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:

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