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

From the diagram, Sin (a + b) = DF/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) =

Substituting (3) and (6) into (7) we get:-

Sin (a + b) =

and cancelling out terms common to top and botton of the equation

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