Cosine of sum of 2 angles

Prove cos(α+β) = cosαcosβ - sinαsinβ

Construct angles AOB (α) and BOC (β)

Let P be any point on OC

From P construct PG perpendicular to OA and G lying on OA

From P construct PE perpendicular to OB and E lying on OB

From E construct EH perpendicular to PG and H lying on PG

From E construct EF perpendicular to OA and F lying on OA

^OGH = ^GFE = 1 rt. angle (construct)

So GH is parallel to FE (complementary angles equal)

^FGH = ^EHP = 1 rt. angle (construct)

So HE is parallel to GF (complementary angles equal)

So HGFE is a rectangle (Opp. sides paral. and all 4 int. angles = 1rt. angle)

Triangle OPG has ^OGP = 1 right angle (construct)

So cos (α+β)= OG/OP .....(1)

HE = GF (Opp. sides of rectangle)

So OG = OF - GF = OF - HE

So cos(α+β) = (OF - HE) / OP .....(2)

Triangle PHE has ^PHE = 1 right angle (construct)

^ODG = (1 rt. ang.) - α (angles of triangle = 2 rt. ang.s)

^HED = ^GOD = α (alternate ang.s FO parallel to HE)

^PEH = 1 rt. ang. - α (^PED = 1 rt. ang. by construct)

^HPE = α (angles of triangle = 2 rt. ang.s)

HE = PEsinα

PE = OPsinβ

OF = OEcosα

OE = OPcosβ

So OG = OEcosα - PEsinα

So OG = OPcosβcosα - OPsinβsin&alpha .....(3)

Rearrange (3) and substitute into (2) gives

cos(α+β) = (OPcoαcosβ - OPsinαsinβ)/OP

Canceling out OP leaves

cos(α+β) = cosαcosβ - sinαsinβ

To see similar derivation when α + β is obtuse click on following button.

Obtuse angle