Sum and Difference of Trigonometric Identities
Prove:
cos(α+β) = cosαcosβ - sinαsinβ
(1,0)
(1,0)(1,0)
(1,0)
A
CO
Unit Circle
OA = OB = OC =OD= 1
cosα= ; sin α=-β
βα
B
= cosα; = sin α
cos(α+β)= ; sin (α+β)=
= cos(α+β); = sin (α+β)
cos-β= ; sin-β =
= cos -β; = sin -β
D
(1,0)
C(1,0)(1,0)
(1,0)
O -β
βα
A(cos(α+β) ,sin(α+β))
B(cosα ,sinα)
D(cos(-β) ,sin(-β))
Distance Formula
and
Since, AC=BD then,
=
Recall
Recalland
Since, AC=BD
Recall: and
WHY?????cos(-β)=β; sin(-β)=-sinβ
Recall:
cos45=
Then,
cos(-45)=
Unit Circle
-β=-45
Β=45
-Y
X
(x,y)
(x,-y)
(x,o)
Prove:
cos(α-β) = cosαcosβ + sinαsinβ
Recall:cos(α+β) = cosαcosβ - sinαsinβ
Let β=- β
cos(α+(-β)) = cosαcos(-β) – sinαsin(-β)
Since,
cos (-β)=cosβ and sin(-β)=- sinβ
cos(α-β) = cosαcosβ + sinαsinβ
Prove:
sin(α+β) = cosαsinβ + sinαcosβ
β
α
α
L
S
R
P
O M
Since, LM=LS+SM
Since, SM=RP
sin(α+β) = cosαsinβ + sinαcosβ
Prove:
sin(α-β) = cosαsinβ - sinαcosβ
Recall:sin(α+β) = cosαsinβ + sinαcosβ
Let β=- β
sin(α+(-β)) = cosαsin(-β) – sinαcos(-β)
Since,
cos (-β)=cosβ and sin(-β)=- sinβ
sin(α-β) = cosαsinβ - sinαcosβ
Prove:
Recall:
Recall: sin(α+β) = cosαsinβ + sinαcosβ
cos(α+β) = cosαcosβ - sinαsinβ
Recall: and
Prove:
Recall:
Recall: sin(α-β) = cosαsinβ - sinαcosβ
cos(α-β) = cosαcosβ + sinαsinβ
andRecall:
THE END …