(B).
sin(π/2 + π/2) = sin(π) = 0 and
sin(π/2 + π/2) ≠ sin(π/2) + sin(π/2) = 1 + 1 = 2.
However, we are able to express sin(A ± B)
in term of the sine and cosine of A and B.

f(A ± B) = f(A) ± f(B).
sin(π/2 + π/2) = sin(π) = 0 and
sin(π/2 + π/2) ≠ sin(π/2) + sin(π/2) = 1 + 1 = 2.
We list the formulas for sum and difference of angles,
for the double-angle, then for the half-angle below.

f(A ± B) = f(A) ± f(B).
sin(π/2 + π/2) = sin(π) = 0 and
sin(π/2 + π/2) ≠ sin(π/2) + sin(π/2) = 1 + 1 = 2.
We list the formulas for sum and difference of angles,
for the double-angle, then for the half-angle below.
We simplify the notation by notating respectively
S(A), C(A) and T(A) for sin(A), cos(A) and tan(A).

Double-Angle Formulas
Sum-Difference of Angles Formulas
S(2A) = 2S(A)C(A)
C(2A) = C2(A) – S2(A)
= 2C2(A) – 1
= 1 – 2S2(A)
1 + C(B)
2
C( ) =
Half-Angle Formulas
±

1 – C(B)
2
S( ) = ±

B
2
B
2
C(A±B) = C(A)C(B) S(A)S(B)–+
S(A±B) = S(A)C(B) ± S(B)C(A)

S(2A) = 2S(A)C(A)
C(2A) = C2(A) – S2(A)
= 2C2(A) – 1
= 1 – 2S2(A)
1 + C(B)
2
C( ) =
Half-Angle Formulas
±

1 – C(B)
2
S( ) = ±

B
2
B
2
C(A±B) = C(A)C(B) S(A)S(B)–+
S(A±B) = S(A)C(B) ± S(B)C(A)
The cosine difference of angles formula
C(A – B) = C(A)C(B) + S(A)S(B)
is the basis for all the other formulas listed above.

Cosine Sum-Difference of Angles Formulas
Given two angles A and B, then
The Cosine Difference-angle Formula
C(A – B) = C(A)C(B) + S(A)S(B)

C(A – B) = C(A)C(B) + S(A)S(B)
Proof:
Let A and B be two angles
with corresponding coordinates
(u, v) and (s, t) on the unit circle as shown.
A(u, v)
B(s, t)
A B
(1, 0)

C(A – B) = C(A)C(B) + S(A)S(B)
Proof:
A(u, v)
B(s, t)
Rotate both angles by –B, so that
B(s, t)→(1, 0)
A B
(1, 0)

C(A – B) = C(A)C(B) + S(A)S(B)
Proof:
A(u, v)
B(s, t)
B(s, t)→(1, 0)
A B
(1, 0)
A(u, v)
B(s, t)
(1, 0)

C(A – B) = C(A)C(B) + S(A)S(B)
Proof:
A(u, v)
B(s, t)
B(s, t)→(1, 0)
A B
(1, 0)
A(u, v)
B(s, t)
(1, 0)
–B

C(A – B) = C(A)C(B) + S(A)S(B)
Proof:
A(u, v)
B(s, t)
B(s, t)→(1, 0) and A(u, v)→(x, y).
A B
(1, 0)
A(u, v)
B(s, t)
(1, 0)
–B

C(A – B) = C(A)C(B) + S(A)S(B)
Proof:
A(u, v)
B(s, t)
B(s, t)→(1, 0) and A(u, v)→(x, y).
A(u, v)
B(s, t)
(1, 0)
(x, y)
–B
A B
–B
A–B
(1, 0)

C(A – B) = C(A)C(B) + S(A)S(B)
Proof:
A(u, v)
B(s, t)
B(s, t)→(1, 0) and A(u, v)→(x, y).
We want to find C(A – B), i.e. x.
A(u, v)
B(s, t)
(1, 0)
(x, y)
–B
A B
–B
A–B
(1, 0)

C(A – B) = C(A)C(B) + S(A)S(B)
Proof:
A(u, v)
B(s, t)
B(s, t)→(1, 0) and A(u, v)→(x, y).
We want to find C(A – B), i.e. x.
Equating the distance D which is
unchanged under the rotation, we´ve
√(u – s)2 + (v – t)2 = √(x – 1)2 + (y – 0)2 .
A(u, v)
B(s, t)
D
(1, 0)
(x, y)
D
–B
A B
–B
A–B
(1, 0)

Removing the root so (u – s)2 + (v – t)2 = (x – 1)2 + y2,
A(u, v)
B(s, t)
D
(1, 0)
A–B
(x, y)
D
–B
A–B

expanding both sides of the equation we obtain:
u2 – 2us + s2 + v2 – 2vt + t2 = x2 – 2x + 1 + y2
A(u, v)
B(s, t)
D
(1, 0)
A–B
(x, y)
D
–B
A–B

u2 – 2us + s2 + v2 – 2vt + t2 = x2 – 2x + 1 + y2
Setting u2 + v2 = 1, s2 + t2 = 1 and x2 + y2 = 1
we have 2 – 2us – 2vt = 2 – 2x
A(u, v)
B(s, t)
D
(1, 0)
A–B
(x, y)
D
–B
A–B

u2 – 2us + s2 + v2 – 2vt + t2 = x2 – 2x + 1 + y2
we have 2 – 2us – 2vt = 2 – 2x or
x = us+ vt.
A(u, v)
B(s, t)
D
(1, 0)
A–B
(x, y)
D
–B
A–B

u2 – 2us + s2 + v2 – 2vt + t2 = x2 – 2x + 1 + y2
we have 2 – 2us – 2vt = 2 – 2x or
x = us+ vt. Writing each coordinate as a trig-value,
x = C(A – B), u = C(A), s = C(B), v = S(A), t = S(B)
we obtain the cosine difference angle formula
C(A – B) = C(A)C(B) + S(A)S(B).
A(u, v)
B(s, t)
D
(1, 0)
A–B
(x, y)
D
–B
A–B

A(u, v)
B(s, t)
D
(1, 0)
A–B
(x, y)
D
–B
u2 – 2us + s2 + v2 – 2vt + t2 = x2 – 2x + 1 + y2
we have 2 – 2us – 2vt = 2 – 2x or
C(A – B) = C(A)C(B) + S(A)S(B).
A–B
Writing C(A + B) = C(A – (–B))

A(u, v)
B(s, t)
D
(1, 0)
A–B
(x, y)
D
–B
u2 – 2us + s2 + v2 – 2vt + t2 = x2 – 2x + 1 + y2
we have 2 – 2us – 2vt = 2 – 2x or
C(A – B) = C(A)C(B) + S(A)S(B).
A–B
= C(A)C(–B) + S(A)S(–B)

A(u, v)
B(s, t)
D
(1, 0)
A–B
(x, y)
D
–B
u2 – 2us + s2 + v2 – 2vt + t2 = x2 – 2x + 1 + y2
we have 2 – 2us – 2vt = 2 – 2x or
C(A – B) = C(A)C(B) + S(A)S(B).
A–B
= C(A)C(–B) + S(A)S(–B)
= C(A)C(B) – S(A)S(B) since
C(–B) = C(B) and that S(–B) = – S(B).

A(u, v)
B(s, t)
D
(1, 0)
A–B
(x, y)
D
–B
u2 – 2us + s2 + v2 – 2vt + t2 = x2 – 2x + 1 + y2
we have 2 – 2us – 2vt = 2 – 2x or
C(A – B) = C(A)C(B) + S(A)S(B).
A–B
= C(A)C(–B) + S(A)S(–B)
= C(A)C(B) – S(A)S(B) since
C(–B) = C(B) and that S(–B) = – S(B).
Hence C(A + B) = C(A)C(B) – S(A)S(B)

cos(A – B) = cos(A)cos(B) + sin(A)sin(B)
cos(A + B) = cos(A)cos(B) – sin(A)sin(B)
or that C(A B) = C(A)C(B) ± S(A)S(B)–+
In summary:

All fractions with denominator 12 may be written as
sums or differences of fractions with denominators
3, 6 and 4.
In summary:

3, 6 and 4. For examples:
3
π
12
π
= – 4
π
In summary:

12
11π = 12
3π
+ 12
8π = 4
π
+ 3
2π
3
π
12
π
= – 4
π
;
In summary:

12
11π = 12
3π
+ 12
8π = 4
π
+ 3
2π
3
π
12
π
= – 4
π
;
Example A. Find cos(11π/12) without a calculator.
In summary:

12
11π = 12
3π
+ 12
8π = 4
π
+ 3
2π
3
π
12
π
= – 4
π
;
C ( )
12
11π = C( ) = C( )C( )
4
π +
3
2π
4
π – S( )
4
π
3
2π
In summary:
3
2π S( )
Cosine-Sum Formulas

12
11π = 12
3π
+ 12
8π = 4
π
+ 3
2π
3
π
12
π
= – 4
π
;
C ( )
12
11π = C( ) = C( )C( )
4
π +
3
2π
4
π – S( )
4
π
3
2π
Cosine-Sum Formulas
= 2
2
(–1)
2 –2
2
3
2 =
In summary:
3
2π S( )

12
11π = 12
3π
+ 12
8π = 4
π
+ 3
2π
3
π
12
π
= – 4
π
;
C ( )
12
11π = C( ) = C( )C( )
4
π +
3
2π
4
π – S( )
4
π
3
2π
Cosine-Sum Formulas
= 2
2
(–1)
2 –2
2
3
2 =–2 – 6
4
 –0.966
In summary:
3
2π S( )

From C(π/2 – A) = S(A) and S(π/2 – A) = C(A), we’ve
sin(A + B) = cos(π/2 – (A+B))
Sine Sum-Difference of Angles Formulas

sin(A + B) = cos(π/2 – (A+B)) = cos((π/2 – A) – B)

= cos(π/2 – A)cos(B)) + sin(π/2 – A)sin(B)
the cosine difference of angles law:

= sin(A)cos(B) – cos(A)sin(B)
the co-relation:

In summary:
sin(A + B) = sin(A)cos(B) + cos(A)sin(B)
the co-relation:

In summary:
By expanding sin(A + (–B)) we have that
sin(A – B) = sin(A)cos(B) – cos(A)sin(B)
the co-relation:

S(A ± B) = S(A)C(B) ± C(A)S(B)
In summary:
By expanding sin(A + (–B)) we have that
sin(A – B) = sin(A)cos(B) – cos(A)sin(B) or that
UDo. Find sin(–π/12) without a calculator.
the co-relation:

From the sum-of-angle formulas, we obtain the
double-angle formulas by setting A = B,
Double Angle Formulas

cos(2A) = cos(A + A) = cos(A)cos(A) – sin(A)sin(A)

cos(2A) = cos2(A) – sin2(A)

(1 – sin2(A)) – sin2(A)
so 1 – 2sin2(A) = cos(2A)

(1 – sin2(A)) – sin2(A) cos2(A) – (1 – cos2(A))
so 2cos2(A) – 1 = cos(2A)so 1 – 2sin2(A) = cos(2A)

sin(2A) = sin(A + A) = sin(A)cos(A) + cos(A)sin(A)

so sin(2A) = 2sin(A)cos(A)

so sin(2A) = 2sin(A)cos(A)
= 1 – 2sin2(A)
= 2cos2(A) – 1
Cosine Double Angle Formulas:
Sine Double Angle Formulas:
sin(2A) = 2sin(A)cos(A)

Example B. Given the angle A in the 2nd quadrant and
that cos(2A) = 3/7, find tan(A). Draw.

Using the formula cos(2A) = 2cos2(A) – 1, we get
3/7 = 2cos2(A) – 1

3/7 = 2cos2(A) – 1
10/7 = 2cos2(A)
5/7 = cos2(A)
±5/7 = cos(A)

3/7 = 2cos2(A) – 1
10/7 = 2cos2(A)
5/7 = cos2(A)
±5/7 = cos(A)
Since A is in the 2nd quadrant, cosine must be negative,
so cos(A) = –5/7 = (–5)/7.
–5
Ay 7

3/7 = 2cos2(A) – 1
10/7 = 2cos2(A)
5/7 = cos2(A)
±5/7 = cos(A)
so cos(A) = –5/7 = (–5)/7. We may use the right
triangle as shown to find tan(A).
–5
Ay 7

3/7 = 2cos2(A) – 1
10/7 = 2cos2(A)
5/7 = cos2(A)
±5/7 = cos(A)
triangle as shown to find tan(A).
y2 + (–5)2 = (7)2
–5
Ay 7

3/7 = 2cos2(A) – 1
10/7 = 2cos2(A)
5/7 = cos2(A)
±5/7 = cos(A)
triangle as shown to find tan(A). We have that
y2 + (–5)2 = (7)2 so y2 = 2
or that y = ±2  y = 2
–5
Ay 7

3/7 = 2cos2(A) – 1
10/7 = 2cos2(A)
5/7 = cos2(A)
±5/7 = cos(A)
triangle as shown to find tan(A). We have that
y2 + (–5)2 = (7)2 so y2 = 2
or that y = ±2  y = 2
Therefore tan(A) = 2
5
–  –0.632
–5
Ay 7

From cos(2A) = 2cos2(A) – 1, we get
1 + cos(2A)
2
cos2(A) =
Half-angle Formulas

1 + cos(2A)
2
cos2(A) =
In the square root form, we get
Half-angle Formulas
1 + cos(2A)
2
cos(A) =±


1 + cos(2A)
2
cos2(A) =
Half-angle Formulas
1 + cos(2A)
2
cos(A) =±

If we replace A by B/2 so that 2A = B,
we have the half-angle formula of cosine:
cos( ) =B
2

1 + cos(2A)
2
cos2(A) =
Half-angle Formulas
1 + cos(2A)
2
cos(A) =±

1 + cos(B)
2
cos( ) = ±

B
2

1 + cos(2A)
2
cos2(A) =
Half-angle Formulas
1 + cos(2A)
2
cos(A) =±

Similarly, we get the half-angle formula of sine:
1 + cos(B)
2
cos( ) = ±

B
2
1 – cos(B)
2
sin( ) = ±

B
2

The half-angle formulas have ± choices
but the double-angle formulas do not.
Half-angle Formulas

This is the case because given
different angles corresponding
to the same position such as
π and –π on the unit circle,
Half-angle Formulas
π
–π

doubling them to 2π and –2π
would result in the same position.
Half-angle Formulas
π
–π
2π
–2π

However if we divide them by 2 as
π/2 and –π/2, the new angles would result in different
positions.
Half-angle Formulas
π
–π
2π
–2π
π/2
–π/2

However if we divide them by 2 as
π/2 and –π/2, the new angles would result in different
positions.
Hence the half-angle formulas require more
information to find the precise answer.
Half-angle Formulas
π
–π
2π
–2π
π/2
–π/2

Example C. Given A where –π < A < –π/2,
and tan(A) = 3/7, draw A and find cos(A/2).
Half-angle Formulas

– 7
Half-angle Formulas
–3
A
Given that tan(A) = 3/7 and A is in
the 3rd quadrant, let angle A be as
shown.

– 7
Half-angle Formulas
–3
A58
shown. Hence the hypotenuse of
the reference triangle is 58.

– 7
Since –π < A < –π /2, so –π/2 < A/2 < –π/4,
Half-angle Formulas
–3
A58

– 7
Since –π < A < –π /2, so –π/2 < A/2 < –π/4, so that
A/2 is in the 4th quadrant and cos(A/2) is +.
Half-angle Formulas
–3
A58

– 7
1 + cos(A)
2
cos( ) =A
2  = 1 – 7/58
2  0.201Hence,
Half-angle Formulas
–3
A58

– 7
1 + cos(A)
2
cos( ) =A
2  = 1 – 7/58
2  0.201Hence,
Half-angle Formulas
–3
A58
We list all these formulas and their deduction in the
following the flow chart.

C(2A) = C2(A) – S2(A)
= 2C2(A) – 1
= 1 – 2S2(A)
S(2A) = 2S(A)C(A)
1 + C(B)
2
C( ) =
Half-Angle Formulas
±
1 – C(B)
2
S( ) = ±

B
2
B
2
C(A – B) = C(A)C(B) + S(A)S(B)
S(A±B) = S(A)C(B) ± S(B)C(A)
Important Trig-relation Formulas
1 + cos(2A)
2
cos(A) =
1 – cos(2A)
2
sin(A) =


±
±
C(A + B) = C(A)C(B) – S(A)S(B)
C(–B) = C(B)
S(–B) = –S(B)

Exercise. A. Find the exact answers.
1. cos(π/12)
9. sin(7π/12)
11. sin(–7π/12)
7. cos(7π/12)
3. cos(–11π/12)
5. cos(–15o)
13. cos(–5π/12)
2. cos(13π/12)
4. cos(105o) 6. sin(π/12)
8. sin(5π/12)
10. sin(–5π/12) 12. cos(5π/12)
14. sin(–5π/12) 15. cos(π/8)
16. sin(π/8) 17. sin(–3π/8) 18. cos(5π/8)
Exercise. B. Find cos(A + B) and cos(A – B).
5. sin(A) = 3/5 and sin(B) = –12/13, A in the 1st quadrant and B in 3rd quad.
3. cos(A) = –1/5 and sin(B) = 3/5, A and B in the 2nd quadrant
1. sin(A) = ½ and cos(B) = ½ , A and B in the 1st quadrant
2. sin(A) = 1/5 and cos(B) = 2/5, A in the 1st quadrant and B in 4th quad.
4. cos(A) = –√3/5 and sin(B) = 2/3, A and B in the 2nd quadrant

Exercise. C. Find sin(A + B)
5. sin(A) = 2/3 and sin(B) = –1/3, A in the 2nd quadrant and B in 4th quad.
3. cos(A) = –8/17 and cos(B) = –3/5, A and B in the 2nd quadrant
1. cos(A) = 3/5 and sin(B) = 5/13 , A and B in the 1st quadrant
2. sin(A) = 1/5 and cos(B) = 2/5, A in the 1st quadrant and B in 4th quad.
4. cos(A) = –√3/5 and sin(B) = 2/3, A and B in the 2nd quadrant
Exercise. D. Half angle formulas
1. Find cos(A/2) given cos(A) = 1/4 with 0 < A < π.
3. Find sin(A/2) given tan(A) = 2 with 0 < A < π.
2. Find sin(A/2) given cos(A) = 1/3 with –π/2 < A < 0.
4. Find cos(A/2) given cot(A) = 4 with –3π/2 < A < –π.

Answers A.
2 + 6
4
1.
9.
− 2 − 6
4
3.
2 + 6
4
− 2 − 6
4
11.2 − 6
4
7.
5.
2 + 6
4
13. 6 − 2
4
15. 2 + 2
2
− 2 + 2
2
17.
B. 16/65 , –56/651. 0, √3/2 3. 5.
4−6 6
25
,
4 + 6 6
25
C. 1. 63/65 3. 5.77/85 4 2 + 5
9
D. 1. 3.
10
4
50 − 10 5
10

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