Maths - RoderickT - Cosine of sum of 2 angles obtuse

Cosine of sum of 2 angles obtuse

Construct angles AOB (α) and BOC (β)

Let P be any point on OC

From P construct PM perpendicular to OA and M lying on OA

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

From E construct EK perpendicular to PM and K lying on PM

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

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

So EK is parallel to FM (complementary angles equal)

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

So PM is parallel to EF (complementary 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)

cos(α+β) = -cos(180-(α+β)) (Cosine in 2nd quadrant is negative)

So cos(α+β) = -OM/OP .....(1)

MF = KE (Opp. sides of rectangle)

OM = MF - OF = KE - OF

So cos(α+β) = -(KE - OF) / OP .....(2)

Triangle PKE has ^PKE = 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)

KE = PEsinα

PE = OPsinβ

So

KE = OPsinβsinα .....(3)

OF = OEcosα

OE = OPcosβ

so

OF = OPcosβcosα .....(4)

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

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

divide top and bottom by OP, rearrange so

cos(α+β) = (cosαcosβ - sinαsinβ)

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

Acute angle