Prove sin(α+β) = sinαcosβ + cosαsinβ

1. α+β < 90°

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 parallel and all 4 internal ang.s = 1rt. ang.)

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

So sin (α+β)= PG/OP                                      .....(1)

HG = EF                                                  (Opp. sides of rectangle)

So PG = PH + HG = PH + EF

So sin(α+β) = (PH + EF)/OP                               .....(2)

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

^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)

EF = OEsinα

OE = OPcosβ

So EF = OPsinαcosβ                                      .....(3)

PH = PEcosα

PE = OPsinβ

So PH = OPcosαsinβ                                      .....(4)

Substitute (3) and (4) into (2) gives

sin(α+β) = (OPsinαcosβ + OPcosαsinβ)/OP

Canceling out OP leaves

sin(α+β) = sinαcosβ + cosαsinβ

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