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After applying product to sum formulae, I got this far $$\frac{\Bigl(1+2\cos(20)\Bigr)\Bigl(\cos(140)+\cos(20)\Bigr)}{8}=\frac{1}{16}.$$ How can I proceed further?

cqfd
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Mathematical Curiosity
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  • Are you saying you've proved it? – Angina Seng Jul 10 '19 at 06:08
  • You can generalise this to $\prod\limits_{k=1}^n \cos\left(\frac{k}{2n+1}180^\circ\right)$ i.e. $\prod\limits_{k=1}^n \cos\left(\frac{k}{2n+1}\pi\right)=2^{-n}$ and here $n=4$. Alternatives with $n=1,2,7,22$ also involve integer degree angles – Henry Jul 10 '19 at 06:23

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If you just want a proof, then ponder this: \begin{align*} &8\sin 20^\circ\cos20^\circ\cos40^\circ\cos80^\circ \\ = \, &4\sin 40^\circ\cos40^\circ\cos80^\circ \\ = \, &2\sin 80^\circ\cos80^\circ \\ = \, &\sin160^\circ=\sin20^\circ. \end{align*}

Theo Bendit
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Angina Seng
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