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Please provide guidance on how to solve the product to sum question. I have also attached my attempt which was unsuccessful...

Show $8 \sin \frac{4 \pi}{9} \sin \frac{2 \pi}{9} \sin \frac{\pi}{9} = \sqrt{3}$.

My attempt

Robin
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racer234
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  • Welcome to MSE. A few notes: please use MathJax to format your posts, make the title of your posts descriptive (not "how do I solve this problem") and don't use images in posts where text is sufficient. – Robin Dec 30 '23 at 06:16
  • Note: $\sin \alpha \sin \beta \sin \gamma = \tfrac{1}{4}[\sin(\alpha+\beta-\gamma)+\sin(\beta+\gamma-\alpha)+\sin(\gamma+\alpha-\beta)-\sin(\alpha+\beta+\gamma)]$. – Anton Vrdoljak Dec 30 '23 at 06:33
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    Are you familiar to identity: $\sin x \cdot \sin (\frac{\pi}{3}+x) \cdot \sin (\frac{\pi}{3}-x)=\frac{1}{4} \sin 3x$ – Dheeraj Gujrathi Dec 30 '23 at 06:34
  • Use https://math.stackexchange.com/questions/1921191/prove-cos3a-cos3120a-cos3240a-frac-34-cos3a or https://math.stackexchange.com/questions/510257/how-to-prove-tan-x-tan-leftx-frac-pi-3-right-tan-leftx-frac2-pi – lab bhattacharjee Dec 30 '23 at 10:12

3 Answers3

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Notice that $$\sin{\left(3\cdot\frac{\pi}{9}\right)} = \sin{\left(3\cdot\frac{2\pi}{9}\right)} = -\sin{\left(3\cdot\frac{4\pi}{9}\right)} = \frac{\sqrt{3}}{2}$$

and so, by the triple angle formula $\sin{3x} = 3\sin{x} - 4\sin^3{x}$,

$\dfrac{\sqrt{3}}{2} = 3t - 4t^3$ has the solutions $t = \sin{\dfrac{\pi}{9}}, \sin{\dfrac{2\pi}{9}}, -\sin{\dfrac{4\pi}{9}}$.

Now rearrange the cubic to $4t^3 - 3t + \dfrac{\sqrt{3}}{2} = 0 = t^3 - \dfrac{3}{4}t + \dfrac{\sqrt{3}}{8}$ and observe the product of the roots; $\left(t - \sin{\dfrac{\pi}{9}}\right)\left(t - \sin{\dfrac{2\pi}{9}}\right)\left(t + \sin{\dfrac{4\pi}{9}}\right) = t^3 - \dfrac{3}{4}t + \dfrac{\sqrt{3}}{8}$

Dstarred
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\begin{gathered} 8\sin \left( {\frac{{4\pi }}{9}} \right)\sin \left( {\frac{{2\pi }}{9}} \right)\sin \left( {\frac{\pi }{9}} \right) \hfill \\ = 4\left( {\cos \left( {\frac{{2\pi }}{9}} \right) - \cos \left( {\frac{{6\pi }}{9}} \right)} \right)\sin \left( {\frac{\pi }{9}} \right) = 4\cos \left( {\frac{{2\pi }}{9}} \right)\sin \left( {\frac{\pi }{9}} \right) + 2\sin \left( {\frac{\pi }{9}} \right) \hfill \\ = 2\left( {\sin \left( {\frac{{3\pi }}{9}} \right) - \sin \left( {\frac{\pi }{9}} \right)} \right) + 2\sin \left( {\frac{\pi }{9}} \right) = 2\sin \left( {\frac{\pi }{3}} \right) - 2\sin \left( {\frac{\pi }{9}} \right) + 2\sin \left( {\frac{\pi }{9}} \right) \hfill \\ = 2.\frac{{\sqrt 3 }}{2} = \sqrt 3 \hfill \\ \end{gathered}

OnTheWay
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Note that:

$$ \begin{align*} &8\sin\dfrac{\pi}{9}\sin\dfrac{2\pi}{9}\sin\dfrac{4\pi}{9}=\sqrt{3}\\ \Leftrightarrow\ &8\sin\dfrac{\pi}{9}\sin\dfrac{2\pi}{9}\cos\dfrac{\pi}{18}=\sqrt{3}\\ \Leftrightarrow\ &8\sin\dfrac{\pi}{9}\sin\dfrac{2\pi}{9}\sin\dfrac{\pi}{18}\cos\dfrac{\pi}{18}=\sqrt{3}\sin\dfrac{\pi}{18}\\ \Leftrightarrow\ &4\sin^2\dfrac{\pi}{9}\sin\dfrac{2\pi}{9}=\sqrt{3}\sin\dfrac{\pi}{18}\\ \Leftrightarrow\ &2\left(1-\cos\dfrac{2\pi}{9}\right)\sin\dfrac{2\pi}{9}=\sqrt{3}\sin\dfrac{\pi}{18}\\ \Leftrightarrow\ &2\sin\dfrac{2\pi}{9}-\sin\dfrac{4\pi}{9}=\sqrt{3}\sin\dfrac{\pi}{18}\\ \Leftrightarrow\ &2\sin\dfrac{2\pi}{9}=\sqrt{3}\sin\dfrac{\pi}{18}+\cos\dfrac{\pi}{18}\\ \Leftrightarrow\ &2\sin\dfrac{2\pi}{9}=2\sin\left(\dfrac{\pi}{18}+\dfrac{\pi}{6}\right) \end{align*} $$

which is obvious. Henceforth, the original proposition is proven.

CaldariNavyFleet
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