Question

geometrically prove that cos(A-B) = cosA.cosB + sinA.sinB

Answer 1

Answer:

Let us draw a horizontal line (the x-axis); mark an origin O. Draw a line from O at an angle αα above the horizontal line and a second line at an angle ββ above that; the angle between the second line and the x-axis is α+βα+β

Then, we place P on the line defined by α+βα+β  at a unit distance from the origin.

Let PQ be a line perpendicular to line defined by angle αα , drawn from point Q on this line to point P

.thereforeOQP is a right angle.

Let QA be a perpendicular from point A on the x-axis to Q and PB be a perpendicular from point B on the x-axis to P.thereforeOAQ and OBP are right angles.

We then draw QR parallel to the x-axis.

Now,OP =1PQ=sinβOQ=cosβOAOQ=cosα ⇒OA =cosαcosβRQPQ=sinα ⇒ RQ=sinαsinβcos(α+β)=OB = OA−BA=OA−RQ=cosαcosβ−sinαsinβBy substituting −β forβ and using symmetry,we getcos(α−β)=cosαcos(−β)−sinαsin(−β)cos(α−β)=cosαcosβ+sinαsinβNow,OP =1PQ=sinβOQ=cosβOAOQ=cosα ⇒OA =cosαcosβRQPQ=sinα ⇒ RQ=sinαsinβcos(α+β)=OB = OA-BA=OA-RQ=cosαcosβ-sinαsinβBy substituting -β forβ and using symmetry,we getcos(α-β)=cosαcos(-β)-sinαsin(-β)cos(α-β)=cosαcosβ+sinαsinβ