# 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 $C\gt 90^{\circ}.$

### Acknowledgment

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