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A Geometric Proof

A Geometric Proof. for Sin(A+B)=sinAcosB+sinBcosA. Take a general triangle, as shown below …. A. B. a. b. h. c1. c2. c. Now consider the area of the triangle as a whole and as a compound area. Area of red triangle = ½ c1 x h. Area of blue triangle = ½ c2 x h. A. B. a. b. h. c1.

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A Geometric Proof

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

  2. Take a general triangle, as shown below … A B a b h c1 c2 c

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

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

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

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