A Geometric Proof for Sin(A+B)=sinAcosB+sinBcosA.

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Presentation transcript:

A Geometric Proof for Sin(A+B)=sinAcosB+sinBcosA

AB a c1c2 b c h Take a general triangle, as shown below …

AB a c1c2 b c h Area of red triangle = ½ c1 x h Area of blue triangle = ½ c2 x h Area of whole triangle = ½ (c1+c2) x h = ½ ab sin(A+B) Now consider the area of the triangle as a whole and as a compound area

AB a c1c2 b c h Area of whole triangle = ½ (c1+c2) x h = ½ ab sin(A+B) But h = a cos(A) = b cos(B) with c1 = a sin(A), c2 = b sin(B)

AB a c1c2 b c h Therefore : Area of whole triangle = ½ (c1+c2) x h = ½ ab sin(A+B) = ½ (asinA +bsinB)h = ½ ab sin(A+B) Substituting values of h gives : absinAcosB + absinBcosA = ab sin(A+B) So finally : sinAcosB + sinBcosA = sin(A+B)