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[SOLVED] Sine function challenge

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anemone

MHB POTW Director
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Feb 14, 2012
3,812
Let $a,\,b$ and $c$ be real numbers such that $\sin a+\sin b+\sin c\ge \dfrac{3}{2}$. Prove that

$\sin \left(a-\dfrac{\pi}{6}\right)+\sin \left(b-\dfrac{\pi}{6}\right)+\sin \left(c-\dfrac{\pi}{6}\right)\ge 0$.
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,812
Assume for contradiction that $\sin \left(a-\dfrac{\pi}{6}\right)+\sin \left(b-\dfrac{\pi}{6}\right)+\sin \left(c-\dfrac{\pi}{6}\right)< 0$

Then by the addition and subtraction formulas, we have

$\dfrac{1}{2}(\cos a+\cos b+\cos c)>\dfrac{\sqrt{3}}{2}(\sin a+\sin b+\sin c)\ge \dfrac{3\sqrt{3}}{4}$

It follows that

$\cos a+\cos b+\cos c>\dfrac{3\sqrt{3}}{2}$

which implies that

$\begin{align*}\sin \left(a+\dfrac{\pi}{3}\right)+\sin \left(b+\dfrac{\pi}{3}\right)+\sin \left(c+\dfrac{\pi}{3}\right)&=\dfrac{1}{2}(\sin a+\sin b+\sin c)+\dfrac{\sqrt{3}}{2}(\cos a+\cos b+\cos c)\\&=\dfrac{1}{2}\cdot \dfrac{3}{2}+\dfrac{\sqrt{3}}{2}\cdot \dfrac{3\sqrt{3}}{2}\\&=3\end{align*}$

which is impossible because $\sin x\le 1$.