Maths - RoderickT - Sine of addition of 2 angles part 2

Sine of addition of 2 angles - part 2 obtuse angle

Construct angles AOB (α) and BOC (β)

Let P be any point on OC

From P construct PM perp. to OA and M lying on OA

From P construct PE perp. to OB and E lying on OB

From E construct EK perp. to PM and K lying on PM

From E construct EF perp. to OA and F lying on OA

^OMK = ^EKP = 1 rt. angle (construct)

So EK is parallel to FM (comp. angles equal)

^EFO = ^OMK = 1 rt. angle (construct)

So PM is parallel to EF (comp. angles equal)

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

Triangle OPM has ^OMP = 1 right angle (construct)

sin(α+β) = sin(180-(α+β)) (Sine in 2nd quadrant is positive)

So Sin (α+β)= PM/OP .....(1)

KM = EF (Opp. sides of rectangle)

So PM = PK + KM = PK + EF

So Sin(α+β) = (PK + EF) / OP .....(2)

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

^PEB = ^OEP = 1 right angle (construct)

^OEK = ^FOE = α (alt. angles EK parallel FM)

^KEP = (1 rt. ang.) - α (^PEO = 1rt. ang. by construct)

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

PK = PEcosα

EF = OEsinα

PE = OPsinβ

so

PK = OPcosαsinβ .....(3)

OE = OPcosβ

so

EF = OPcosβsinα .....(4)

Substitute (3) and (4) in (2)

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

divide top and bottom by OP, rearrange so

sin(α+β) = sinαcosβ + cosαsinβ

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

Sin(α+β) when 90°<(α+β)<180°

Acute angle