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$f(x)$ is defined such that: $$f(x)=\prod_{n=1}^{999}\cos(nx)$$

Given that solve for $k$ (which is integer):

$$f\left(\frac{2\pi}{1999}\right)=\frac{1}{2^k}$$

I tried many different methods including euler's formula, trig identities and expressing each cosine in terms of $\cos x$ using De Moivre's theorem. Then I tried to multiply these polynomials of powers of cosine and I reached some interesting results. For example first term's power will be $\dfrac{(n+1)!}{(n-1)!2!}$ and its coefficient - $a$ - will be $\dfrac{n!}{(n-2)!2!}$ with term itself $a\cos^n(x)$. Then powers are in descending order, each power equal to previous one minus 2. I wasn't able to find a rule for coefficients except the first and last one, though they are always in the form of binomial coefficients or sum of them such that highest power chooses 2 or 3 or other number. After few terms it always became really cumbersome so I couldn't continue solving it that way with just pen and paper.

All that didn't lead me even close to solving the equation though. I solved it numerically using Java app and answer was $k=999$ if I remember correctly.

Is there any way to solve it algebraically?

StubbornAtom
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Convaly
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    also using $i$ in sums and products is in some contexts not very good as people can confuse it with imaginary number $\sqrt{-1}$. But interesting question. +1 – mathreadler Sep 11 '16 at 13:22
  • http://math.stackexchange.com/questions/324426/finding-prod-n-1999-sin-left-fracn-pi1999-right – lab bhattacharjee Sep 12 '16 at 16:10

1 Answers1

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Since $$ z^2-2\cos\left(\frac{2k\pi}{1999}\right)z+1=\left(z-e^{i\frac{2k\pi}{1999}}\right)\left(z-e^{-i\frac{2k\pi}{1999}}\right)\tag{1} $$ After multiplying both sides of $(1)$ from $k=1$ to $999$ and multiplying by $z-1$, we then set $z=i$. This yields $$ \begin{align} \overbrace{\ \ (i-1)\ \ \vphantom{\prod_{k=1}^{999}}}^{z-1}\overbrace{(-2i)^{999}\prod_{k=1}^{999}\cos\left(\frac{2k\pi}{1999}\right)}^{\text{product for $k=1$ to $999$}} &=\overbrace{\left.\prod_{j=-999}^{999}\left(z-e^{i\frac{2j\pi}{1999}}\right)\right|_{z=i}}^{\text{$(z-1)\,\times\,$product for $k=1$ to $999$}}\\ &=\left.z^{1999}-1\right|_{z=i}\\[9pt] &=-i-1\tag{2} \end{align} $$ Therefore, solving $(2)$ for the product we want $$ \begin{align} \prod_{k=1}^{999}\cos\left(\frac{2k\pi}{1999}\right) &=\frac{-i-1}{i-1}\frac{i^{999}}{2^{999}}\\ &=\frac1{2^{999}}\tag{3} \end{align} $$

robjohn
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  • Thank you, it looks like it answers the problem. Could you though please explain how did you get from $\left(z-e^{i\frac{2k\pi}{1999}}\right)\left(z-e^{-i\frac{2k\pi}{1999}}\right)$ to $(i-1)(-2i)^{999}\prod_{k=1}^{999}\cos\left(\frac{2k\pi}{1999}\right)$ and how exactly is that equal to the right side then? If you could just give a little more detailed working, that would be very helpful. – Convaly Sep 11 '16 at 18:54
  • I have added a bit of explanation. Let me know if you still have questions. – robjohn Sep 12 '16 at 06:27