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

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

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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

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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)

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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)

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© T Madas. The Cosine Rule © T Madas A B C a b c a2a2 = b2b2 + c2c2 – 2 b ccosA b2b2 = a2a2 + c2c2 – 2 a ccosB c2c2 = a2a2 + b2b2 – 2 a bcosC The cosine.

© T Madas. The Cosine Rule © T Madas A B C a b c a2a2 = b2b2 + c2c2 – 2 b ccosA b2b2 = a2a2 + c2c2 – 2 a ccosB c2c2 = a2a2 + b2b2 – 2 a bcosC The cosine.

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