Aim: How do we find the cosine of the
difference and sum of two angles?
Do Now:
If mA = 35º and mB = 20º show which
of the following identities is true?
cos(A – B) = cos A – cos B
cos(35º – 20º) = cos 35º – cos 20º
cos(15º) = .8191520443 – .9396926208
.9659258263 = -.1205405765
cos(A – B) = cosAcosB +sinASinB
.9659258263 = cos 35•cos 20 + sin 35•sin20
.9659258263 = .9659258263
Aim: Cosine of Two Angles 
Course: Alg. 2 & Trig.
Outline of Proof – Cos(A – B)
y
1
P(cos B, sin B)
Q(cos A, sin A)
(x,y)
A–B
B
-1
O
A
1
x
Find the length of PQ using both the
Law of Cosines and the distance formula,
equating the two to arrive at:
The Cosine of the Difference of 2 Angles
cos (A – B) = cos A cos B + sin A sin B
Aim: Cosine of Two Angles 
Course: Alg. 2 & Trig.
Model Problem
Use the identity for the cosine of the difference
of two angle measures to prove that
cos (180º – x) = -cos x.
cos (A – B) = cos A cos B + sin A sin B
Substitute 180 for A and x for B:
cos (180 – x) = cos 180 cos x + sin 180 sin x
simplify:
cos (180 – x) = -1 • cos x + 0 • sin x
cos (180 – x) = -cos x
Aim: Cosine of Two Angles 
Course: Alg. 2 & Trig.
Model Problem
If sin A = 3/5 and A is in QII, and
cos B = 5/13 and B is in QI, find cos (A – B).
cos (A – B) = cos A cos B + sin A sin B
to use this you need to know both sine and
cosine values for both A and B.
HOW?
Pythagorean Identity
sin2 A + cos2 A = 1 sin2 B + cos2 B = 1
(3/5)2 + cos2 A = 1 sin2 B + (5/13)2 = 1
9/25 + cos2 A = 1 sin2 B + 25/169 = 1
sin2 B = 144/169
cos2 A = 16/25
sin B = 12/13
cos A = 4/5
QII cos A = -4/5
QI
Aim: Cosine of Two Angles 
Course: Alg. 2 & Trig.
Model Problem (con’t)
If sin A = 3/5 and A is in QII, and
cos B = 5/13 and B is in QI, find cos (A – B).
A: sin A = 3/5, cos A = -4/5
B: cos B = 5/13, sin B = 12/13
cos (A – B) = cos A cos B + sin A sin B
Substitute & simplify:
cos (A – B) = (-4/5)(5/13) + (3/5)(12/13)
cos (A – B) = (-20/65) + (36/65)
cos (A – B) = (16/65)
Aim: Cosine of Two Angles 
Course: Alg. 2 & Trig.
Cosine of Sum of 2 Angles
Prove:
cos (A + B) = cos A cos B – sin A sin B
cos (A + B) = cos (A - (-B))
cos (A + B) = cos A cos(-B) + sin A sin(-B)
cos (-x) = cos (x) sin (-x) = -sin x
cos (A + B) = cos A(cos B) + sin A(-sinB)
cos (A + B) = cos A cos B – sin A sinB
The Cosine of the Sum of 2 Angles:
cos (A + B) = cos A cos B – sin A sin B
ex. Show that cos 90 = 0 by using cos(60 + 30)
cos (60 + 30) = cos 60 cos 30 – sin 60 sin 30
1
3
3 1
 ( )(
)(
)( ) = 0
2 2
2 2
Aim: Cosine of Two Angles 
Course: Alg. 2 & Trig.
Model Problem
Find the exact value of
cos45ºcos15º – sin45ºsin15º
The Cosine of the Sum of 2 Angles:
cos (A + B) = cos A cos B – sin A sin B
cos(45 + 15) =cos45ºcos15º – sin45ºsin15º
cos(60) = 1/2
Find the exact value of cos 75 by using cos (45 + 30)
cos (45 + 30) = cos 45 cos 30 – sin 45 sin 30
2
3
2 1
(
)( )  (
)( )
2
2
2 2
6
2
6 2



4
4
4
Aim: Cosine of Two Angles 
Course: Alg. 2 & Trig.
More Proofs
Prove:
cos (-x) = cos x
cos (A – B) = cos A cos B + sin A sin B
cos (0 – x) = cos 0 cos (x) + sin 0 sin (x)
cos (0 – x) = 1 • cos (-x) + 0 • sin (-x)
cos (-x) = cos (x)
Prove:
sin (-x) = -sin x
the sine of an angle
equals the cosine
sin  = cos(90 – ), let  = -x
of its complement
sin (-x) = cos(90 – (-x)) =
= cos (90 + x) = cos (x + 90) = cos (x – (-90)
= cos x cos (-90) + sin x sin (-90)
= cos x • 0 + sin x • -1
= -sin x
Aim: Cosine of Two Angles 
Course: Alg. 2 & Trig.
Model Problem
Aim: Cosine of Two Angles 
Course: Alg. 2 & Trig.
Regents Question
3
24
If sin A  and cos B 
and
5
25
A and B are both in Quadrant I,
what is the exact value of cos(A  B )?
3
4
44
117
1)
2)
3)
4)
5
5
125
125
Aim: Cosine of Two Angles 
Course: Alg. 2 & Trig.
The Product Rule
Aim: Cosine of Two Angles 
Course: Alg. 2 & Trig.