The document discusses trigonometric sum and difference angle formulas. It states that trig functions are not additive, providing the example that sin(π/2 + π/2) ≠ sin(π/2) + sin(π/2). It then derives the cosine difference angle formula C(A - B) = C(A)C(B) + S(A)S(B) through a geometric proof involving rotating points on a unit circle. This formula is the basis for other sum and difference formulas listed, including the cosine sum formula C(A + B) = C(A)C(-B) + S(A)S(-B).

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

2. A function f(x) is said to be additive if
f(A ± B) = f(A) ± f(B).
Sum and Difference Angle Formulas
Double-Angle and Half-Angle Formulas

3. A function f(x) is said to be additive if
f(A ± B) = f(A) ± f(B).
Trig-functions are not additive.
Sum and Difference Angle Formulas
Double-Angle and Half-Angle Formulas

4. A function f(x) is said to be additive if
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.
Sum and Difference Angle Formulas
Double-Angle and Half-Angle Formulas

5. A function f(x) is said to be additive if
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.
Hence in general, sin(A ± B) ≠ sin(A) ± sin(B).
Sum and Difference Angle Formulas
Double-Angle and Half-Angle Formulas

6. A function f(x) is said to be additive if
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.
Hence in general, sin(A ± B) ≠ sin(A) ± sin(B).
However, we are able to express sin(A ± B)
in term of the sine and cosine of A and B.
Sum and Difference Angle Formulas
Double-Angle and Half-Angle Formulas

7. A function f(x) is said to be additive if
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.
Hence in general, sin(A ± B) ≠ sin(A) ± sin(B).
However, we are able to express sin(A ± B)
in term of the sine and cosine of A and B.
We list the formulas for sum and difference of angles,
for the double-angle, then for the half-angle below.
Sum and Difference Angle Formulas
Double-Angle and Half-Angle Formulas

8. A function f(x) is said to be additive if
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.
Hence in general, sin(A ± B) ≠ sin(A) ± sin(B).
However, we are able to express sin(A ± B)
in term of the sine and cosine of A and B.
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).
Sum and Difference Angle Formulas
Double-Angle and Half-Angle Formulas

9. 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)
Sum and Difference Angle Formulas
Double-Angle and Half-Angle Formulas

10. 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)
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.
Sum and Difference Angle Formulas
Double-Angle and Half-Angle Formulas

11. 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)

12. 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)
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)

13. 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)
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)
Rotate both angles by –B, so that
B(s, t)→(1, 0)
A B
(1, 0)

14. 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)
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)
Rotate both angles by –B, so that
B(s, t)→(1, 0)
A B
(1, 0)
A(u, v)
B(s, t)
(1, 0)

15. 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)
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)
Rotate both angles by –B, so that
B(s, t)→(1, 0)
A B
(1, 0)
A(u, v)
B(s, t)
(1, 0)
–B

16. 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)
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)
Rotate both angles by –B, so that
B(s, t)→(1, 0) and A(u, v)→(x, y).
A B
(1, 0)
A(u, v)
B(s, t)
(1, 0)
–B

17. 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)
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)
Rotate both angles by –B, so that
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)

18. 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)
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)
Rotate both angles by –B, so that
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)

19. 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)
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)
Rotate both angles by –B, so that
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)

20. Removing the root so (u – s)2 + (v – t)2 = (x – 1)2 + y2,
Cosine Sum-Difference of Angles Formulas
A(u, v)
B(s, t)
D
(1, 0)
A–B
(x, y)
D
–B
A–B

21. Removing the root so (u – s)2 + (v – t)2 = (x – 1)2 + y2,
expanding both sides of the equation we obtain:
u2 – 2us + s2 + v2 – 2vt + t2 = x2 – 2x + 1 + y2
Cosine Sum-Difference of Angles Formulas
A(u, v)
B(s, t)
D
(1, 0)
A–B
(x, y)
D
–B
A–B

22. Removing the root so (u – s)2 + (v – t)2 = (x – 1)2 + y2,
expanding both sides of the equation we obtain:
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
Cosine Sum-Difference of Angles Formulas
A(u, v)
B(s, t)
D
(1, 0)
A–B
(x, y)
D
–B
A–B

23. Removing the root so (u – s)2 + (v – t)2 = (x – 1)2 + y2,
expanding both sides of the equation we obtain:
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 or
x = us+ vt.
Cosine Sum-Difference of Angles Formulas
A(u, v)
B(s, t)
D
(1, 0)
A–B
(x, y)
D
–B
A–B

24. Removing the root so (u – s)2 + (v – t)2 = (x – 1)2 + y2,
expanding both sides of the equation we obtain:
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 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).
Cosine Sum-Difference of Angles Formulas
A(u, v)
B(s, t)
D
(1, 0)
A–B
(x, y)
D
–B
A–B

25. 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
expanding both sides of the equation we obtain:
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 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–B
Writing C(A + B) = C(A – (–B))
Cosine Sum-Difference of Angles Formulas

26. 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
expanding both sides of the equation we obtain:
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 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–B
Writing C(A + B) = C(A – (–B))
= C(A)C(–B) + S(A)S(–B)
Cosine Sum-Difference of Angles Formulas

27. 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
expanding both sides of the equation we obtain:
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 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–B
Writing C(A + B) = C(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).
Cosine Sum-Difference of Angles Formulas

28. 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
expanding both sides of the equation we obtain:
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 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–B
Writing C(A + B) = C(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)
Cosine Sum-Difference of Angles Formulas

29. 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:
Cosine Sum-Difference of Angles Formulas

30. cos(A – B) = cos(A)cos(B) + sin(A)sin(B)
cos(A + B) = cos(A)cos(B) – sin(A)sin(B)
All fractions with denominator 12 may be written as
sums or differences of fractions with denominators
3, 6 and 4.
or that C(A B) = C(A)C(B) ± S(A)S(B)–+
In summary:
Cosine Sum-Difference of Angles Formulas

31. cos(A – B) = cos(A)cos(B) + sin(A)sin(B)
cos(A + B) = cos(A)cos(B) – sin(A)sin(B)
All fractions with denominator 12 may be written as
sums or differences of fractions with denominators
3, 6 and 4. For examples:
3
π
12
π
= – 4
π
or that C(A B) = C(A)C(B) ± S(A)S(B)–+
In summary:
Cosine Sum-Difference of Angles Formulas

32. cos(A – B) = cos(A)cos(B) + sin(A)sin(B)
cos(A + B) = cos(A)cos(B) – sin(A)sin(B)
All fractions with denominator 12 may be written as
sums or differences of fractions with denominators
3, 6 and 4. For examples:
12
11π = 12
3π
+ 12
8π = 4
π
+ 3
2π
3
π
12
π
= – 4
π
or that C(A B) = C(A)C(B) ± S(A)S(B)–+
;
In summary:
Cosine Sum-Difference of Angles Formulas

33. cos(A – B) = cos(A)cos(B) + sin(A)sin(B)
cos(A + B) = cos(A)cos(B) – sin(A)sin(B)
All fractions with denominator 12 may be written as
sums or differences of fractions with denominators
3, 6 and 4. For examples:
12
11π = 12
3π
+ 12
8π = 4
π
+ 3
2π
3
π
12
π
= – 4
π
or that C(A B) = C(A)C(B) ± S(A)S(B)–+
;
Example A. Find cos(11π/12) without a calculator.
In summary:
Cosine Sum-Difference of Angles Formulas

34. cos(A – B) = cos(A)cos(B) + sin(A)sin(B)
cos(A + B) = cos(A)cos(B) – sin(A)sin(B)
All fractions with denominator 12 may be written as
sums or differences of fractions with denominators
3, 6 and 4. For examples:
12
11π = 12
3π
+ 12
8π = 4
π
+ 3
2π
3
π
12
π
= – 4
π
or that C(A B) = C(A)C(B) ± S(A)S(B)–+
;
Example A. Find cos(11π/12) without a calculator.
C ( )
12
11π = C( ) = C( )C( )
4
π +
3
2π
4
π – S( )
4
π
3
2π
In summary:
Cosine Sum-Difference of Angles Formulas
3
2π S( )
Cosine-Sum Formulas

35. cos(A – B) = cos(A)cos(B) + sin(A)sin(B)
cos(A + B) = cos(A)cos(B) – sin(A)sin(B)
All fractions with denominator 12 may be written as
sums or differences of fractions with denominators
3, 6 and 4. For examples:
12
11π = 12
3π
+ 12
8π = 4
π
+ 3
2π
3
π
12
π
= – 4
π
or that C(A B) = C(A)C(B) ± S(A)S(B)–+
;
Example A. Find cos(11π/12) without a calculator.
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:
Cosine Sum-Difference of Angles Formulas
3
2π S( )

36. cos(A – B) = cos(A)cos(B) + sin(A)sin(B)
cos(A + B) = cos(A)cos(B) – sin(A)sin(B)
All fractions with denominator 12 may be written as
sums or differences of fractions with denominators
3, 6 and 4. For examples:
12
11π = 12
3π
+ 12
8π = 4
π
+ 3
2π
3
π
12
π
= – 4
π
or that C(A B) = C(A)C(B) ± S(A)S(B)–+
;
Example A. Find cos(11π/12) without a calculator.
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:
Cosine Sum-Difference of Angles Formulas
3
2π S( )

37. 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

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

39. From C(π/2 – A) = S(A) and S(π/2 – A) = C(A), we’ve
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)
Sine Sum-Difference of Angles Formulas
the cosine difference of angles law:

40. From C(π/2 – A) = S(A) and S(π/2 – A) = C(A), we’ve
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)
= sin(A)cos(B) – cos(A)sin(B)
Sine Sum-Difference of Angles Formulas
the cosine difference of angles law:
the co-relation:

41. From C(π/2 – A) = S(A) and S(π/2 – A) = C(A), we’ve
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)
= sin(A)cos(B) – cos(A)sin(B)
In summary:
sin(A + B) = sin(A)cos(B) + cos(A)sin(B)
Sine Sum-Difference of Angles Formulas
the cosine difference of angles law:
the co-relation:

42. From C(π/2 – A) = S(A) and S(π/2 – A) = C(A), we’ve
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)
= sin(A)cos(B) – cos(A)sin(B)
In summary:
sin(A + B) = sin(A)cos(B) + cos(A)sin(B)
By expanding sin(A + (–B)) we have that
sin(A – B) = sin(A)cos(B) – cos(A)sin(B)
Sine Sum-Difference of Angles Formulas
the cosine difference of angles law:
the co-relation:

43. S(A ± B) = S(A)C(B) ± C(A)S(B)
From C(π/2 – A) = S(A) and S(π/2 – A) = C(A), we’ve
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)
= sin(A)cos(B) – cos(A)sin(B)
In summary:
sin(A + B) = sin(A)cos(B) + cos(A)sin(B)
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.
Sine Sum-Difference of Angles Formulas
the cosine difference of angles law:
the co-relation:

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

45. 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)

46. 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)

47. 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)

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

49. 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) cos2(A) – (1 – cos2(A))
sin(2A) = sin(A + A) = sin(A)cos(A) + cos(A)sin(A)
so 2cos2(A) – 1 = cos(2A)so 1 – 2sin2(A) = cos(2A)

50. 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) cos2(A) – (1 – cos2(A))
sin(2A) = sin(A + A) = sin(A)cos(A) + cos(A)sin(A)
so sin(2A) = 2sin(A)cos(A)
so 2cos2(A) – 1 = cos(2A)so 1 – 2sin2(A) = cos(2A)

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

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

53. 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
Double Angle Formulas

54. 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
10/7 = 2cos2(A)
5/7 = cos2(A)
±5/7 = cos(A)
Double Angle Formulas

55. 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
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.
Double Angle Formulas
–5
Ay 7

56. 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
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. We may use the right
triangle as shown to find tan(A).
Double Angle Formulas
–5
Ay 7

57. 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
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. We may use the right
triangle as shown to find tan(A).
y2 + (–5)2 = (7)2
Double Angle Formulas
–5
Ay 7

58. 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
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. We may use the right
triangle as shown to find tan(A). We have that
y2 + (–5)2 = (7)2 so y2 = 2
or that y = ±2  y = 2
Double Angle Formulas
–5
Ay 7

59. 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
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. We may use the right
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
Double Angle Formulas
–5
Ay 7

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

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

62. From cos(2A) = 2cos2(A) – 1, we get
1 + cos(2A)
2
cos2(A) =
In the square root form, we get
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

63. From cos(2A) = 2cos2(A) – 1, we get
1 + cos(2A)
2
cos2(A) =
In the square root form, we get
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:
1 + cos(B)
2
cos( ) = ±

B
2

64. From cos(2A) = 2cos2(A) – 1, we get
1 + cos(2A)
2
cos2(A) =
In the square root form, we get
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:
Similarly, we get the half-angle formula of sine:
1 + cos(B)
2
cos( ) = ±

B
2
1 – cos(B)
2
sin( ) = ±

B
2

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

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

67. The half-angle formulas have ± choices
but the double-angle formulas do not.
This is the case because given
different angles corresponding
to the same position such as
π and –π on the unit circle,
doubling them to 2π and –2π
would result in the same position.
Half-angle Formulas
π
–π
2π
–2π

68. The half-angle formulas have ± choices
but the double-angle formulas do not.
This is the case because given
different angles corresponding
to the same position such as
π and –π on the unit circle,
doubling them to 2π and –2π
would result in the same position.
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

69. The half-angle formulas have ± choices
but the double-angle formulas do not.
This is the case because given
different angles corresponding
to the same position such as
π and –π on the unit circle,
doubling them to 2π and –2π
would result in the same position.
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

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

71. Example D. Given A where –π < A < –π/2,
and tan(A) = 3/7, draw A and find cos(A/2).
– 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.

72. Example D. Given A where –π < A < –π/2,
and tan(A) = 3/7, draw A and find cos(A/2).
– 7
Half-angle Formulas
–3
A58
Given that tan(A) = 3/7 and A is in
the 3rd quadrant, let angle A be as
shown. Hence the hypotenuse of
the reference triangle is 58.

73. Example D. Given A where –π < A < –π/2,
and tan(A) = 3/7, draw A and find cos(A/2).
– 7
Since –π < A < –π /2, so –π/2 < A/2 < –π/4,
Half-angle Formulas
–3
A58
Given that tan(A) = 3/7 and A is in
the 3rd quadrant, let angle A be as
shown. Hence the hypotenuse of
the reference triangle is 58.

74. Example D. Given A where –π < A < –π/2,
and tan(A) = 3/7, draw A and find cos(A/2).
– 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
Given that tan(A) = 3/7 and A is in
the 3rd quadrant, let angle A be as
shown. Hence the hypotenuse of
the reference triangle is 58.

75. Example D. Given A where –π < A < –π/2,
and tan(A) = 3/7, draw A and find cos(A/2).
– 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 +.
1 + cos(A)
2
cos( ) =A
2  = 1 – 7/58
2  0.201Hence,
Half-angle Formulas
–3
A58
Given that tan(A) = 3/7 and A is in
the 3rd quadrant, let angle A be as
shown. Hence the hypotenuse of
the reference triangle is 58.

76. Example D. Given A where –π < A < –π/2,
and tan(A) = 3/7, draw A and find cos(A/2).
– 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 +.
1 + cos(A)
2
cos( ) =A
2  = 1 – 7/58
2  0.201Hence,
Half-angle Formulas
–3
A58
Given that tan(A) = 3/7 and A is in
the 3rd quadrant, let angle A be as
shown. Hence the hypotenuse of
the reference triangle is 58.
We list all these formulas and their deduction in the
following the flow chart.

77. Double-Angle Formulas
Sum-Difference of Angles Formulas
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)