Scalene Simplifying

Geometry Level 3

If A A , B B , and C C are the angles of a scalene triangle, simplify

sin ⁡ A cos ⁡ B cos ⁡ C + sin ⁡ B cos ⁡ C cos ⁡ A + sin ⁡ C cos ⁡ A cos ⁡ B \sin A \cos B \cos C + \sin B \cos C \cos A + \sin C \cos A \cos B

3 sin ⁡ A cos ⁡ 2 B 3 \sin A \cos^2 B sin ⁡ A sin ⁡ B sin ⁡ C \sin A \sin B \sin C 3 sin ⁡ A − 3 sin ⁡ A sin ⁡ 2 B 3 \sin A - 3 \sin A \sin^2 B 3 3 cos ⁡ A cos ⁡ B cos ⁡ C 3 \sqrt{3} \cos A \cos B \cos C

David Vreken by David Vreken , 42, USA

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2 solutions

David Vreken
Jul 17, 2019

First,

sin ⁡ C \sin C

= sin ⁡ ( 180 ° − ( A + B ) ) = \sin(180° - (A + B))

= sin ⁡ ( A + B ) = \sin(A + B)

= sin ⁡ A cos ⁡ B + sin ⁡ B cos ⁡ A = \sin A \cos B + \sin B \cos A

Second,

cos ⁡ C \cos C

= cos ⁡ ( 180 ° − ( A + B ) ) = \cos(180° - (A + B))

= − cos ⁡ ( A + B ) = -\cos(A + B)

= − cos ⁡ A cos ⁡ B + sin ⁡ A sin ⁡ B = -\cos A \cos B + \sin A \sin B .

Therefore,

sin ⁡ A cos ⁡ B cos ⁡ C + sin ⁡ B cos ⁡ C cos ⁡ A + sin ⁡ C cos ⁡ A cos ⁡ B \sin A \cos B \cos C + \sin B \cos C \cos A + \sin C \cos A \cos B

= cos ⁡ C ( sin ⁡ A cos ⁡ B + sin ⁡ B cos ⁡ A ) + sin ⁡ C cos ⁡ A cos ⁡ B = \cos C (\sin A \cos B + \sin B \cos A) + \sin C \cos A \cos B

= cos ⁡ C ( sin ⁡ C ) + sin ⁡ C cos ⁡ A cos ⁡ B = \cos C (\sin C) + \sin C \cos A \cos B

= sin ⁡ C ( cos ⁡ C + cos ⁡ A cos ⁡ B ) = \sin C (\cos C + \cos A \cos B)

= sin ⁡ C ( ( − cos ⁡ A cos ⁡ B + sin ⁡ A sin ⁡ B ) + cos ⁡ A cos ⁡ B ) = \sin C ((-\cos A \cos B + \sin A \sin B) + \cos A \cos B)

= sin ⁡ C ( sin ⁡ A sin ⁡ B ) = \sin C (\sin A \sin B)

= sin ⁡ A sin ⁡ B sin ⁡ C = \boxed{\sin A \sin B \sin C}

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The given expression equals SinAcosBcosC+CosA(SinBCosC+CosBSinC)=SinA(CosBCosC+CosA)=SinASinBSinC

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