Trigonometric Identity for the Pi Day

Problem

Trigonometric Identity for the Pi Day

Solution 1

Setting $\displaystyle x=\frac{A}{3},$ $\displaystyle y=\frac{B}{3},$ $\displaystyle z=\frac{C}{3},$ we have $\displaystyle x+y+z=\frac{\pi}{3}$ and

$\displaystyle \begin{align} &\sum_{cycl}\sin\left(\frac{\pi}{3}-3x\right)\sin\left(y-z\right)=0&\Leftrightarrow\\ &\sum_{cycl}\cos\left(\frac{\pi}{3}-3x-y+z\right)=\sum_{cycl}\cos\left(\frac{\pi}{3}-3x+y-z\right). \end{align}$

But $\displaystyle \left(\frac{\pi}{3}-3x-y+z\right)+\left(\frac{\pi}{3}-3z+x-y\right)=0,$ etc. We are done.

Solution 2

Using the identity $2\sin x\cos x=\sin(x+y)+\sin(x-y),$

$\displaystyle\begin{align} &2\sin\left(\frac{B-C}{3}\right)\cos\left(A+\frac{\pi}{6}\right)\\ &\qquad\qquad=\sin\left(\frac{B-C}{3}+A+\frac{\pi}{6}\right)+\sin\left(\frac{B-C}{3}-A-\frac{\pi}{6}\right)\\ &\qquad\qquad=\sin\left(\frac{3B-C+7A}{6}\right)+\sin\left(\frac{B-3C-7A}{6}\right)\\ &\qquad\qquad=\sin\left(\frac{\pi}{2}-\frac{4(C-A)}{6}\right)+\sin\left(-\frac{\pi}{2}+\frac{4(B-A)}{6}\right)\\ &\qquad\qquad=\cos\left(\frac{2(C-A)}{3}\right)-\cos\left(\frac{2(B-A)}{3}\right). \end{align}$

Similarly,

$\displaystyle \begin{align} 2\sin\left(\frac{C-A}{3}\right)\cos\left(B+\frac{\pi}{6}\right)&=\cos\left(\frac{2(A-B)}{3}\right)-\cos\left(\frac{2(C-B)}{3}\right)\\ 2\sin\left(\frac{A-B}{3}\right)\cos\left(C+\frac{\pi}{6}\right)&=\cos\left(\frac{2(B-C)}{3}\right)-\cos\left(\frac{2(A-C)}{3}\right) \end{align}$

Since $\cos x=\cos (-x),$ adding the three identities up gives $0.$

Solution 3

$f=\cos \left(B+\frac{\pi }{6}\right) \sin \left(\frac{C-A}{3}\right)+\cos \left(C+\frac{\pi }{6}\right) \sin \left(\frac{A-B}{3}\right)+\cos \left(A+\frac{\pi }{6}\right) \sin \left(\frac{B-C}{3}\right)$

We can rewrite as:

$\displaystyle \begin{align} f\,&=\frac{1}{4} i \left(e^{-i A-\frac{i \pi }{6}}+e^{i A+\frac{i \pi }{6}}\right) \left(e^{-\frac{1}{3} i (B-C)}-e^{\frac{1}{3} i (B-C)}\right)\\ &\qquad +\frac{1}{4} i \left(e^{-i B-\frac{i \pi }{6}}+e^{i B+\frac{i \pi }{6}}\right) \left(e^{-\frac{1}{3} i (C-A)}-e^{\frac{1}{3} i (C-A)}\right)\\ &\qquad+\frac{1}{4} i \left(e^{-i C-\frac{i \pi }{6}}+e^{i C+\frac{i \pi }{6}}\right) \left(e^{-\frac{1}{3} i (A-B)}-e^{\frac{1}{3} i (A-B)}\right)\\ f\,&=-\frac{1}{4} i\prod_{cycl}\left(e^{\frac{2 i B}{3}}-e^{\frac{2 i A}{3}}\right)\cdot e^{-\frac{1}{6} i (6 A+6 B+6 C+\pi )} \left(1+e^{\frac{1}{3} i (2 A+2B+2 C+\pi )}\right). \end{align}$

And, by Euler's identity,

$\displaystyle 1+e^{\frac{1}{3} i (2 A+2 B+2 C+\pi )}= 1+e^{ i \pi }=0$

Solution 4

$\displaystyle \begin{align} &2\cos\left(A+\frac{\pi}{6}\right)\sin\left(\frac{B-C}{3}\right)=-2\cos\left(B+C-\frac{\pi}{6}\right)\sin\left(\frac{B-C}{3}\right) \\ &=\sin\left(\frac{2B}{3}+\frac{4C}{3}-\frac{\pi}{6}\right)-\sin\left(\frac{4B}{3}+\frac{2C}{3}-\frac{\pi}{6}\right) \\ &=\sin\left(\frac{2\pi}{3}-\frac{2A}{3}+\frac{2C}{3}-\frac{\pi}{6}\right)-\sin\left(\frac{2\pi}{3}-\frac{2A}{3}+\frac{2B}{3}-\frac{\pi}{6}\right)\\ &=\sin\left(\frac{\pi}{2}-\frac{2A}{3}+\frac{2C}{3}\right)-\sin\left(\frac{\pi}{2}-\frac{2A}{3}+\frac{2B}{3}\right) \\ &=\cos\left[\frac{2}{3}(A-C)\right]-\cos\left[\frac{2}{3}(A-B)\right]. \end{align}$

On adding the cyclic terms,

$\displaystyle \begin{align} &2\sum_{cyc}\cos\left(A+\frac{\pi}{6}\right)\sin\left(\frac{B-C}{3}\right) \\ &=\cos\left[\frac{2}{3}(A-C)\right] -\cos\left[\frac{2}{3}(A-B)\right] +\cos\left[\frac{2}{3}(B-A)\right] \\ &-\cos\left[\frac{2}{3}(B-C)\right] +\cos\left[\frac{2}{3}(C-B)\right] -\cos\left[\frac{2}{3}(C-A)\right] \\ &=\cos\left[\frac{2}{3}(C-A)\right] -\cos\left[\frac{2}{3}(B-A)\right] +\cos\left[\frac{2}{3}(B-A)\right] \\ &-\cos\left[\frac{2}{3}(B-C)\right] +\cos\left[\frac{2}{3}(B-C)\right] -\cos\left[\frac{2}{3}(C-A)\right]=0. \end{align}$

Acknowledgment

This problem was posted by Cezar Lozada at the Olimpiada pe Scoala (The School Yard Olympiad) facebook group and kindly communicated to me by Leo Giugiuc, along with a solution of his (Solution 1). Solution 2 is by Antreas Hatzipolakis; Solution 3 is by N. N. Taleb; Solution 4 is by Amit Itagi.

 

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