A Curious Identity in Triangle

Let $A,$ $B,$ $C$ denote the internal angles of $\Delta ABC.$ Then

$\begin{align} \;&\sin A (|\cos A| - |\cos B\cos C|)\\ +& \sin B (|\cos B| - |\cos C\cos A|)\\ +&\sin C (|\cos C| - |\cos A\cos B|)\\ =&\sin A\sin B\sin C. \end{align}$

The identity is the invention of Dan Sitaru and Leo Giugiuc.

Lemma 1

$\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C.$

For a proof, see a separate page.

Lemma 2

$-\sin 2A+\sin 2B+\sin 2C=4\sin A\cos B\cos C.$

(Two additional relations are obtained cyclically.)

Lemma 3

$\begin{align} \;&\sin A (\cos A - \cos B\cos C)\\ +& \sin B (\cos B - \cos C\cos A)\\ +& \sin C (\cos C - \cos A\cos B)\\ =& \sin A\sin B\sin C. \end{align}$

Proof of Lemma 3

$\displaystyle\sum_{cyc}\sin A(\cos A-\cos B\cos C)=\sin A\sin B\sin C$ is equivalent to

$\begin{align}\displaystyle 4\sin A\sin B\sin C &= \sum_{cyc}\sin A\cos A -4\sum_{cyc}\sin A\cos B\cos C\\ &=2\sum_{cyc}\sin 2A-\sum_{cyc}(-\sin 2A+\sin 2B+\sin 2C)\\ &=\sum_{cyc}\sin 2A, \end{align}$

which is true by Lemma 1 while on the penultimate state we applied Lemma 2.

Proof of the claim

For an acute angled triangle the statement reduces to Lemma 3. Thus, let's, without loss of generality, assume that $C\gt 90^{\circ}.$ Under this assumption, we shall prove that

$\begin{align} \;&\sin A (\cos A + \cos C \cos B)\\ +&\sin B (\cos B + \cos A \cos C)\\ -&\sin C (\cos C + \cos B \cos A)\\ =&\sum_{cyc}\sin C (\cos C - \cos A \cos B). \end{align}$

This is indeed so because the difference of the two sides equals

$\begin{align}\displaystyle &{\;}{\;}{-2} \sin C \cos C + 2 \sin B \cos A \cos C + 2 \sin A \cos B \cos C\\ &=- \sin 2C + \frac{1}{2}(- \sin 2A + \sin 2B + \sin 2C + \sin 2A - \sin 2B + \sin 2C)\\ & = 0 \end{align}$

The claim follows because $\cos C=-|\cos C|,$ for $C\gt 90^{\circ}.$


The identity and the proof have been communicated to me by Leo Giugiuc.

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Copyright © 1996-2017 Alexander Bogomolny


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