If A,B,C are three internal angles of a triangle, show that

\sin^3 A \cos (B – C) + \sin^3 B \cos (C – A) + \sin^3 C \cos (A – B) = 3 \sin A \sin B \sin C

Collected in the board: Trigonometry

Steven Zheng posted 4 hours ago


Since A,B,C are three internal angles of a triangle

A = \pi-(B+C)

\sin A = \sin(B+C)

Apply the product to sum identity

\sin(B+C)\cos (B – C) = \dfrac{1}{2} [\sin(B+C+B-C ) +\sin(B+C - B-C) ]

Then we get

\sin(B+C)\cos (B – C) = \dfrac{1}{2} (\sin 2B+\sin 2C)


\sin A \cos (B – C) = \dfrac{1}{2} (\sin 2B+\sin 2C)

Apply the double angle identity for sines function. Then,

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

So the first term of the expression on the left hand side is

\sin^3 A \cos (B – C) = \sin^2 A(\sin B\cos B+\sin C\cos C)

Similarly, we can deduce similar equations for the second and third terms.

\sin^3 B \cos (C – A) = \sin^2 B(\sin C\cos C+\sin A\cos A)
\sin^3 C \cos (A – B) = \sin^2 C(\sin A\cos A+\sin B\cos B)

Now apply the Laws of Sines, and let k be the ratio of sines to sides

\dfrac{\sin A}{a}= \dfrac{\sin B}{b} = \dfrac{\sin C}{c} =k

Then we can convert sines function to the multiples of sides.

\sin A = ka
\sin B = kb
\sin C = kc

So LHS is the sum of (5), (6), (7)

LHS = \sin^2 A(\sin B\cos B+\sin C\cos C)+ \sin^2 B(\sin C\cos C+\sin A\cos A)+ \sin^2 C(\sin A\cos A+\sin B\cos B)

=k^2a^2(kb\cos B+kc\cos C)+k^2b^2(kc\cos C+ka\cos A)+k^2c^2(ka\cos A+kb\cos B)

=k^3a^2(b\cos B+c\cos C)+k^3b^2(c\cos C+a\cos A)+k^3c^2(a\cos A+b\cos B)

Expand the brackets, regroup the terms and then factor out common factors

LHS = k^3[ab(a\cos B+b\cos A)+ac(a\cos C+c\cos A)+(b\cos C+c\cos B)]

Apply the projection law formulas

a=b\cos C+c\cos B

b=a\cos C+c\cos A

c=b\cos A+a\cos B

Projection Law of a triangle

LHS = 3k^3abc

On the other hand, substitute (8), (9),(10) to RHS

RHS = 3 \sin A \sin B \sin C = 3k^3abc

which shows both sides are equal.

Steven Zheng posted 4 hours ago

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