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Let $A,B,C$ by the angles of a triangle, and let $a,b,c$ be the length of the corresponding opposite sides.

How can you prove that

$$a^3\cos(B-C) + b^3\cos(C-A) +c^3\cos(A-B) = 3abc?$$

I divided both sides by $abc$ and then tried to open the cosine function but nothing worked. I also took the cube of sine in the triple angle identity but that too just made the function more complicated and yielded no result.

Mike Earnest
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Harsh Sharma
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    What is your reason for thinking the equation is true? – Gerry Myerson Jun 10 '16 at 07:02
  • @GerryMyerson Because if the equation would be true, after dividing both sides by $abc$ the and solving the LHS, the answer would have come out to be 3. I assume the the equation is true and if the answer doesn't come equal to 3, that would have been in contrast to our assumption. – Harsh Sharma Jun 10 '16 at 07:29
  • See http://www.askiitians.com/forums/Trigonometry/in-a-triangle-abc-prove-that-a-3-cos-b-c-b-3-cos_123839.htm – lab bhattacharjee Jun 10 '16 at 09:08
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    My question was, why do you assume the equation is true? Though it hardly matters any more, now that you have an answer. – Gerry Myerson Jun 10 '16 at 09:10

2 Answers2

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Let $2R=k$

$a=k\sin A, b=k\sin B,c=k\sin C$ (sine rule)

$$ a^3 \cos (B-C)= a^2\cdot a\cos (B-C)=$$ $$= a^2\cdot k\sin A\cos (B-C)= ka^2\sin (180-(B+C))\cos (B-C)=$$ $$= ka^2\sin (B+C)\cos (B-C)=\frac{ka^2}{2} (\sin 2B+\sin 2C) = $$ $$=\frac{ka^2}{2} (2\sin B\cos B+2\sin C\cos C) = ka^2(\sin B\cos B+\sin C\cos C) = $$ $$= a^2((k\sin B)\cos B+(k\sin C)\cos C)) =$$$$= a^2 (b\cos B+c\cos C)$$

The original expression now looks like this: $$a^2(b\cos B+c\cos C)+b^2(c\cos C+a\cos A)+c^2(a\cos A+b\cos B) = $$ $$= a^2b\cos B + a^2c\cos C+b^2c\cos C+b^2a\cos A+c^2a\cos A+c^2b\cos B$$ $$= a^2b\cos B + b^2a\cos A + b^2c\cos C + c^2b\cos B + a^2c\cos C + c^2a\cos A $$ $$=ab(a\cos B+b\cos A)+bc(b\cos C+c\cos B)+ca(a\cos C+c\cos A) =$$

Triangle $$= abc+bca+cab = 3abc$$

Akiiino
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shraman shraman
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Using $\cos{(x-y)}=\frac{\sin{x}\cos{x}+\sin{y}\cos{y}}{\sin{(x+y)}}$, we obtain \begin{align*} \cos{(B-C)} &=\frac{\sin{B}\cos{B}+\sin{C}\cos{C}}{\sin{(B+C)}} \\ &=\frac{b\cos{B}+c\cos{C}}{a} \;\;\; \text{[By extended Law of Sines]} \end{align*}

\begin{align*} & \therefore \; \sum_{cyc} {a^3\cos{(B-C)}} \\ &= \sum_{cyc} {a^2(b\cos{B}+c\cos{C}) } \\ &= a^2(b\cos{B}+c\cos{C})+b^2(c\cos{C}+a\cos{A})+c^2(a\cos{A}+b\cos{B})\\ &= ab(a\cos{B}+b\cos{A})+bc(b\cos{C}+c\cos{B})+ca(c\cos{A}+a\cos{C}) \\ &= abc+abc+abc \;\;\; \text{[Using Projection Formulas]} \\ &= 3abc \end{align*} $QED.$

Entrepreneur
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