To prove Sin (a + b) = Sin a Cos b + Cos a Sin b

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From the diagram, Sin (a + b) = DF/AD = EF + DE/AD
Therefore it is necesary to find DE, EF & AD in terms of 'a' and 'b'
Because AB and CE are two parallel lines, angle ACE is 'a', DCE is 90 - 'a' therefore CDE is 'a'.
In triangle CDE, Cos a = DE/CD therefore DE = CD Cos a --- (1)
In triangle ACD, Sin b = CD/AD therefore CD = AD Sin b --- (2)
Substituting (2) into (1) we get:- DE = AD Cos a Sin b --- (3)
In triangle ABC, Sin a = BC/AC therefore BC = AC Sin a --- (4)
In triangle ACD, Cos b = AC/AD therefore AC = AD Cos b --- (5)
Substituting (5) into (4) we get:- BC = AD Sin a Cos b --- (6)
From the diagram, Sin (a + b) = EF + DE/AD and EF = BC, therefore Sin (a + b) = BC + DE/AD --- (7)
Substituting (3) and (6) into (7) we get:-
Sin (a + b) = AD Sin a Cos b + AD Cos a Sin b/AD
and cancelling out terms common to top and botton of the equation
--- Sin (a + b)= Sin a Cos b + Cos a Sin b --- QED

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Sin (a + b)