Prove geometrically that sin(A+B)=SinA.CosB+CosA.SinB Share with your friends Share 1 Lovina Kansal answered this Dear student Taking alpha and beta instead of A and B. Draw a horizontal line the x-axis;mark an origin O.Draw a line from O atan angle α above the horizontal line and a second line at an angle β that;the anglebetween the second line and the x-axis is α+β.Place P on the line defined by α+β at a unit distance from the origin.Let PQ be a line prependicular to line defined by angle α,drawn from pont Q on thisline to point P.∴OQP is a right angle.Let QA be a prependicular from point A on the x-axis to Q and PB be a prependicular from point B on the x-axis to P.∴OAQ and OBP are right angles.Draw R on PB so that QR is parallel to the x-axis.Now ∠RPQ=αbecause OQA=90-α,making RQO=α,RQP=90-α and RPQ=αRPQ=π2-RQP=π2-π2-RQO=RQO=αOP=1PQ=sinβOQ=cosβAQOQ=sinα, so AQ=sinα cosβand PRPQ=cosα, so PR=cosα sinβThus,sin(α+β)=PB=RB+PR=AQ+PR=sinα cosβ+cosα sinβ Regards 4 View Full Answer
